26:0:0: CONDUCTION OF ELECTRICITY THROUGH MATERIALS:
Material such as metal conduct electricity. They allow electric charge to flow through them.
OHM'S LAW:
Ohm's law states that the current that flows through a conductor is directly proportional to the voltage that is applied across the terminals of the conductor, provided other factors remain constant.
Ohm's law shows the relationship that exist between the voltage that is applied across a conductor and the current that flows through the conductor the time the voltage is applied.
FORMULA OF OHM'S LAW:
From the statement of ohm's law, current is directly proportional to voltage applied across the conductor.
Mathematically, Voltage V ∞ current I . Voltage V = Resistance R * Current I.
. V = I * R
We can get different formula from the above formula by making I and R subject of the formula one after the other.
Therefore, I = V / R and R = V / I .
APPLICATION OF OHM'S LAW:
EXAMPLES:
Calculate the current produced when 250 volt cell is connected across 12 ohms resistor.
SOLUTION:
Data given in the question:
Voltage V = 250 volt, Resistance R = 12 ohms, current I = ?
Formula: Voltage V = Current I * Resistance R
Substitution: 250. = current I * 12
Make current the subject of the formula: current I = 250 / 12. . Current I = 20.83 Amperes
Determine the value of resistance that will cause a voltage of 125 volt to generate 2.5 amperes of current.
SOLUTION:
Data given in the question:
Voltage V = 125 volt, current I = 2.5 amperes, resistance R = ?
Formula: Voltage V = Current I * Resistance R
Substitution: 125. = 2.5 * resistance R
Make resistance R the subject of formula:
125 / 2.5. = resistance R . Resistance R = 50 ohms
EXPERIMENTAL VERIFICATION OF OHM'S LAW:
AIM:
To verify ohm's law
APPARATUS:
Ammeter, voltmeter, rheostat, standard resistor, cell, connection wires and switch.
SETUP DIAGRAM:
PROCEDURES:
The circuit diagram is connected as shown above.
Close the switch so that current flow in the circuit.
Read and record the values of current that flow in the circuit and voltage across the resistor recorded by the ammeter and voltmeter respectively.
Adjust the rheostat to a new position so as to change the resistance of the circuit. Read and record the new values of current I, and voltage,V recorded by the ammeter and voltmeter respectively. Repeat the process three more times. Read and record the corresponding values of I and V recorded by the ammeter and voltmeter.
Tables of values:
S/n. Rheostat position P. Current I ( A ). Voltage V (volt)
1 P1. I1. V1
2 P2. I2. V2
3 P3. I3. V3
4 P4. I4. V4
5 P5. I5. V5
GRAPH OF VOLTAGE V AGAINST CURRENT I:
The different values of Voltages V that were obtained are plotted on the vertical axis against different values of current I on the horizontal axis. A straight line graph which passes through the origin is obtained as shown below.
SLOPE OF THE GRAPH OF VOLTAGE V AGAINST CURRENT I:
Slope S or gradient = ∆ Voltage / ∆ current
Slope s or gradient = change in voltage / change in current
Slope s /gradient = ( Voltage V2 – Voltage V1 ) / ( current I2 – current I1 )
Slope s / gradient = V2 - V1 / I2 - I1
Slope s / gradient = Voltage / current
Recall that resistance R = voltage V / current I.
Therefore, slope = resistance.
The slope of the grape of the graph represent the resistance of the circuit.
PRECAUTIONS:
Make sure the contacts are tight to prevent leakages.
Off the circuit after each experiment so as not to run down the cell
Take the reading when the ammeter and voltmeter pointers are steady
Avoid error due to parallax when taking the readings.
CONNECTION OF RESISTORS:
Resistors can be connected in two ways in a circuit. The connections are :
Connection of resistors in series
Connection of resistors in parallel
1..CONNECTION OF RESISTORS IN SERIES:
In series connection of resistors, the resistors are connected end to end to form a linear network as shown in the circuit diagram below.
TAKE NOTE:
In series connection of resistors:
The same amount of current flows through each of the resistors, one after the other.
The voltage drops across each of the resistors are different, but if the resistances of the resistors are of the same value,then the voltage drop across each of the resistors would be the same.
FORMULA FOR CALCULATING TOTAL RESISTANCE OF RESISTORS CONNECTED IN SERIES:
Assuming three resistors R , R , R are connected in series as shown in the diagram below, the total resistance is calculated by the formula
Total resistance Rt = R1 + R2 + R3
DERIVATION OF THE FORMULA FOR CALCULATING TOTAL RESISTANCE OF RESISTORS CONNECTED IN SERIES:
Circuit diagram for connection of resistors in series is as shown by the your below..
From the circuit diagram, the voltage V is connected across the three resistors this voltage is shared across each of the resistor.
Remember that the voltages drop across each of the resistors are different because the resistances of the resistors are different. Therefore, Let the Voltage of the cell connected across the three resistors be V.
Let the voltage drop across resistor R1 be V1.
Let the voltage drop across resistor R2 be V2, also,
Let the voltage drop across resistor R3 be V3.
Therefore,
Vt = V1 + V2 + V3
From ohm's law which states that, V = I * R
For Rt, we have Vt = I*Rt, for R1, V1 = I*R1, for R2, V2 = I*R2 and for R3, V3 = I*R3
Now, We will substitute for Vt, V1, V2 and V3 in the formula above. Then we will get,
Vt = V1 + V2 + V3
I*Rt = I*R1 + I*R2 + I*R3
Let us factorie I : I*Rt = I ( R1 + R2 + R3 )
Divide both side by I: I*Rt / I = I ( R1 + R2 + R3 ) / I. Then we will get,
Rt = R1 + R2 + R3
APPLICATION OF FORMULA TO SOLE PROBLEMS:
EXAMPLES:
1. From the diagrams below, calculate the total resistance of the circuits.Tfrf
Solution:
Data given in the figure:
From the figure, R1 = 5ohms, R2 = 8 ohms, R 3 = 12 ohms.
Formula:
Rt = R1 + R2 + R3
Substitution: Rt = 5 + 8 + 12
➡ Total resistance of the circuit Rt = 25 ohms
2. What is the equivalent resistance of the circuit containing three resistors connected
in series?
Solution:
Data given in the question:
R1 = 1 ohm, R2 = 2 ohms, R3 = 3 ohms
Formula: Rt = R1 + R2 + R3
Substitution: Rt = 1 + 2 + 3
Total resistance of the circuit Rt = 6 ohms.
3. 4 ohms, 8 ohms and R resistors are connected in series. If the total resistance of the circuit is 23, calculate the value of R.
Solution:
Data given in the question:
R1 = 4 ohms, R2 = 8 ohms, R3 = ?
Formula: Rt = R1 + R2 + R3
Substitution: 23 = 4 + 8 + R. ➡ 23 = 12 + R
Make R the subject: R = 23 - 12. ➡ R = 11 ohms
2..CONNECTION OF RESISTORS IN PARALLEL:
In parallel connection of resistors, the positive terminals of all the resistors are connected to the same positve terminal and the negative terminals of the resistors are connected to the same negative terminal. Parallel connection of resistors always form a network of two or more rooms depending on the numbers of resistors used in the connection. The circuit diagram is as shown in the figure by your right.
TAKE NOTE:
In parallel connection of resistors:
The voltages across the resistors are the same or the same voltage act across each of the resistors.
The current that flow through each of the resistors are different because the resistances of each of the resistors are different.
If the resistance of each of the resistors are the same, then the current that will flow through each of the resistors would be the same.
FORMULA FOR CALCULATING TOTAL RESISTANCE OF RESISTORS CONNECTED IN PARALLEL:
Assuming three resistors R1 , R2 , R3 are connected in parallel as shown in the diagram below, the total resistance is calculated by the formula:
diagram is as shown in the figure by your right. 1 / Rt = 1/R1 + 1/R2 + 1/R3
DERIVATION OF THE FORMULA FOR CALCULATING TOTAL RESISTANCE OF RESISTORS CONNECTED IN SERIES:Circuit diagram:
From the circuit diagram, it is the voltage V that is connected across the three resistors. I is the total current generated in the circuit. I1,I2 and I3 are the current that will flow through the resistors, R1, R2, and R3 respectively while the voltages across the resistors are the same.
Remember that different values of current will flow through each of the resistors because the resistances of the resistors are different.
Therefore,
Let the Voltage of the cell connected across the three resistors be V.
Let the current that flow through resistor R1 be I1.
Let the current that flow through resistor R2 be I2
Let the current that flow through resistor R3 be I3
Therefore, total current Iohm
It = I1 + I2 + I3
Recall from ohm's law that, V = I*R. Let us make I the subject of the formula. Then we will get,
I = V / R, then,
For It, we get It V/Rt; for I1, we get I1 = V/R1; for I2, we get I2 = V/R2; for I3, we get I3 = V/R3.
We will substitute for It, I1,I2 and I3 in the formula above. Then, we will get:
V/Rt = V/R1 + V/R2 + V/R3
Let us factorie V: V/Rt = V ( 1/R1 + 1/R2 + 1/R3 )
Divide both sides of the equation by V: V/Rt ÷V = V ( 1/R1 + 1/R2 + 1/R3 )
V cancels V. Then we will get: 1/Rt = 1/R1 + 1/R2 + 1/R3
which is the formula for calculating total resistance of resistors connected in parallel.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
1. Three resistors 5 ohms, 8 ohms and 3 ohms are connected in parallel. Calculate the total resistance of the circuit.
Solution:
Data given in the question:
R1 =3 ohms, R2 = 8 ohms, R3 = 5 ohms
Formula: 1/Rt = 1/R1 + 1/R2 + 1/R3
Substitution: 1/Rt = 1/3 + 1/8 + 1/5
Lc of the fraction = 3 * 5 * 8 = 120
1/Rt = 1(120)/3 + 1(120)/5 + 1(120)/8
1/Rt = 40 + 24 + 15. ➡ 1/Rt = 79 ➡ Rt = 1/ 79 ➡ Rt = 0.013 ohm
2. From the figure, calculate the effective resistance of the circuit.
Solution:
Data given in the question:
R1 = 6 ohms, R2 = 4 ohms, R3 = 2 ohms
Formula: 1/Rt = 1/R1 + 1/R2 + 1/R3
Substitution: 1/Rt = 1/6 + 1/4 + 1/2
LCM of the fractions = 12
Therefore, 1/Rt = 1(12)/6 + 1(12)/4 + 1(12)/2
1/Rt = 2 + 3 + 6 ➡ 1/Rt = 11➡ Rt = 1/11
Effective resistance Rt = 0.09 ohm
FACTORS THAT AFFECT THE RESISTANCE OF A CONDUCTOR:
The following factors affect the resistance of a conductor:
Let us look at these factors one after the other. We will build a formula that are very useful in calculations:
1..LENGTH OF CONDUCTOR:
The resistance R of a conductor is directly proportional to the length of the conductor. This means that the resistance of a conductor increases as the length of the conductor increases and vice versa.
Mathematically:
Resistance R is directly proportional to length of conductor L
Resistance R ∞ length of conductor L
R ∞ L
R = K*L
Make K the subject of the formula.: K = R / L
If there is any change in R or L, there will be a corresponding change in L or R.
Therefore,
we have the formula, R1/L1 = R2/ L2 or R1*L2 = R2*L1
R1 and L1 are the initial resistance and length of the conductor. If there is a change in R1 to R2, then L1 will also change from L1 to L2 so that the constant will always remain constant.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
The resistance of a conductor of length 25cm is 0.25 ohm. Calculate the resistance of the conductor when its length is increased by 20%.
SOLUTION:
Data given in the question:
Resistance R1 = 0.25 ohm, Length L1 = 25 cm, Resistance R2 = ?, Length L2 is increased by 20% of L1.
Therefore,
20% of L1 = 20% of 25cm = 20/100 * 25 = 5
L2 = 5 + L1 = 5 + 25 = 30cm. L2 = 30cm
Formula : R1 * L2 = R2 * L1
Substitution: 0.25 * 30 = R2 * 25
Make R2 the subject: R2 = 0.25 * 30 / 25 R2 = 81.25 ohms
A conductor has a resistance of 30ohms when its length is 120cm. Find the length of the conductor when its resistance is one third its original value.
SOLUTION:
Data given in the question:
Resistance R1 = 30 ohms, length L1 = 120 cm, resistance R2 = one- third of R1, length L2 = ?
Resistance R2 = ⅓ of R1 = ⅓ * 30 = 10.then, R2 = 10 ohms
Formula: R1 * L2 = R2 * L1
Substitution: 30 * L2 = 10 * 120
Make L2 the subject of the formula: L2 = 10 * 120 / 30
L2 = 40 cm
The resistance of a wire is R ohms when its length is 1.25 m. If the length becomes 1.75m when the resistance increased to ( R+ 2 ) ohms, find the value of R.
SOLUTION:
Data given in the question:
Resistance R1 = R, length L1 = 1.25m, resistance R2 = (R + 2) ohms, length L2 = 1.75m
Formula: R1 * L2 = R2 * L1
Substituting: R * 1.75 = (R + 2) * 1.25
Open the bracket: R * 1.75 = 1.25R + 2.50
Collect like terms: 1.75R – 1.25R = 2.50
0.50R = 2.50
Make R the subject of the formula: R = 2.50 / 0.50
R = 5 ohms
Therefore, R1 = 5 ohms, R2 = R + 2 = 5 + 2 = 7 ohms
2..AREA OF CONDUCTOR:
The resistance of a conductor is inversely proportional to the cross sectional area of the conductor provided all other factors remain constant. This means that the resistance of a conductor will increase if the cross sectional area of the conductor is reduced or decreased. And vice versa.
Mathematically:
Resistance R is inversely proportional to cross sectional area of conductor
Resistance R ∞ 1/A
R = K / A
Make K the subject of the formula: K = R * A
If there is any change in R or A, there will be a corresponding change in A or R.
Therefore, we have the formula, R1 * A1 = R2 * A2 or R1/A2 = R2/A1
R1 and A1 are the initial resistance and area of the conductor. If there is a change in R1 to R2, then A1 will also change from A1 to A2 so that the constant will always remain constant.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
*The area of a wire is A m² when its resistance is 1.25 ohms. If the resistance becomes 1.75ohms when the area is increased to ( A+ 2 ) m², find the value of A.
SOLUTION:
Data given in the question:
Area A1 = A, resistance R1 = 1.25 ohms, area A2 = (A + 2) m², resistance R2 = 1.75 ohms
Formula: R1 * A1 = R2 * A2
Substituting: 1.25 * A = 1.75 * (A + 2)
Open the bracket: 1.25 * A = 1.75R + 3.50
Collect like terms: 3.50 = 1.75A – 1.25A
= 0.5A
Make A the subject of the formula: A = 3.50 / 0.50
A = 7 m²
Therefore, A1 = 7m², A2 = A + 2 = 7 + 2 = 9 m²
*A conductor has a resistance of 30ohms when its area is 12.0cm². Find the area of the conductor when its resistance is one third its original value.
SOLUTION:
Data given in the question:
Resistance R1 = 30 ohms, area A1 = 12.0 cm², resistance R2 = one- third of R1, area A2 = ?
Resistance R2 = ⅓ of R1 = ⅓ * 30 = 10.then, R2 = 10 ohms
Formula: R1 * A1 = R2 * A2
Substitution: 30 * 12.0 = 10 * A2
Make A2 the subject of the formula: A2 = 30 * 12.0 / 10. . A2 = 36 cm²
*The resistance of a conductor of area 25cm² is 0.25 ohm. Calculate the resistance of the conductor when its area is increased by 20%.
SOLUTION:
Data given in the question:
Resistance R1 = 0.25 ohm, area A1 = 25 cm², Resistance R2 = ?, Area A2 is increased by 20% of A1.
Therefore,
20% of A1 = 20% of 25cm² = 20/100 * 25 = 5
A2 = 5 + A1 = 5 + 25 = 30cm. A2 = 30cm
Formula : R1 * A1 = R2 * A2
Substitution: 0.25 * 25 = R2 * 30
Make R2 the subject: R2 = 0.25 * 25 / 30. R2 = 0.21 ohm
3..TEMPERATURE OF CONDUCTOR:
The resistance R of a conductor is directly proportional to the temperature, T of the conductor provided all other factors remain constant. This means that the resistance R of a conductor will increase if the temperature T of the conductor is increased or vice versa.
Mathematically,
Resistance R is directly proportional to temperature T of conductor
Resistance R ∞ Temperature T
R ∞ T
R = KT
Make K the subject of the formula: K = R / T
If there is any change in R or T, there will be a corresponding change in T or R.
Therefore, we have the formula, R1 / T1 = R2 / T2 or R1 * T2 = R2 * T1
R1 and T1 are the initial resistance and temperature of the conductor. If there is a change in R1 to R2, then T will also change from T1 to T2 so that the constant will always remain constant.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
A conductor has a resistance of 30ohms when its temperature is 12.0°C. Find the temperature of the conductor when its resistance is one third its original value.
SOLUTION:
Data given in the question:
Resistance R1 = 30 ohms, temperature T1 = 12.0 °C, resistance R2 = one- third of R1, temperature T2 = ?
Resistance R2 = ⅓ of R1 = ⅓ * 30 = 10.then, R2 = 10 ohms
Formula: R1 * T2 = R2 * T1
Substitution: 30 * T2 = 10 * 12.0
Make A2 the subject of the formula: T2 = 10 * 12.0 / 30. . T2 = 4 °C
*The resistance of a conductor of temperature 25 °C is 0.25 ohm. Calculate the resistance of the conductor when its temperature is increased by 20%.
SOLUTION:
Data given in the question:
Resistance R1 = 0.25 ohm, temperature T1 = 25 °C, Resistance R2 = ?, temperature T2 is increased by 20% of T1.
Therefore,
20% of T1 = 20% of 25 °C = 20/100 * 25 = 5
T2 = 5 + T1 = 5 + 25 = 30 °C. T2 = 30 °C
Formula : R1 * T2 = R2 * T1
Substitution: 0.25 * 30 = R2 * 25
Make R2 the subject: R2 = 0.25 * 30 / 25. R2 = 0.30 ohm
RESISTIVITY OF A CONDUCTOR:
The formula of resistivity is obtained by combining the three factors which affect the resistance of a conductor that I discussed above together. Thus:
The resistance R of a conductor is directly proportional to the length L and inversely proportional to the cross sectional area A of the conductor, all other factors remain constant.
Mathematically,
Resistance R is directly proportional to length L and inversely proportional to cross sectional area A of conductor.
Resistance R ∞ Length / Area A
R ∞ L / A
R = K*L / A
Make K the subject if the formula: K = R * A / L.
K is constant of proportionality called resistivity. Its unit is ohm-meter
If there is any change in any one of these ( R, L and A), there will be a corresponding change in the other two.
Therefore, we get the formula, R1 * A1 / L1 = R2 * A2 / L2. Or. R1 * A1 * L2 = R2 * A2 * L1
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
The resistance of a wire is 2 ohms when its length and cross sectional area are 1.5m and 0.5m². What would be the resistance of the wire when its length becomes 1.3m and its cross sectional area change to 1.7m²?
SOLUTION:
Data given in the question:
Resistance R1 = 2 ohms, Length L1 = 1.5m , Area A1 = 0.5m² , Resistance R2= ? , Length L2 = 1.3m, Area A2 = 1.7m²
Formula: R1 * A1 / L1 = R2 * A2 / L2
Substitution: 2 * 0.5 / 1.5 = R2 * 1.7 / 1.3
Make R2 the subject of the formula: R2 = ( 2 * 0.5 * 1.3 ) / (1.5 * 1.7 )
R2 = 1.3 / 2.55. R2 = 0.51 ohm.
*At a length of 12 cm, the resistance of a wire of cross sectional area of 5.0cm² is 35ohms. If the resistance of the wire remain constant when its length increased to 20 cm, find the area of the conductor.
SOLUTION:
Data given in the question:
Resistance R1 = 35ohms, Length L1 = 12 cm, area A1 = 5.0 cm², R2 remains constant, R2 = R1 = 35ohms, L2 = 20cmm,
A2 = ?
Formula: R1 * A1 / L1 = R2 * A2 / L2
Substitution: 35 * 5.0 / 12 = 35 * A2 / 20
Make A2 subject if the formula: A2 = ( 35 * 5.0 * 20 ) / ( 12 * 35 )
A2 = 3500 / 420. . A2 = 8.33 m²
*The worked examples are to show you how to solve problems using the formula that instate above. More questions will be in the exercises below.
ELECTRICAL WORK DONE IN A CIRCUIT:
In an electric circuit, work is done when electricity flow from a point of higher potential to a point of lower potential. The unit of work done is joule, (J).
FORMULA FOR WORK DONE IN ELECTRIC CIRCUIT:
If Q Coulomb of electricity flows from one point to another whose difference in potential is V volt, then the formula for work done in the circuit is
Work done donein joule = charge Q in Coulomb * voltage V in volt
W = Q * V
Using their units units, Then joule = coulomb * volt
Recall that, Q = I * t. Let us substitute for Q in the equation above,
Then Work done = Q * V . Work done = I * t * V
Also, recall from ohm's law, V = I * R. Let us substitute for V in the equation above. Then we will get,
Work done W = I * t * I * R. W.D = I² * t * R.
Again from ohm's law, we make I the subject of the formula. Then we will get, I = V / R.
Let us substitute for I in the formula above, then we will get,
Work done = I * t * V work done = V / R * t * V. work done = V² * t / R
Where I is current, t is time and v is voltage.
CONVERSION OF ELECTRICAL ENERGY AND DEVICES USED:
Electrical energy can be converted into other forms of energy as I explain below:
Electrical energy can be converted into heat energy by the use of electric iron, electric kettle, immerse heater.
The amount of heat energy can be calculated using the formula:
Work done = current * voltage * time
Work done = I * V * t
Work done = I² * R * t
Electrical energy can be converted into mechanical energy by the movement of the blades of electric fan, electrica motor.
Electrical energy can be converted into light energy by bulbs, lambs.
Electrical energy that can be converted into electrical by telephone ear piece.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
Calculate the heat energy generated when 2.5 A of current flow through a resistor of resistance 8.0 Ohms for
¾ minutes.
SOLUTION:
Data given in the question:
Current I = 2.5 A, time = ¾minutes = ¾ * 60 seconds = 45 seconds, resistance R = 8.0 ohms.
Formula: Work done = I² * R * t
Substitution: work done = 2.5² * 8 * 45. work done = 2250 joules
Given that the work done in electric circuit is 334 joules, determine the current in the voltage if the resistance of the circuit is 5 ohms and runs for 6 minutes.
SOLUTION:
Data given in the question:
Work done = 334 joules, resistance R = 5 ohms, time = 6 minutes = 6 * 60 seconds = 360 seconds, voltage v = ?
Let me do some explanations hear. Ok!
Formula: Work done = Q * V
From the question, Q is not given. So, we must recall that Q = I * t and substitute it for Q in the formula above. Then we will get,
work done = Q * V work done = I * t * V
At this point, if you look at the above formula, you will notice that current I is present but it is not given in the question. So we must recall ohms law which says V = I * R. We have to make I the subject. then I = V / R. Now we have to substitute for I in the formula, work done = I * t * V. The formula will become, work done = V / R * t * V.,
work done = V² * t / R. This the formula we will use for the calculation because it contains R, t and V that we want to calculate.
Let us substitute for w, t and R: work done = V² * t / R
334 = V² * 360 / 5.
Make V the subject of the formula: 334 * 5 = V² * 360
* 5 / 360 = V²
= V²
V = √4.64
V = 2.15 volt
ELECTRIC POWER:
Electric power is the time rate of electric work done in a circuit. It is the ratio of electric work done to time taken to do the work. The unit of electric power is watt.
FORMULA FOR CALCULATING ELECTRIC POWER:
Power = work done in joule / time taken in second
Recall that electric work done = I * V * t. Then we substitute for work done in the above formula.
Therefore,
Power = work done / time . Power = I * V * t / t
t divides t: power = I * V
You have to recall that, from ohm's law, V = I * R and you have to substitute for V in the above equation. Then,
Power = I * V. power = I * I * R = I² * R. power = I² * R
DEFINITION OF WATT:
Watt is defined as one joule per second. It is the power that is consumed in an electric circuit when one joule of work is done in one second.
KILOWATT (KW) AND MEGAWATT (MW):
Kilowatt and megawatt are larger units of power in which power can also be measured.
VALUES OF KILOWATT AND MEGAWATT:
1 KW. = 1000 W. = 1 * 10³ W
1 MW = 1000000 W. = 1 * 10²+⁴ W
CONVERSION OF WATT TO KILOWATT, MEGAWATT AND VICE VERSA:
TAKE NOTE:
To change power in watt to kilowatt, you divide the watt by 1000.
To change power in watt to megawatt, you divide the watt by 1000000.
To change kilowatt to megawatt, you divide the kilowatt by 1000.
To change kilowatt to watt, you multiply the 1000.
To change megawatt to watt, you multiply the megawatt by 1000000.
To change megawatt to kilowatt, you multiply the megawatt by 1000.
EXAMPLES:
What is the value of 12¾ kilowatt in watt?
SOLUTION:
Data given in the question:
Kw = 12¾ kw.
1000W = 1 KW. . 12¾ KW = 12¾ * 1000 = 51/4 * 1000 = 798 Watts
What is the value of 0.25 MW in kilowatt?
SOLUTION:
Data given in the question:
Power = 0.25 MW
1000 KW = 1 MW . to change megawatt to kilowatt, multiply the megawatt by 1000
0.25 MW = 0.25 * 1000 KW = 250 kilowatts
III. Convert 724 KW to megawatt.
SOLUTION:
Data given in the question:
Power = 724 KW
1000 KW = 1 MW. to change KW to megawatt, you divide the kilowatt by 1000
724 KW = 724 ÷ 1000 MW = 0.724 megawatts
Calculate the value of 21⅔ MW in watt.
SOLUTION: Data given in the question:
Power = 21⅔ MW
1 MW = 1000000 W. to change megawatt to Watt, you multiply megawatt by 1000000
21⅔ MW = 21⅔ * 1000000 = 65/3 * 1000000 = 21666666.67 watts
RATING OF ELECTRICAL APPLIANCES:
Every electrical appliances are rated to indicate their power in watt and voltage in volt. The current consumption of the device can be calculated from the formula,
power = current I * voltage V
Rating electrical appliances tells one how the appliances are to be used to avoid under use or over use which will damage the appliances. Electrical appliances function best over a reasonable period of time where they best fit in a circuit.
EXPLANATION:
Assuming a mark of 110 W, 200 V is written on the body of an electrical appliance, it means that the electrical appliance consume a power of 110 watts ( or it consume energy of 110 joules in one second ) when it is connected to a circuit that maintains a voltage of 200 volt.
The current consumed by the electrical appliance can be calculated using the formula,
power = current I * voltage V.
110 = I * 200. . I = 100 / 220. . I = 0.55 Ampere
BUYING AND SELLING OF ELECTRICAL POWER:
The electrical energy that our appliances consumed is measured and sold by National Electric Power Authority ( NEPA ) in kilowatt-hour ( KWh)
DEFINITION OF KILOWATT-HOUR (KWh):
Kilowatt hour is defined as the electrical energy that is consumed by an electrical appliance when the appliance consume one kilowatt of energy in one hour.
VALUE OF ONE KILOWATT HOUR ( 1 kWh ) IN JOULE:
1 KWh = 1 KW of energy * 1 hour. . 1 KWh = 1000 watts * 60 seconds * 60 minutes 1 KWh = 3600000 joules
1 KWh = 3.6 * 10^6 joules
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
An electrical appliance rated 5 A, 110V is used for 10 hours. What is the cost of operating the appliance at the cost of 10 kobo per KWh?
SOLUTION:
Data given in the question:
Current = 5 A, voltage = 110V, time = 10 hours = 10 hours, cost = 10 Kobo per KWh
First, calculate the power of the appliance:
power = current * voltage . Power = 5 * 110 power = 550 watts
Second, change the power in watt to kilowatt by dividing it by 1000:
550 watts = 550 / 1000 = 0.550 kilowatt
Third: calculate the energy consumed:
Energy = power * time. energy = 0.55 * 10 energy = 5.5 kwh
Fourth: calculate the cost of consuming 5.5 kWh:
1 kWh cost 10 kobo, then 5.5 kWh will cost 5.5 * 10 = 55 kobo = 55 / 100 = #0.55
Heating Effect of Current:
When electricity or electric current flows through a conductor, the resistance of the conductor convert electrical energy into heat energy. It is this heating effect of electric current that is utilized in devices such as electric iron, electric toaster, electric cooker, hair drier, and electric bulb. The quantity of heat generated depends on the value of current.
Formula For Calculating Heat Energy of Electric Current:
The following factors determine the quantity of heat energy that is generated or produced in a given circuit:
The quantity of heat generated is directly proportional to the square of current that flow in the circuit. This means that the quantity of heat will increase if the square of current is increased and vice versa.
Mathematically:
Quantity of heat Q is directly proportional to the square of current I
Q ∞ I²
Q = K * I²
Make k the subject of formula: K = Q / I²
If there is any change in Q or I, there will be a corresponding change in I or Q. That is if Q change from Q1 to Q2 , I will also change from I1 to I2.
Therefore, we get the formula:
Q1 / (I1)²= Q2 / (I2)² or Q1 * (I2)² = Q2 * (I1)²
Application of Formula for Solving Problems:
Examples:
1..When 2.5 amperes of current flow in a wire, 35J of heat is produced. Calculate the quantity of heat that will be generated if the current is 1.25 A.
Solution:
Data given in the question:
Q1 = 35 J, I1 = 2.5 A, Q2 = ? and I2 = 1.25J
Formula: Q1 * (I²)2 = Q2 * (I²)1
Substitution: 35 * (2.5)² = Q2 * (1.25)²
Make Q2 the subject of the formula: Q2 = 35 * (2.5)² / (1.25)² Q2 = 140 Joules
2..470 J of heat is generated when the value of current in a circuit is I ampere. If 300 Joules of heat is generated when the current becomes ( I + 3 ) amperes, find the value of I1 and I2.
Solution:
Data given in the question:
Q1 = 470 J, I1 = I ampere, Q2 = 300 J and I2 = ( I + 3) amperes
Formula: Q1 * (I²)2 = Q2 * (I²)1
Substitution: 470 * ( I + 3 )² = 300 * (I)²
Expand the power first: 470 (( I + 3 )*( I + 3 )) = 300 * (I)²
470 ( I² + 6I + 9 ) = 300 * (I)²
( I² + 6I + 9 ) = 300 * (I)² / 470
. I² + 6I + 9 = 0.64(I)²
Collect like terms: I² - 0.64I² + 6I + 9 = 0
0.36I² + 6I + 9 = 0
Solve the equation using general formula: I = - b ± √ ( b² - 4ac ) / 2a.
Then, a = 0.36, b = 6 and c = 9
Substitution: I = - 6 ± √ ( 6² - 4 * 0.36 * 9 ) / 2 * 0.36
I = ( - 6 ± √ ( 36 - 3.24 )) / 0.72
I = ( -6 ± √ 32.76 ) / 0.72
I = ( - 6 ± 5.72 ) / 0.72
I1 = ( - 6 + 5.72 ) / 0.72 = 0.28 / 0.72 = 0.39 ampere
I2 = ( - 6 – 5.72 ) / 0.72 = - 11.72 / 0.72 = - 16.28 amperes
We will take the positive value and ignore the negative the value because current can not be negative.
Then, I = 0.39 A. I1 = I = 0.39Ampere, I2 = ( I + 3 ) = 0.39 + 3 = 3.39Amperes
Please take your time to follow and understand the working.
The quantity of heat energy that is generated in a conductor is directly proportional to the resistance of the conductor. This means that the quantity of heat generated will increase of the resistance of the conductor is increased and vice versa.
Mathematically:
Quantity of heat Q is directly proportional to the resistance R of the conductor
Q ∞ R
Q = K * R
Make k the subject of formula: K = Q / R
If there is any change in Q or R, there will be a corresponding change in R or Q. That is if Q change from Q1 to Q2 , R will also change from R1 toIR2. Therefore, we get the formula:
Q1 / R1 = Q2 / R2 or Q1 * R2 = Q2 * R1
Application of Formula to Solve Problems:
Examples:
*720 J of heat is generated when the resistance of the circuit is R ohms. If 300 Joules of heat is generated when it becomes 50 ohms, find the value of R1.
Solution:
Data given in the question:
Q1 = 720 Joules, R1 = R, Q2 = 300 Joules and R2 = 50 ohms
Formula: Q1 / R1 = Q2 / R2 or Q1 * R2 = Q2 * R1 ( you can use any of the formulae)
Substitution: 720 / R = 300 / 50
Make R the subject: R = 720 * 50 / 300
R = 120 ohms
*When the resistance of a wire is 2.5 ohms, 35J of heat is produced. Calculate the quantity of heat that will be generated if the resistance is 1.25 ohms
Solution:
Data given in the question:
Q1 = 35 J, R1 = 2.5 ohms, Q2 = ? and R2 = 1.25 ohms
Formula: Q1 * R2 = Q2 * R1
Substitution: 35 * 1.25 = Q2 * 2.5
Make Q2 the subject of the formula: Q2 = 35 * 1.25 / 2.5. Q2 = 17.5 Joules
The quantity of heat energy that is generated in a conductor is directly proportional to the time for which the current flow through the conductor. This means that the quantity of heat energy will increase if the time for which the current flow is increased and vice versa.
Mathematically:
Quantity of heat Q is directly proportional to the time T which the current flow through the conductor
Q ∞ T
Q = K * T
Make T the subject of formula: K = Q / T
If there is any change in Q or R, there will be a corresponding change in R or Q. That is if Q change from Q1 to Q2 , R will also change from R1 toIR2. Therefore, we get the formula:
Q1 / T1 = Q2 / T2 or Q1 * T2 = Q2 * T1
Application of Formula to Solve Problems:
Examples:
*520 J of heat is generated when the of current flow in a circuit is 120 seconds. If 300 Joules of heat is generated when it becomes y seconds, find the value of y.
Solution:
Data given in the question:
Q1 = 520 Joules, T1 = 120 seconds, Q2 = 300 Joules and T2 = ?
Formula: Q1 / T1 = Q2 / T2 or Q1 * T2 = Q2 * T1 ( you can use any of the formulae)
Substitution: 520 / 120 = 300 / T2
Make T2 the subject: T2 = 300 * 120 / 520
. T2 = 69.23 seconds
*A wire generated 3000 joules of heat energy when 2.35 A of current flow in the wire for 2 minutes. What would be heat generated if the same current flow for 1⅓ minutes?
Solution:
Data given in the question:
Q1 = 3000 joules , T1 = 2 minutes = 2 * 60 seconds = 120 seconds, Q2 = ?, T2 = 1⅓ minutes= 1⅓ * 60 seconds = ⁴/3 * 60 sec
T2 = 80 seconds.
Formula: Q1 / T1 = Q2 / T2
Substitution: 3000 * 120 = Q2 * 80
Make Q2 the subject of the formula: Q2 = 3000 * 120 / 80
Q2 = 4500 Joules.
Formula For Calculating Quantity of Heat Generated in an Electrical Circuit:
Now i will now obtain a single formula that is used to calculate the quantity of heat energy generated in an electric circuit by combining the three factors that discussed above together.
thus:
The quantity of heat energy generated in an electric circuit is directly proportional to the square of current that flow in the conductor, the resistance of the conductor and the time for which the current flow.
Mathematically:
Quantity of heat Q is directly proportional to square of current, resistance of wire and time current flow
Quantity of heat Q ∞ I² * R * T
Q ∞ I²RT. . Q = KI²RT
K is constant of proportionality. If k = 1 , . Q = I²RT
Application Of The Formula To Solve Problems:
Examples:
*Determine the quantity of heat generated in a coil of resistance 25 ohms when a current of 7.2A flow for 2 minutes.
Solution:
Data given in the question:
R = 25 ohms, Current I = 7.2 A , time T = 2 minutes = 2 * 60 seconds = 120 seconds
Formula: Q = I²RT
Substitution: Q = 7.2² * 25 * 120 . Q = 155520 joules
*Calculate the current that will generate 350 Joules of energy in a wire of resistance 50 ohms if it flow for 320 seconds.
Solution:
Data given in the question:
Quantity of heat Q = 350 J, resistance R = 50 ohms, Current I = ? , time T = 320 seconds.
Formula: Q = I²RT
Substitution: 350 = I² * 50 * 320
Make I the subject of the formula: I² = 350 / ( 50 * 320 )
I² = 350 / 16000
I² = 0.022. . I = √ 0.022 I = 0.148 ampere
*What would be the resistance of a wire whose current is 0.05% of 200 joule of heat energy generated for 85 seconds.
Solution:
Data given in the question:
Quantity of heat Q = 200 Joules, time T = 85 seconds, current I = 0.05% of Q = 0.05% * 200 = 0.05/100 * 200
I = 0.10 Ampere.
Formula: Q = I²RT
Substitution: 200 = 0.1² * R * 85
Make R the subject of the formula: R = 200 /( 0.1² * 85 )
R = 200 / 0.85 . R = 235.29 ohms
Electric Filament Lamps:
In electric filament lamps, electrical energy is converted into light and heat energy. Light is produced when current flow into the filament wire, the filament is heated up due to its resistance to a high temperature which cause the filament to glow and give out light.
The filament is enclosed in an Evacuated enclosure which contain inert gas ( argon ) at low pressure. This is to prevent the hot filament from evaporating and enable the lamp to be run at high temperature without being damaged.
Fuse:
A fuse is an electrical safety device that protect other devices from being damaged by excess current. It is a short length of wire of low melting point that is connected to the live line of an electric circuit.
Function of a Fuse:
The function of a fuse is to automatically open the circuit when excess current that might destroy the circuit or appliances flow in the circuit. The excess current generate enough heat which melt the fuse and brake the circuit.
Fuse Rating:
Fuse rating is the maximum safe current that is permitted to flow through the fuse before the fuse brake.
Diagram of a Fuse:
Continuity Tester:
It is a simple tester that is used to detect fault in an electric circuit. It is used to test for the continuity of a circuit in order to detect an open circuit fault in the circuit.
Simple continuity tester circuit:
A simple continuity tester contains a cell, a bulb, and connection wires. The circuit diagram is as shown below:
PRINCIPLE OF OPERATION OF A CONTINUITY TESTER:
When the ends A and B of the wire are connected, the circuit is completed and the lamp light up.
HOW TO USE A CONTINUITY TESTER:
When the ends A and B of the wires are connected across a component in a circuit, if the lamp will light up. This show that the component is good component or is not faulty. If the lamp do not light up , it means that the component is a faulty component.
EXERCISES:
*thee 3.0 ohms resistors are connected in parallel. What is the equivalent resistance?
*Two 40 ohms resistors are connected in parallel across 12 volt cell. If the current in the circuit is 1.2A, calculate the energy developed in 5 minutes.
*The headlamp of a car takes a current of 0.4A from a 12V supply. The energy developed in 5 minutes is ...
*An electric lamp is marked 240 V, 60 W. What is its resistance when it is operated at a constant voltage?
*define the following: (I) electric field (ii) electric line of force (iii) uniform field (iv) variable field
*Sketch the pattern of line of force around (I) an isolated positive charge (ii) an isolated negative charge ( iii) tow positive charges (iv) two negative charges
(V) positive and negative charge
*List four properties of electric line of force. Why they nor intersect?
*define the following: (I) electric current (ii) potential difference (iii) electromotive force (iv) open circuit ( v) closed circuit (vi) short circuit
* state ohm's law and define resistance. Describe an experiment to explain ohms law. How can you obtain the resistance of a short length of wire using ohms law?
*Four resistors of resistance 1 ohm, 2 ohms , 3 ohms and 5 ohms, are connected in series and afterwards connected in parallel., calculate the total resistance in each case.
*Calculate the total resistance of the following combinations: (I) 3&2 ohms in series with 5 & 10 ohms in parallel (ii) 2 ohm in series with 3, 4 & 5 ohms parallel connection in series with 6 ohms (iii) 4 ohms in series with 1 & 2 ohms parallel connection in series with 3, 5 & 5 ohms parallel connection (iv) 2 & 3 ohms parallel connection in series with 5 ohms in series with 2 & 4 ohms parallel connection.
*Describe two defect of a simple cell. How can you prevent the defects?
* Define volt, watt and kilowatt- hour. An electric lamp iks makes “100W, 220V”. What do you understand by thee? What current is taken by the lamp when it is connected to a 220 V mains supply? What is the resistance of the lamp?
*A television ( 65W), a refrigerator ( 1.2KW), an electric kettle (650W ) and 10 lamps ( 40 W each ) are connected in series in a house. How much will it cost the housekeeper to switch on all the appliances for 24 hours if the cost of electricity is 15 kobo per kWh?
*What is a fuse? How dose it prove t a circuit?
*Define the terms thermoelectric effect, thermocouple. Differentiate between conductors and insulators. Give 3 examples for each.
Material such as metal conduct electricity. They allow electric charge to flow through them.
OHM'S LAW:
Ohm's law states that the current that flows through a conductor is directly proportional to the voltage that is applied across the terminals of the conductor, provided other factors remain constant.
Ohm's law shows the relationship that exist between the voltage that is applied across a conductor and the current that flows through the conductor the time the voltage is applied.
FORMULA OF OHM'S LAW:
From the statement of ohm's law, current is directly proportional to voltage applied across the conductor.
Mathematically, Voltage V ∞ current I . Voltage V = Resistance R * Current I.
. V = I * R
We can get different formula from the above formula by making I and R subject of the formula one after the other.
Therefore, I = V / R and R = V / I .
APPLICATION OF OHM'S LAW:
EXAMPLES:
Calculate the current produced when 250 volt cell is connected across 12 ohms resistor.
SOLUTION:
Data given in the question:
Voltage V = 250 volt, Resistance R = 12 ohms, current I = ?
Formula: Voltage V = Current I * Resistance R
Substitution: 250. = current I * 12
Make current the subject of the formula: current I = 250 / 12. . Current I = 20.83 Amperes
Determine the value of resistance that will cause a voltage of 125 volt to generate 2.5 amperes of current.
SOLUTION:
Data given in the question:
Voltage V = 125 volt, current I = 2.5 amperes, resistance R = ?
Formula: Voltage V = Current I * Resistance R
Substitution: 125. = 2.5 * resistance R
Make resistance R the subject of formula:
125 / 2.5. = resistance R . Resistance R = 50 ohms
EXPERIMENTAL VERIFICATION OF OHM'S LAW:
AIM:
To verify ohm's law
APPARATUS:
Ammeter, voltmeter, rheostat, standard resistor, cell, connection wires and switch.
SETUP DIAGRAM:
PROCEDURES:
The circuit diagram is connected as shown above.
Close the switch so that current flow in the circuit.
Read and record the values of current that flow in the circuit and voltage across the resistor recorded by the ammeter and voltmeter respectively.
Adjust the rheostat to a new position so as to change the resistance of the circuit. Read and record the new values of current I, and voltage,V recorded by the ammeter and voltmeter respectively. Repeat the process three more times. Read and record the corresponding values of I and V recorded by the ammeter and voltmeter.
Tables of values:
S/n. Rheostat position P. Current I ( A ). Voltage V (volt)
1 P1. I1. V1
2 P2. I2. V2
3 P3. I3. V3
4 P4. I4. V4
5 P5. I5. V5
GRAPH OF VOLTAGE V AGAINST CURRENT I:
The different values of Voltages V that were obtained are plotted on the vertical axis against different values of current I on the horizontal axis. A straight line graph which passes through the origin is obtained as shown below.
SLOPE OF THE GRAPH OF VOLTAGE V AGAINST CURRENT I:
Slope S or gradient = ∆ Voltage / ∆ current
Slope s or gradient = change in voltage / change in current
Slope s /gradient = ( Voltage V2 – Voltage V1 ) / ( current I2 – current I1 )
Slope s / gradient = V2 - V1 / I2 - I1
Slope s / gradient = Voltage / current
Recall that resistance R = voltage V / current I.
Therefore, slope = resistance.
The slope of the grape of the graph represent the resistance of the circuit.
PRECAUTIONS:
Make sure the contacts are tight to prevent leakages.
Off the circuit after each experiment so as not to run down the cell
Take the reading when the ammeter and voltmeter pointers are steady
Avoid error due to parallax when taking the readings.
CONNECTION OF RESISTORS:
Resistors can be connected in two ways in a circuit. The connections are :
Connection of resistors in series
Connection of resistors in parallel
1..CONNECTION OF RESISTORS IN SERIES:
In series connection of resistors, the resistors are connected end to end to form a linear network as shown in the circuit diagram below.
TAKE NOTE:
In series connection of resistors:
The same amount of current flows through each of the resistors, one after the other.
The voltage drops across each of the resistors are different, but if the resistances of the resistors are of the same value,then the voltage drop across each of the resistors would be the same.
FORMULA FOR CALCULATING TOTAL RESISTANCE OF RESISTORS CONNECTED IN SERIES:
Assuming three resistors R , R , R are connected in series as shown in the diagram below, the total resistance is calculated by the formula
Total resistance Rt = R1 + R2 + R3
DERIVATION OF THE FORMULA FOR CALCULATING TOTAL RESISTANCE OF RESISTORS CONNECTED IN SERIES:
Circuit diagram for connection of resistors in series is as shown by the your below..
From the circuit diagram, the voltage V is connected across the three resistors this voltage is shared across each of the resistor.
Remember that the voltages drop across each of the resistors are different because the resistances of the resistors are different. Therefore, Let the Voltage of the cell connected across the three resistors be V.
Let the voltage drop across resistor R1 be V1.
Let the voltage drop across resistor R2 be V2, also,
Let the voltage drop across resistor R3 be V3.
Therefore,
Vt = V1 + V2 + V3
From ohm's law which states that, V = I * R
For Rt, we have Vt = I*Rt, for R1, V1 = I*R1, for R2, V2 = I*R2 and for R3, V3 = I*R3
Now, We will substitute for Vt, V1, V2 and V3 in the formula above. Then we will get,
Vt = V1 + V2 + V3
I*Rt = I*R1 + I*R2 + I*R3
Let us factorie I : I*Rt = I ( R1 + R2 + R3 )
Divide both side by I: I*Rt / I = I ( R1 + R2 + R3 ) / I. Then we will get,
Rt = R1 + R2 + R3
APPLICATION OF FORMULA TO SOLE PROBLEMS:
EXAMPLES:
1. From the diagrams below, calculate the total resistance of the circuits.Tfrf
Solution:
Data given in the figure:
From the figure, R1 = 5ohms, R2 = 8 ohms, R 3 = 12 ohms.
Formula:
Rt = R1 + R2 + R3
Substitution: Rt = 5 + 8 + 12
➡ Total resistance of the circuit Rt = 25 ohms
2. What is the equivalent resistance of the circuit containing three resistors connected
Solution:
Data given in the question:
R1 = 1 ohm, R2 = 2 ohms, R3 = 3 ohms
Formula: Rt = R1 + R2 + R3
Substitution: Rt = 1 + 2 + 3
Total resistance of the circuit Rt = 6 ohms.
3. 4 ohms, 8 ohms and R resistors are connected in series. If the total resistance of the circuit is 23, calculate the value of R.
Solution:
Data given in the question:
R1 = 4 ohms, R2 = 8 ohms, R3 = ?
Formula: Rt = R1 + R2 + R3
Substitution: 23 = 4 + 8 + R. ➡ 23 = 12 + R
Make R the subject: R = 23 - 12. ➡ R = 11 ohms
2..CONNECTION OF RESISTORS IN PARALLEL:
In parallel connection of resistors, the positive terminals of all the resistors are connected to the same positve terminal and the negative terminals of the resistors are connected to the same negative terminal. Parallel connection of resistors always form a network of two or more rooms depending on the numbers of resistors used in the connection. The circuit diagram is as shown in the figure by your right.
TAKE NOTE:
In parallel connection of resistors:
The voltages across the resistors are the same or the same voltage act across each of the resistors.
The current that flow through each of the resistors are different because the resistances of each of the resistors are different.
If the resistance of each of the resistors are the same, then the current that will flow through each of the resistors would be the same.
FORMULA FOR CALCULATING TOTAL RESISTANCE OF RESISTORS CONNECTED IN PARALLEL:
Assuming three resistors R1 , R2 , R3 are connected in parallel as shown in the diagram below, the total resistance is calculated by the formula:
diagram is as shown in the figure by your right. 1 / Rt = 1/R1 + 1/R2 + 1/R3
DERIVATION OF THE FORMULA FOR CALCULATING TOTAL RESISTANCE OF RESISTORS CONNECTED IN SERIES:
Circuit diagram:From the circuit diagram, it is the voltage V that is connected across the three resistors. I is the total current generated in the circuit. I1,I2 and I3 are the current that will flow through the resistors, R1, R2, and R3 respectively while the voltages across the resistors are the same.
Remember that different values of current will flow through each of the resistors because the resistances of the resistors are different.
Therefore,
Let the Voltage of the cell connected across the three resistors be V.
Let the current that flow through resistor R1 be I1.
Let the current that flow through resistor R2 be I2
Let the current that flow through resistor R3 be I3
Therefore, total current Iohm
It = I1 + I2 + I3
Recall from ohm's law that, V = I*R. Let us make I the subject of the formula. Then we will get,
I = V / R, then,
For It, we get It V/Rt; for I1, we get I1 = V/R1; for I2, we get I2 = V/R2; for I3, we get I3 = V/R3.
We will substitute for It, I1,I2 and I3 in the formula above. Then, we will get:
V/Rt = V/R1 + V/R2 + V/R3
Let us factorie V: V/Rt = V ( 1/R1 + 1/R2 + 1/R3 )
Divide both sides of the equation by V: V/Rt ÷V = V ( 1/R1 + 1/R2 + 1/R3 )
V cancels V. Then we will get: 1/Rt = 1/R1 + 1/R2 + 1/R3
which is the formula for calculating total resistance of resistors connected in parallel.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
1. Three resistors 5 ohms, 8 ohms and 3 ohms are connected in parallel. Calculate the total resistance of the circuit.
Solution:
Data given in the question:
R1 =3 ohms, R2 = 8 ohms, R3 = 5 ohms
Formula: 1/Rt = 1/R1 + 1/R2 + 1/R3
Substitution: 1/Rt = 1/3 + 1/8 + 1/5
Lc of the fraction = 3 * 5 * 8 = 120
1/Rt = 1(120)/3 + 1(120)/5 + 1(120)/8
1/Rt = 40 + 24 + 15. ➡ 1/Rt = 79 ➡ Rt = 1/ 79 ➡ Rt = 0.013 ohm
2. From the figure, calculate the effective resistance of the circuit.
Solution:
Data given in the question:
R1 = 6 ohms, R2 = 4 ohms, R3 = 2 ohms
Formula: 1/Rt = 1/R1 + 1/R2 + 1/R3
Substitution: 1/Rt = 1/6 + 1/4 + 1/2
LCM of the fractions = 12
Therefore, 1/Rt = 1(12)/6 + 1(12)/4 + 1(12)/2
1/Rt = 2 + 3 + 6 ➡ 1/Rt = 11➡ Rt = 1/11
Effective resistance Rt = 0.09 ohm
FACTORS THAT AFFECT THE RESISTANCE OF A CONDUCTOR:
The following factors affect the resistance of a conductor:
- The length of the conductor
- The cross sectional area of the conductor
- The temperature of the conductor
- The nature of the material of which the conductor is made.
Let us look at these factors one after the other. We will build a formula that are very useful in calculations:
1..LENGTH OF CONDUCTOR:
The resistance R of a conductor is directly proportional to the length of the conductor. This means that the resistance of a conductor increases as the length of the conductor increases and vice versa.
Mathematically:
Resistance R is directly proportional to length of conductor L
Resistance R ∞ length of conductor L
R ∞ L
R = K*L
Make K the subject of the formula.: K = R / L
If there is any change in R or L, there will be a corresponding change in L or R.
Therefore,
we have the formula, R1/L1 = R2/ L2 or R1*L2 = R2*L1
R1 and L1 are the initial resistance and length of the conductor. If there is a change in R1 to R2, then L1 will also change from L1 to L2 so that the constant will always remain constant.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
The resistance of a conductor of length 25cm is 0.25 ohm. Calculate the resistance of the conductor when its length is increased by 20%.
SOLUTION:
Data given in the question:
Resistance R1 = 0.25 ohm, Length L1 = 25 cm, Resistance R2 = ?, Length L2 is increased by 20% of L1.
Therefore,
20% of L1 = 20% of 25cm = 20/100 * 25 = 5
L2 = 5 + L1 = 5 + 25 = 30cm. L2 = 30cm
Formula : R1 * L2 = R2 * L1
Substitution: 0.25 * 30 = R2 * 25
Make R2 the subject: R2 = 0.25 * 30 / 25 R2 = 81.25 ohms
A conductor has a resistance of 30ohms when its length is 120cm. Find the length of the conductor when its resistance is one third its original value.
SOLUTION:
Data given in the question:
Resistance R1 = 30 ohms, length L1 = 120 cm, resistance R2 = one- third of R1, length L2 = ?
Resistance R2 = ⅓ of R1 = ⅓ * 30 = 10.then, R2 = 10 ohms
Formula: R1 * L2 = R2 * L1
Substitution: 30 * L2 = 10 * 120
Make L2 the subject of the formula: L2 = 10 * 120 / 30
L2 = 40 cm
The resistance of a wire is R ohms when its length is 1.25 m. If the length becomes 1.75m when the resistance increased to ( R+ 2 ) ohms, find the value of R.
SOLUTION:
Data given in the question:
Resistance R1 = R, length L1 = 1.25m, resistance R2 = (R + 2) ohms, length L2 = 1.75m
Formula: R1 * L2 = R2 * L1
Substituting: R * 1.75 = (R + 2) * 1.25
Open the bracket: R * 1.75 = 1.25R + 2.50
Collect like terms: 1.75R – 1.25R = 2.50
0.50R = 2.50
Make R the subject of the formula: R = 2.50 / 0.50
R = 5 ohms
Therefore, R1 = 5 ohms, R2 = R + 2 = 5 + 2 = 7 ohms
2..AREA OF CONDUCTOR:
The resistance of a conductor is inversely proportional to the cross sectional area of the conductor provided all other factors remain constant. This means that the resistance of a conductor will increase if the cross sectional area of the conductor is reduced or decreased. And vice versa.
Mathematically:
Resistance R is inversely proportional to cross sectional area of conductor
Resistance R ∞ 1/A
R = K / A
Make K the subject of the formula: K = R * A
If there is any change in R or A, there will be a corresponding change in A or R.
Therefore, we have the formula, R1 * A1 = R2 * A2 or R1/A2 = R2/A1
R1 and A1 are the initial resistance and area of the conductor. If there is a change in R1 to R2, then A1 will also change from A1 to A2 so that the constant will always remain constant.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
*The area of a wire is A m² when its resistance is 1.25 ohms. If the resistance becomes 1.75ohms when the area is increased to ( A+ 2 ) m², find the value of A.
SOLUTION:
Data given in the question:
Area A1 = A, resistance R1 = 1.25 ohms, area A2 = (A + 2) m², resistance R2 = 1.75 ohms
Formula: R1 * A1 = R2 * A2
Substituting: 1.25 * A = 1.75 * (A + 2)
Open the bracket: 1.25 * A = 1.75R + 3.50
Collect like terms: 3.50 = 1.75A – 1.25A
= 0.5A
Make A the subject of the formula: A = 3.50 / 0.50
A = 7 m²
Therefore, A1 = 7m², A2 = A + 2 = 7 + 2 = 9 m²
*A conductor has a resistance of 30ohms when its area is 12.0cm². Find the area of the conductor when its resistance is one third its original value.
SOLUTION:
Data given in the question:
Resistance R1 = 30 ohms, area A1 = 12.0 cm², resistance R2 = one- third of R1, area A2 = ?
Resistance R2 = ⅓ of R1 = ⅓ * 30 = 10.then, R2 = 10 ohms
Formula: R1 * A1 = R2 * A2
Substitution: 30 * 12.0 = 10 * A2
Make A2 the subject of the formula: A2 = 30 * 12.0 / 10. . A2 = 36 cm²
*The resistance of a conductor of area 25cm² is 0.25 ohm. Calculate the resistance of the conductor when its area is increased by 20%.
SOLUTION:
Data given in the question:
Resistance R1 = 0.25 ohm, area A1 = 25 cm², Resistance R2 = ?, Area A2 is increased by 20% of A1.
Therefore,
20% of A1 = 20% of 25cm² = 20/100 * 25 = 5
A2 = 5 + A1 = 5 + 25 = 30cm. A2 = 30cm
Formula : R1 * A1 = R2 * A2
Substitution: 0.25 * 25 = R2 * 30
Make R2 the subject: R2 = 0.25 * 25 / 30. R2 = 0.21 ohm
3..TEMPERATURE OF CONDUCTOR:
The resistance R of a conductor is directly proportional to the temperature, T of the conductor provided all other factors remain constant. This means that the resistance R of a conductor will increase if the temperature T of the conductor is increased or vice versa.
Mathematically,
Resistance R is directly proportional to temperature T of conductor
Resistance R ∞ Temperature T
R ∞ T
R = KT
Make K the subject of the formula: K = R / T
If there is any change in R or T, there will be a corresponding change in T or R.
Therefore, we have the formula, R1 / T1 = R2 / T2 or R1 * T2 = R2 * T1
R1 and T1 are the initial resistance and temperature of the conductor. If there is a change in R1 to R2, then T will also change from T1 to T2 so that the constant will always remain constant.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
A conductor has a resistance of 30ohms when its temperature is 12.0°C. Find the temperature of the conductor when its resistance is one third its original value.
SOLUTION:
Data given in the question:
Resistance R1 = 30 ohms, temperature T1 = 12.0 °C, resistance R2 = one- third of R1, temperature T2 = ?
Resistance R2 = ⅓ of R1 = ⅓ * 30 = 10.then, R2 = 10 ohms
Formula: R1 * T2 = R2 * T1
Substitution: 30 * T2 = 10 * 12.0
Make A2 the subject of the formula: T2 = 10 * 12.0 / 30. . T2 = 4 °C
*The resistance of a conductor of temperature 25 °C is 0.25 ohm. Calculate the resistance of the conductor when its temperature is increased by 20%.
SOLUTION:
Data given in the question:
Resistance R1 = 0.25 ohm, temperature T1 = 25 °C, Resistance R2 = ?, temperature T2 is increased by 20% of T1.
Therefore,
20% of T1 = 20% of 25 °C = 20/100 * 25 = 5
T2 = 5 + T1 = 5 + 25 = 30 °C. T2 = 30 °C
Formula : R1 * T2 = R2 * T1
Substitution: 0.25 * 30 = R2 * 25
Make R2 the subject: R2 = 0.25 * 30 / 25. R2 = 0.30 ohm
RESISTIVITY OF A CONDUCTOR:
The formula of resistivity is obtained by combining the three factors which affect the resistance of a conductor that I discussed above together. Thus:
The resistance R of a conductor is directly proportional to the length L and inversely proportional to the cross sectional area A of the conductor, all other factors remain constant.
Mathematically,
Resistance R is directly proportional to length L and inversely proportional to cross sectional area A of conductor.
Resistance R ∞ Length / Area A
R ∞ L / A
R = K*L / A
Make K the subject if the formula: K = R * A / L.
K is constant of proportionality called resistivity. Its unit is ohm-meter
If there is any change in any one of these ( R, L and A), there will be a corresponding change in the other two.
Therefore, we get the formula, R1 * A1 / L1 = R2 * A2 / L2. Or. R1 * A1 * L2 = R2 * A2 * L1
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
The resistance of a wire is 2 ohms when its length and cross sectional area are 1.5m and 0.5m². What would be the resistance of the wire when its length becomes 1.3m and its cross sectional area change to 1.7m²?
SOLUTION:
Data given in the question:
Resistance R1 = 2 ohms, Length L1 = 1.5m , Area A1 = 0.5m² , Resistance R2= ? , Length L2 = 1.3m, Area A2 = 1.7m²
Formula: R1 * A1 / L1 = R2 * A2 / L2
Substitution: 2 * 0.5 / 1.5 = R2 * 1.7 / 1.3
Make R2 the subject of the formula: R2 = ( 2 * 0.5 * 1.3 ) / (1.5 * 1.7 )
R2 = 1.3 / 2.55. R2 = 0.51 ohm.
*At a length of 12 cm, the resistance of a wire of cross sectional area of 5.0cm² is 35ohms. If the resistance of the wire remain constant when its length increased to 20 cm, find the area of the conductor.
SOLUTION:
Data given in the question:
Resistance R1 = 35ohms, Length L1 = 12 cm, area A1 = 5.0 cm², R2 remains constant, R2 = R1 = 35ohms, L2 = 20cmm,
A2 = ?
Formula: R1 * A1 / L1 = R2 * A2 / L2
Substitution: 35 * 5.0 / 12 = 35 * A2 / 20
Make A2 subject if the formula: A2 = ( 35 * 5.0 * 20 ) / ( 12 * 35 )
A2 = 3500 / 420. . A2 = 8.33 m²
*The worked examples are to show you how to solve problems using the formula that instate above. More questions will be in the exercises below.
ELECTRICAL WORK DONE IN A CIRCUIT:
In an electric circuit, work is done when electricity flow from a point of higher potential to a point of lower potential. The unit of work done is joule, (J).
FORMULA FOR WORK DONE IN ELECTRIC CIRCUIT:
If Q Coulomb of electricity flows from one point to another whose difference in potential is V volt, then the formula for work done in the circuit is
Work done donein joule = charge Q in Coulomb * voltage V in volt
W = Q * V
Using their units units, Then joule = coulomb * volt
Recall that, Q = I * t. Let us substitute for Q in the equation above,
Then Work done = Q * V . Work done = I * t * V
Also, recall from ohm's law, V = I * R. Let us substitute for V in the equation above. Then we will get,
Work done W = I * t * I * R. W.D = I² * t * R.
Again from ohm's law, we make I the subject of the formula. Then we will get, I = V / R.
Let us substitute for I in the formula above, then we will get,
Work done = I * t * V work done = V / R * t * V. work done = V² * t / R
Where I is current, t is time and v is voltage.
CONVERSION OF ELECTRICAL ENERGY AND DEVICES USED:
Electrical energy can be converted into other forms of energy as I explain below:
Electrical energy can be converted into heat energy by the use of electric iron, electric kettle, immerse heater.
The amount of heat energy can be calculated using the formula:
Work done = current * voltage * time
Work done = I * V * t
Work done = I² * R * t
Electrical energy can be converted into mechanical energy by the movement of the blades of electric fan, electrica motor.
Electrical energy can be converted into light energy by bulbs, lambs.
Electrical energy that can be converted into electrical by telephone ear piece.
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
Calculate the heat energy generated when 2.5 A of current flow through a resistor of resistance 8.0 Ohms for
¾ minutes.
SOLUTION:
Data given in the question:
Current I = 2.5 A, time = ¾minutes = ¾ * 60 seconds = 45 seconds, resistance R = 8.0 ohms.
Formula: Work done = I² * R * t
Substitution: work done = 2.5² * 8 * 45. work done = 2250 joules
Given that the work done in electric circuit is 334 joules, determine the current in the voltage if the resistance of the circuit is 5 ohms and runs for 6 minutes.
SOLUTION:
Data given in the question:
Work done = 334 joules, resistance R = 5 ohms, time = 6 minutes = 6 * 60 seconds = 360 seconds, voltage v = ?
Let me do some explanations hear. Ok!
Formula: Work done = Q * V
From the question, Q is not given. So, we must recall that Q = I * t and substitute it for Q in the formula above. Then we will get,
work done = Q * V work done = I * t * V
At this point, if you look at the above formula, you will notice that current I is present but it is not given in the question. So we must recall ohms law which says V = I * R. We have to make I the subject. then I = V / R. Now we have to substitute for I in the formula, work done = I * t * V. The formula will become, work done = V / R * t * V.,
work done = V² * t / R. This the formula we will use for the calculation because it contains R, t and V that we want to calculate.
Let us substitute for w, t and R: work done = V² * t / R
334 = V² * 360 / 5.
Make V the subject of the formula: 334 * 5 = V² * 360
* 5 / 360 = V²
= V²
V = √4.64
V = 2.15 volt
ELECTRIC POWER:
Electric power is the time rate of electric work done in a circuit. It is the ratio of electric work done to time taken to do the work. The unit of electric power is watt.
FORMULA FOR CALCULATING ELECTRIC POWER:
Power = work done in joule / time taken in second
Recall that electric work done = I * V * t. Then we substitute for work done in the above formula.
Therefore,
Power = work done / time . Power = I * V * t / t
t divides t: power = I * V
You have to recall that, from ohm's law, V = I * R and you have to substitute for V in the above equation. Then,
Power = I * V. power = I * I * R = I² * R. power = I² * R
DEFINITION OF WATT:
Watt is defined as one joule per second. It is the power that is consumed in an electric circuit when one joule of work is done in one second.
KILOWATT (KW) AND MEGAWATT (MW):
Kilowatt and megawatt are larger units of power in which power can also be measured.
VALUES OF KILOWATT AND MEGAWATT:
1 KW. = 1000 W. = 1 * 10³ W
1 MW = 1000000 W. = 1 * 10²+⁴ W
CONVERSION OF WATT TO KILOWATT, MEGAWATT AND VICE VERSA:
TAKE NOTE:
To change power in watt to kilowatt, you divide the watt by 1000.
To change power in watt to megawatt, you divide the watt by 1000000.
To change kilowatt to megawatt, you divide the kilowatt by 1000.
To change kilowatt to watt, you multiply the 1000.
To change megawatt to watt, you multiply the megawatt by 1000000.
To change megawatt to kilowatt, you multiply the megawatt by 1000.
EXAMPLES:
What is the value of 12¾ kilowatt in watt?
SOLUTION:
Data given in the question:
Kw = 12¾ kw.
1000W = 1 KW. . 12¾ KW = 12¾ * 1000 = 51/4 * 1000 = 798 Watts
What is the value of 0.25 MW in kilowatt?
SOLUTION:
Data given in the question:
Power = 0.25 MW
1000 KW = 1 MW . to change megawatt to kilowatt, multiply the megawatt by 1000
0.25 MW = 0.25 * 1000 KW = 250 kilowatts
III. Convert 724 KW to megawatt.
SOLUTION:
Data given in the question:
Power = 724 KW
1000 KW = 1 MW. to change KW to megawatt, you divide the kilowatt by 1000
724 KW = 724 ÷ 1000 MW = 0.724 megawatts
Calculate the value of 21⅔ MW in watt.
SOLUTION: Data given in the question:
Power = 21⅔ MW
1 MW = 1000000 W. to change megawatt to Watt, you multiply megawatt by 1000000
21⅔ MW = 21⅔ * 1000000 = 65/3 * 1000000 = 21666666.67 watts
RATING OF ELECTRICAL APPLIANCES:
Every electrical appliances are rated to indicate their power in watt and voltage in volt. The current consumption of the device can be calculated from the formula,
power = current I * voltage V
Rating electrical appliances tells one how the appliances are to be used to avoid under use or over use which will damage the appliances. Electrical appliances function best over a reasonable period of time where they best fit in a circuit.
EXPLANATION:
Assuming a mark of 110 W, 200 V is written on the body of an electrical appliance, it means that the electrical appliance consume a power of 110 watts ( or it consume energy of 110 joules in one second ) when it is connected to a circuit that maintains a voltage of 200 volt.
The current consumed by the electrical appliance can be calculated using the formula,
power = current I * voltage V.
110 = I * 200. . I = 100 / 220. . I = 0.55 Ampere
BUYING AND SELLING OF ELECTRICAL POWER:
The electrical energy that our appliances consumed is measured and sold by National Electric Power Authority ( NEPA ) in kilowatt-hour ( KWh)
DEFINITION OF KILOWATT-HOUR (KWh):
Kilowatt hour is defined as the electrical energy that is consumed by an electrical appliance when the appliance consume one kilowatt of energy in one hour.
VALUE OF ONE KILOWATT HOUR ( 1 kWh ) IN JOULE:
1 KWh = 1 KW of energy * 1 hour. . 1 KWh = 1000 watts * 60 seconds * 60 minutes 1 KWh = 3600000 joules
1 KWh = 3.6 * 10^6 joules
APPLICATION OF FORMULA TO SOLVE PROBLEMS:
EXAMPLES:
An electrical appliance rated 5 A, 110V is used for 10 hours. What is the cost of operating the appliance at the cost of 10 kobo per KWh?
SOLUTION:
Data given in the question:
Current = 5 A, voltage = 110V, time = 10 hours = 10 hours, cost = 10 Kobo per KWh
First, calculate the power of the appliance:
power = current * voltage . Power = 5 * 110 power = 550 watts
Second, change the power in watt to kilowatt by dividing it by 1000:
550 watts = 550 / 1000 = 0.550 kilowatt
Third: calculate the energy consumed:
Energy = power * time. energy = 0.55 * 10 energy = 5.5 kwh
Fourth: calculate the cost of consuming 5.5 kWh:
1 kWh cost 10 kobo, then 5.5 kWh will cost 5.5 * 10 = 55 kobo = 55 / 100 = #0.55
Heating Effect of Current:
When electricity or electric current flows through a conductor, the resistance of the conductor convert electrical energy into heat energy. It is this heating effect of electric current that is utilized in devices such as electric iron, electric toaster, electric cooker, hair drier, and electric bulb. The quantity of heat generated depends on the value of current.
Formula For Calculating Heat Energy of Electric Current:
The following factors determine the quantity of heat energy that is generated or produced in a given circuit:
- The Current that Flow:
The quantity of heat generated is directly proportional to the square of current that flow in the circuit. This means that the quantity of heat will increase if the square of current is increased and vice versa.
Mathematically:
Quantity of heat Q is directly proportional to the square of current I
Q ∞ I²
Q = K * I²
Make k the subject of formula: K = Q / I²
If there is any change in Q or I, there will be a corresponding change in I or Q. That is if Q change from Q1 to Q2 , I will also change from I1 to I2.
Therefore, we get the formula:
Q1 / (I1)²= Q2 / (I2)² or Q1 * (I2)² = Q2 * (I1)²
Application of Formula for Solving Problems:
Examples:
1..When 2.5 amperes of current flow in a wire, 35J of heat is produced. Calculate the quantity of heat that will be generated if the current is 1.25 A.
Solution:
Data given in the question:
Q1 = 35 J, I1 = 2.5 A, Q2 = ? and I2 = 1.25J
Formula: Q1 * (I²)2 = Q2 * (I²)1
Substitution: 35 * (2.5)² = Q2 * (1.25)²
Make Q2 the subject of the formula: Q2 = 35 * (2.5)² / (1.25)² Q2 = 140 Joules
2..470 J of heat is generated when the value of current in a circuit is I ampere. If 300 Joules of heat is generated when the current becomes ( I + 3 ) amperes, find the value of I1 and I2.
Solution:
Data given in the question:
Q1 = 470 J, I1 = I ampere, Q2 = 300 J and I2 = ( I + 3) amperes
Formula: Q1 * (I²)2 = Q2 * (I²)1
Substitution: 470 * ( I + 3 )² = 300 * (I)²
Expand the power first: 470 (( I + 3 )*( I + 3 )) = 300 * (I)²
470 ( I² + 6I + 9 ) = 300 * (I)²
( I² + 6I + 9 ) = 300 * (I)² / 470
. I² + 6I + 9 = 0.64(I)²
Collect like terms: I² - 0.64I² + 6I + 9 = 0
0.36I² + 6I + 9 = 0
Solve the equation using general formula: I = - b ± √ ( b² - 4ac ) / 2a.
Then, a = 0.36, b = 6 and c = 9
Substitution: I = - 6 ± √ ( 6² - 4 * 0.36 * 9 ) / 2 * 0.36
I = ( - 6 ± √ ( 36 - 3.24 )) / 0.72
I = ( -6 ± √ 32.76 ) / 0.72
I = ( - 6 ± 5.72 ) / 0.72
I1 = ( - 6 + 5.72 ) / 0.72 = 0.28 / 0.72 = 0.39 ampere
I2 = ( - 6 – 5.72 ) / 0.72 = - 11.72 / 0.72 = - 16.28 amperes
We will take the positive value and ignore the negative the value because current can not be negative.
Then, I = 0.39 A. I1 = I = 0.39Ampere, I2 = ( I + 3 ) = 0.39 + 3 = 3.39Amperes
Please take your time to follow and understand the working.
- The Resistance of the Conductor or wire:
The quantity of heat energy that is generated in a conductor is directly proportional to the resistance of the conductor. This means that the quantity of heat generated will increase of the resistance of the conductor is increased and vice versa.
Mathematically:
Quantity of heat Q is directly proportional to the resistance R of the conductor
Q ∞ R
Q = K * R
Make k the subject of formula: K = Q / R
If there is any change in Q or R, there will be a corresponding change in R or Q. That is if Q change from Q1 to Q2 , R will also change from R1 toIR2. Therefore, we get the formula:
Q1 / R1 = Q2 / R2 or Q1 * R2 = Q2 * R1
Application of Formula to Solve Problems:
Examples:
*720 J of heat is generated when the resistance of the circuit is R ohms. If 300 Joules of heat is generated when it becomes 50 ohms, find the value of R1.
Solution:
Data given in the question:
Q1 = 720 Joules, R1 = R, Q2 = 300 Joules and R2 = 50 ohms
Formula: Q1 / R1 = Q2 / R2 or Q1 * R2 = Q2 * R1 ( you can use any of the formulae)
Substitution: 720 / R = 300 / 50
Make R the subject: R = 720 * 50 / 300
R = 120 ohms
*When the resistance of a wire is 2.5 ohms, 35J of heat is produced. Calculate the quantity of heat that will be generated if the resistance is 1.25 ohms
Solution:
Data given in the question:
Q1 = 35 J, R1 = 2.5 ohms, Q2 = ? and R2 = 1.25 ohms
Formula: Q1 * R2 = Q2 * R1
Substitution: 35 * 1.25 = Q2 * 2.5
Make Q2 the subject of the formula: Q2 = 35 * 1.25 / 2.5. Q2 = 17.5 Joules
- The time which the current flow in the wire:
The quantity of heat energy that is generated in a conductor is directly proportional to the time for which the current flow through the conductor. This means that the quantity of heat energy will increase if the time for which the current flow is increased and vice versa.
Mathematically:
Quantity of heat Q is directly proportional to the time T which the current flow through the conductor
Q ∞ T
Q = K * T
Make T the subject of formula: K = Q / T
If there is any change in Q or R, there will be a corresponding change in R or Q. That is if Q change from Q1 to Q2 , R will also change from R1 toIR2. Therefore, we get the formula:
Q1 / T1 = Q2 / T2 or Q1 * T2 = Q2 * T1
Application of Formula to Solve Problems:
Examples:
*520 J of heat is generated when the of current flow in a circuit is 120 seconds. If 300 Joules of heat is generated when it becomes y seconds, find the value of y.
Solution:
Data given in the question:
Q1 = 520 Joules, T1 = 120 seconds, Q2 = 300 Joules and T2 = ?
Formula: Q1 / T1 = Q2 / T2 or Q1 * T2 = Q2 * T1 ( you can use any of the formulae)
Substitution: 520 / 120 = 300 / T2
Make T2 the subject: T2 = 300 * 120 / 520
. T2 = 69.23 seconds
*A wire generated 3000 joules of heat energy when 2.35 A of current flow in the wire for 2 minutes. What would be heat generated if the same current flow for 1⅓ minutes?
Solution:
Data given in the question:
Q1 = 3000 joules , T1 = 2 minutes = 2 * 60 seconds = 120 seconds, Q2 = ?, T2 = 1⅓ minutes= 1⅓ * 60 seconds = ⁴/3 * 60 sec
T2 = 80 seconds.
Formula: Q1 / T1 = Q2 / T2
Substitution: 3000 * 120 = Q2 * 80
Make Q2 the subject of the formula: Q2 = 3000 * 120 / 80
Q2 = 4500 Joules.
Formula For Calculating Quantity of Heat Generated in an Electrical Circuit:
Now i will now obtain a single formula that is used to calculate the quantity of heat energy generated in an electric circuit by combining the three factors that discussed above together.
thus:
The quantity of heat energy generated in an electric circuit is directly proportional to the square of current that flow in the conductor, the resistance of the conductor and the time for which the current flow.
Mathematically:
Quantity of heat Q is directly proportional to square of current, resistance of wire and time current flow
Quantity of heat Q ∞ I² * R * T
Q ∞ I²RT. . Q = KI²RT
K is constant of proportionality. If k = 1 , . Q = I²RT
Application Of The Formula To Solve Problems:
Examples:
*Determine the quantity of heat generated in a coil of resistance 25 ohms when a current of 7.2A flow for 2 minutes.
Solution:
Data given in the question:
R = 25 ohms, Current I = 7.2 A , time T = 2 minutes = 2 * 60 seconds = 120 seconds
Formula: Q = I²RT
Substitution: Q = 7.2² * 25 * 120 . Q = 155520 joules
*Calculate the current that will generate 350 Joules of energy in a wire of resistance 50 ohms if it flow for 320 seconds.
Solution:
Data given in the question:
Quantity of heat Q = 350 J, resistance R = 50 ohms, Current I = ? , time T = 320 seconds.
Formula: Q = I²RT
Substitution: 350 = I² * 50 * 320
Make I the subject of the formula: I² = 350 / ( 50 * 320 )
I² = 350 / 16000
I² = 0.022. . I = √ 0.022 I = 0.148 ampere
*What would be the resistance of a wire whose current is 0.05% of 200 joule of heat energy generated for 85 seconds.
Solution:
Data given in the question:
Quantity of heat Q = 200 Joules, time T = 85 seconds, current I = 0.05% of Q = 0.05% * 200 = 0.05/100 * 200
I = 0.10 Ampere.
Formula: Q = I²RT
Substitution: 200 = 0.1² * R * 85
Make R the subject of the formula: R = 200 /( 0.1² * 85 )
R = 200 / 0.85 . R = 235.29 ohms
Electric Filament Lamps:
In electric filament lamps, electrical energy is converted into light and heat energy. Light is produced when current flow into the filament wire, the filament is heated up due to its resistance to a high temperature which cause the filament to glow and give out light.
The filament is enclosed in an Evacuated enclosure which contain inert gas ( argon ) at low pressure. This is to prevent the hot filament from evaporating and enable the lamp to be run at high temperature without being damaged.
Fuse:
A fuse is an electrical safety device that protect other devices from being damaged by excess current. It is a short length of wire of low melting point that is connected to the live line of an electric circuit.
Function of a Fuse:
The function of a fuse is to automatically open the circuit when excess current that might destroy the circuit or appliances flow in the circuit. The excess current generate enough heat which melt the fuse and brake the circuit.
Fuse Rating:
Fuse rating is the maximum safe current that is permitted to flow through the fuse before the fuse brake.
Diagram of a Fuse:
Continuity Tester:
It is a simple tester that is used to detect fault in an electric circuit. It is used to test for the continuity of a circuit in order to detect an open circuit fault in the circuit.
Simple continuity tester circuit:
A simple continuity tester contains a cell, a bulb, and connection wires. The circuit diagram is as shown below:
PRINCIPLE OF OPERATION OF A CONTINUITY TESTER:
When the ends A and B of the wire are connected, the circuit is completed and the lamp light up.
HOW TO USE A CONTINUITY TESTER:
When the ends A and B of the wires are connected across a component in a circuit, if the lamp will light up. This show that the component is good component or is not faulty. If the lamp do not light up , it means that the component is a faulty component.
EXERCISES:
*thee 3.0 ohms resistors are connected in parallel. What is the equivalent resistance?
*Two 40 ohms resistors are connected in parallel across 12 volt cell. If the current in the circuit is 1.2A, calculate the energy developed in 5 minutes.
*The headlamp of a car takes a current of 0.4A from a 12V supply. The energy developed in 5 minutes is ...
*An electric lamp is marked 240 V, 60 W. What is its resistance when it is operated at a constant voltage?
*define the following: (I) electric field (ii) electric line of force (iii) uniform field (iv) variable field
*Sketch the pattern of line of force around (I) an isolated positive charge (ii) an isolated negative charge ( iii) tow positive charges (iv) two negative charges
(V) positive and negative charge
*List four properties of electric line of force. Why they nor intersect?
*define the following: (I) electric current (ii) potential difference (iii) electromotive force (iv) open circuit ( v) closed circuit (vi) short circuit
* state ohm's law and define resistance. Describe an experiment to explain ohms law. How can you obtain the resistance of a short length of wire using ohms law?
*Four resistors of resistance 1 ohm, 2 ohms , 3 ohms and 5 ohms, are connected in series and afterwards connected in parallel., calculate the total resistance in each case.
*Calculate the total resistance of the following combinations: (I) 3&2 ohms in series with 5 & 10 ohms in parallel (ii) 2 ohm in series with 3, 4 & 5 ohms parallel connection in series with 6 ohms (iii) 4 ohms in series with 1 & 2 ohms parallel connection in series with 3, 5 & 5 ohms parallel connection (iv) 2 & 3 ohms parallel connection in series with 5 ohms in series with 2 & 4 ohms parallel connection.
*Describe two defect of a simple cell. How can you prevent the defects?
* Define volt, watt and kilowatt- hour. An electric lamp iks makes “100W, 220V”. What do you understand by thee? What current is taken by the lamp when it is connected to a 220 V mains supply? What is the resistance of the lamp?
*A television ( 65W), a refrigerator ( 1.2KW), an electric kettle (650W ) and 10 lamps ( 40 W each ) are connected in series in a house. How much will it cost the housekeeper to switch on all the appliances for 24 hours if the cost of electricity is 15 kobo per kWh?
*What is a fuse? How dose it prove t a circuit?
*Define the terms thermoelectric effect, thermocouple. Differentiate between conductors and insulators. Give 3 examples for each.
Pls I need help with number 2 exercise question pls can you help me send it. You can send the solution to my email
ReplyDeleteadeolaadekunle721@gmail.com this is my email
ReplyDeleteI'm looking for capacitor
ReplyDelete