Alternating Current (AC.)
A.C. means alternating current. It is a current that flow in the positive and negative direction periodically.
Waveform of Alternating current:
The waveform of alternating current is sinusoidal waveform.
Diagram of a.c waveform:
Alternating Current Circuit:
Equation of A.C current:
I = Io Sin@
In the production of A.C voltage or current, the coil is in a magnetic field and sweep through angle @.
Therefore,
I = Io Sin @
W = @/t. If we make @ the subject, we will have, @ = w*t. Understand?
If we substitute for @ in the equation, I = Io Sin @
Then the equation will become,
I = Io sin wt.
The next thing is:
When the coil sweep through one revolution or one circle, angle @ = 360°. We will change 360° to radian
Therefore,
@ = 360° = 2π radian. Therefore,
W = @/t = 360°/t = 2π/t. Therefore, w = 2π/t.
At this point, remember the formula that connect t and f,
t = 1/f. You will now substitute for t in the formula,
w = 2π/t thus:
W = 2π/t = 2π ÷ 1/f = 2π x f/1 = 2πf,
therefore
W = 2πf, where f is frequency. It is measured in hertz.
At this point, we will substitute for w in I = Io Sin wt.
Therefore,
I = Io sinwt = Io sin 2πf * t. That is, I = Io sin 2πft
Therefore,
the general equation for a.c current is
I = Io sin 2πft, where
f is the frequency. it is measured in hertz
t is the time. It is measured in second
Io is the peak current. It is measured in ampere
I is the instantaneous current, measured in ampere.
W is angular velocity measured in radian per second.
Equation of A.C current:
Every thing that we did above is applicable to a.c voltage. Therefore the general equation for a.c voltage is the same as that of a.c current. The only difference is that letter V is used instead of Letter I.
Therefore,
V = Vo sin @ , V = sin wt, V = Vo sin 2πft. That is all.
Application of general equation of a.c. current/voltage to solve problems
An a.c. voltage is represented by the equation, V = 4Sin900πt.
Calculate the: I) the peak voltage. II) the frequency. III) the angular velocity.
Solution:
You have to compare each parts of the give equation with the general equation thus:
General equation is I = Io Sin 2πft
The given equation is V = 4 Sin 900πt
I) Peak voltage is Vo = 4volt
II) To find frequency, you compare by saying,
Sin 2πft = Sin 900πt
Sin will cancel Sin, π will cancel π and t will also cancel t.
Therefore we have,
2f = 900, make f the subject of formula, then
f = 900/2 = 450 hertz
III) To find angular velocity, w, you will use
W = 2πf, therefore,
W = 2 x π x f = 2 x π x 450 , w = 900π radian per second.
Or, w = 2 x 22/7 x 450 = 2828.57 rad/sec
Peak Current, Io / peak Voltage, Vo
It is the maximum value of the a.c current or a.c Voltage.
Root Mean Square (r.m.s) of current/ voltage
It is the steady part or d.c part of the alternating current that will produce the same quantity of heat if it flow through the same resistance in the same amount of time.
Relationship between peak current (Io) and I(r.m.s)
Let us derive the formula that connect Io and I(r.m.s).
Firstly,
The rate at which heat is produced is directly proportional to the square of the current.
The I(r.m.s)² is the square of the average of Io².
Therefore,
I(r.m.s)² = average of Io²
I(r.m.s)² = Io²/2. We will square both sides . then we will have,
I(r.m.s)² = Io²/2, if we make I(r.m.s) the subject, we will get,
I(r.m.s) = square root of Io²/2 = √(Io²/2) = √Io²/√2 = Io/√2
Therefore,
I(r.m.s) = Io/√2. Or I(r.m.s) = 0.7070*Io, Io = √2*I(r.m.s)
Io is the peak current while I(r.m.s) is the root mean square of the current.
Relationship between peak voltage(Vo) and V(r.m.s):
Every thing we did above is also applicable to voltage. Therefore, the formula is the for voltage. The only difference is that letter V is used instead of letter I.
Therefore,
V(r.m.s) = Vo/√2. Or V(r.m.s) = 0.7070*Vo Or, Vo = √2*V(r.m.s)
Where Go is the peak voltage, V(r.m.s) is the root mean square of the voltage.
Note:
Why is it that when a voltmeter or ammeter is connected across a.c source, the voltmeter or ammeter give a steady reading?
Answer:
When a voltmeter or ammeter is connected across a.c source, the voltmeter or ammeter give a steady reading because the instrument is only measuring the steady or d.c Part of the a.c current or voltage and not the a.c part.
It is the d.c part of the a.c current/voltage that is powering the electrical devices, not the a.c part. If not so, the devices would power on and off repeatedly for as long as it is powered by ac current/voltage.
Alternating Current Circuit
These are circuit which are powered by alternating current or alternating voltage.
Types of a.c circuits
Resistive a.c circuit:
Circuit diagram:
A resistive circuit is an a.c. circuit that contains only resistor.
Resistance of a R- circuit:
Since the circuit contains only resistor, the circuit therefore obey ohm’s law. Therefore,
V = I*R ( ohm’s law)
Remember that for a.c, voltage and current,
V = Vo Sin wt, and I = Io Sin wt . we will put (substitute) this into the ohm’s law that we stated above.
Therefore, the formula will become,
V = I*R
Vo Sin wt = (Io Sin wt)* R.
We will make R the subject, then we will get,
R = (Vo Sin wt)/(Io Sin wt). ( Sin wt will cancel sin wt, then we will get)
R = Vo/Io (for peak voltage and current)
R = V(r.m.s)/I(r.m.s) ( for r.m.s voltage and current)
Phase Relationship between Current and Voltage in Resistive Circuit
In a resistive circuit, the current is in phase with the voltage. This means that current and voltage start their journey from the same point at the same time, reach their maximum and minimum points at the same time.
Diagram that illustrate
Vector or Phasor diagram of R- Circuit
Worked Examples:
Capacitive a.c Circuit
Diagram of C- circuit:
A capacitive circuit is an a.c circuit that contains only capacitor as the active components.
Phase Relationship between Current and Voltage in Capacitive Circuit
In a capacitive circuit, current leads the voltage by 90°or π/2 radian or ¼ circle. Or voltage lag behind current by 90° or π/2 radian or ¼ circle. The reason is that voltage always wait behind to charge the capacitor before it will start to flow in the circuit while current start to flow immediately in the circuit.
Current and voltage are out of phase. This means that current and voltage start their journey from different points at different time and reach their maximum and minimum points at different time.
Diagram to show phase relationship:
Vector or phasor diagram of C- Circuit:
Phase Difference between voltage and current in c- circuit
This is the amount in degree by which the current leads the voltage in a capacitive circuit. The phase difference is 90° or π/2 or ¼ circle.
Therefore,
V = Vo Sin wt , and I = Io Sin ( wt + π/2 ) .
Note that π/2 is added because current leads the voltage in capacitive circuit.
Capacitive Reactance ( Xc) of a capacitor in a capacitive circuit
The resistance of a capacitor in a capacitive circuit is called capacitive reactance. The symbol is Xc. Its unit is ohm.
Definition of capacitive reactance Xc:
Capacitive reactance is the resistance that a capacitor offers to the flow of an a.c. current.
Formula for calculating capacitive reactance Xc
Xc = 1/wC.
Remember that w = 2πf ( I had derived this formula above)
Therefore,
Xc = 1/2πfC .
where f is the frequency of current or voltage, measured in hertz. C is the capacitance of the capacitor, measured in microfarad.
At this point, we will put ( substitute for ) Xc in the formula of ohm’s law. Therefore,
From ohm’s law, V = I*R. Where R is represent capacitive reactance Xc.
Therefore,
V = I*R = I * 1/2πfC = I/2πfC
Worked Examples:
A 2microfarad capacitor is connected across 150V(r.m.s.), 60 Hz a.c source. Calculate: I) the r.m.s value if the current ii) the peak value.
Solution:
From the question, V(r.m.s) = 150Volt, f = 60Hz,
C = 2microfarad = 2x10~6Farad
Note that:
we are to calculate the r.m.s value of the current. i.e I(r.m.s)
The only formula that we can use to calculate the r.m.s value of current is Xc = V(r.m.s)/I(r.m.s)
Xc is not given in the question. You will notice that C and f are given in the question. So we must first of all calculate Xc and then use the calculated value of Xc to calculate I(r.m.s).
Therefore, we use the formula:
I. Xc = 1/2πfC, to calculate Xc
Xc = 1/ 2*3.142*60*0.000002 = 1/ 0.00075408
Xc = 1326.11924 ohms
At this point, we use the calculated value of Xc to calculate the current. Therefore,
Xc = V(r.m.s) / I(r.m.s) , make I(r.m.s) the subject of the formula,
I(r.m.s) = V(r.m.s) / Xc = 150 / 1326.11924 = 0.113112 Amps
II. The peak value of the current.
To calculate the peak value of the current, we use the formula:
I(r.m.s) = Io/√2 . Let us substitute
0.113112 = Io/√2, 0.113112 * √2 = Io
Io = 0.113112 * 1.4142 = 0.1600 Amps.
A.C Circuit that contain only Inductor L:
Circuit diagram of inductive circuit
An inductive circuit is an a.c. circuit that contains only inductor. The resistance of the inductor is called inductance. It is measured in Henry. The inductance of an inductor is represented by letter L.
Phase Relationship between Current and Voltage in an Inductive Circuit
In an inductive circuit, the voltage leads the current by 09° or π/2 radian or ¼ circle. Or we say the current lag behind the voltage by 90° , or π/2 radian or ¼ circle. The reason is that current always wait behind to build a magnetic field around the inductor before it flow in the circuit while voltage start to flow in the circuit immediately.
Diagram to show phase relationship between V and I in inductive- circuit
Phase Difference between voltage and current in inductive - circuit
This is the amount in degree by which the voltage leads the current or the amount in degree by which current lag behind voltage. The phase difference is 90° or π/2 radian or ¼ circle.
Now, since current lag behind voltage, the formula of a.c current will become,
I = Io Sin ( wt – π/2 ) while V = Vo Sin wt. ( no change).
Inductive Reactance XL of inductive circuit
This is the resistance that an inductor offer to the flow of a.c current. Its unit is ohm. Its symbol is L.
Formula for calculating inductive reactance XL
XL = wL.
Remember that w = 2πf. Therefore,
XL = 2πfL.
Where L is the inductance of the coil, f is frequency of current or voltage. Π is 22/7.
At this point, we will put ( substitute for ) XL in the formula of ohm’s law. Therefore,
From ohm’s law, V = I*R. Where R is represent inductive reactance XL.
Therefore,
V = I*R = I * 2πfL = I*2πfL. Therefore,
V = I*2πfL
Worked Examples
Find the impedance ( inductive reactance) across an inductor of 0.2H when an a.c voltage of 60Hz is applied across it, if the voltage is given by V = 150 Sin 120πt, calculate the r.m.s and the peak value of the current.
Solution:
Note: impedance is another name for inductive reactance.
From the question,
L = 0.2 henry, and V = 150 Sin 120πt.
If you compare the equation with the general equation for a.c. voltage or current
( V = Vo Sin wt), you will agree that,
Vo = 150 volt,
If you look at the question, you will notice that frequency is not given. Ok! Then you have to calculate the frequency using the method I used in worked example above under general equation of a.c current or voltage.
Therefore, general equation is V = Vo Sin 2πft while in the question,
v = 150 Sin 120πt, then Vo = 150 V, compare the sin parts.
Sin 2πft = Sin 120πt. The sin,π and t will cancel one another.
Then,
2f = 120.
If we make f the subject of formula, then f = 120/2 = 60Hz. Let us bring all the data together:
Vo = 150V, f = 60 Hz and L = 0.2 Henry
i) Impedance Z: XL = 2πfL = 2*3.142*60*0.2
Impedance Z( XL) = 75.408 ohms
ii) We must first calculate Io before we can calculate I(r.m.s) because the value Vo is given in the question. Also, we can not calculate R.m.s value of current now because the value of V(r.m.s) is not given in the question. Also, note that you can’t use Vo and V(r.m.s) in the same formula. You can only use Vo and Io together in the same formula or V(r.m.s) and I(r.m.s).
We will use the formula, Vo = Io * XL,
Vo = 150 V , XL = 75.408 ohms, Io = ?
therefore, if we substitute the values in the formula, then we will get:
150 = Io * 75.408, then, Io = 150/ 75.408 , Io = 1.989 Amps.
iii) Now that we have calculated peak current(Io), we can now calculate r.m.s current by using the formula I(r.m.s) = Io/√2.
Therefore,
Io = 1.989 Amps., then , I(r.m.s) = 1.989/1.4142 = 1.406 Amps.
Therefore, r.m.s current = 1.406 amps.
Series R-C circuit:
A series R-C circuit is an AC circuit that contains a resistor and a capacitor connected in series. The circuit connection is as shown below:
Circuit diagram:
Impedance of R-C Circuit , Z:
It is the opposition that the two components offer to the flow of an AC current. Its unit is ohm.
Vector or Phasor diagram of R-C circuit:
Formula for calculating the impedance of R-C circuit
From the vector/ phasor diagram, we will obtain the formula for calculating the impedance of R-C circuit. Thus:
Vector diagram here:
Using Pythagoras theorem,
Impedance, Z² = R² + Xc², Z = √(R² + Xc²). Recall that Xc = 1/2πfc. Then, Z = √(R² + (1/2πfc)²). Where Xc is the capacitive reactance. R is the resistance. c is the capacitance of the capacitor. F is the frequency of the ac voltage or current.
At this point, we will bring in the formula into ohm's law.
Therefore, V = I * R, where R represent Z. i.e R = Z. Then
V = I *R ; V = I * Z;
V = I*√(R² + (1/2πfc)²)
Formula for calculating phase angel in R- C circuit:
Using shocahtoa concept, tan @ = opposite / adjacent, tan @ = Xc / R. If we make A the subject of the formula, then we will have,
@ = tan~ ¹(Xc/R). Where R is the resistance of the resistor measured in ohms. Xc is the capacitive reactance of the capacitor measured in ohms. A is the phase angle by which current leads the voltage.
Worked Examples:
Series R-L circuit:
Is an ac circuit that contains a resistor and an inductor connected in series. The circuit is as shown below.
Series R-L circuit diagram:
Impedance of R- L series circuit(Z):
This is the opposition/ resistance that a resistor and an inductoroffer to the flow of an ac current or in an ac circuit.
Vector / Phasor diagram of R-L series circuit:
Vector diagram here:
From the vector / phasor above,
Tan @ = XL / R. If we make A he subject, then we will get,
@ = tan~¹ (XL / R). Recall that XL = 2πfL. Therefore,
@ = tan~¹(2πfL/R). R is the resistance measured in ohm, f is the frequency of the ac voltage / current measured in hertz, L is the inductance of the inductor measured in henry. @ is the phase angle between current and voltage.
Formula of impedance (Z) in R-L series circuit:
From the vector / phasor diagram above, we will obtain the formula for calculating the impedance of R-L series circuit thus:
Z² = R² + XL² ; Z = √(R² + XL²). Recall that XL = 2πfL, then we will get,
Z = √( R² + (2πfL)² ).
At this point, we will bring in the formula into ohm's law .
From ohm’s law ,
V = I * R. Where R represent impedance, Z. I.e R = Z. Therefore,
V = I * Z; V = I * √(R²+XL²) ; V = I * √(R² + (2πfL)²).
Note: you can equally use the voltage vector diagram to get the impedance.
Worked Examples:
Series R-L-C circuit:
This circuit contains resistor, inductor and capacitor connected in series as shown in the diagram below.
Impedance of R-L-C series circuit(Z):
This is the resistance that resistor, inductor and a capacitor offer to the flow of ac current in an a. Circuit.
Vector or phasor diagram of R-L-C series circuit:
Phase angle of R-L-C series circuit:
From the vector diagram,
If we use voltage vector diagram, Tan @ = (VL – VC) / Vr
Also, if we use reactance vector diagram, we will get,
Tan @ = ( XL – XC ) / R. If we makeanglew @ the subject, we will get,
@ = Tan~1 ( XL – XC ) / R. Where XL is the inductive reactance of the inductor, XC is the capacitive reactance of the capacitor and R is the resistance of the resistor. @ is the phase angle of lead or lag between V and I.
Recall that XL = 2π*f*L and XC = 1 /2π*f*C. have it in mind that these formulae can be substituted for in the formula above. Then we will get, @ = tan~1 ( 2πfL + 1/2πfC )
Formula for calculating impedance in R-L-C series circuit:
From the vector diagram above, if we use the voltage vector diagram and we will use pythagoras theorem, then, we will get ,
(Vr.m.s)² = Vr² + ( VL - VC)²; Recall that V = I*R, then Vr.m.s = I*Z, VL = I*XL, VC = I*Xc and Vr = I*R. We will now substitute for all of them in the above formula . then we will get,
Vr.m.s² =Vr² + ( VL – VC)² ;
(I*Z)² = ( I*R)² + ( I*XL – I*XC)².if we factorise I², then we will get,
I²Z² = I² ( R² + ( XL – XC )² ). I² at the right hand side cancels I² at the left hand side. Then we will get,
Z² = R² + ( XL – XC )².if we make Z the subject of the formula. Then we will get,
Impedance, Z = √( R² + ( XL – XC )² )
At this point, we will substitute for Z in ohm's law, V = I * R, where R = Z,
Then, V = I * Z ,
V = I * √( R² + ( XL – XC )² ). If we make I the subject of the formula, then we will get,
I = V / ( √( R² + ( XL – XC )² )
Worked Example:
From the circuit below, calculate the capacitive, Xc, inductive reactance XL, impedance, the current and power.
Circuit diagram:
Solution:
R = 100 ohms, L = 3 henry, C = 4 microfarad, V = 100 volt, f = 160/π hertz.
1. Inductive reactance, XL
XL = 2πfL , f = 160/π Hz, L = 3 H
XL = 2×π×160/π×3
XL = 2×160×3 ( the π cancels each other)
XL = 960 ohms.
2. Capacitive reactance, Xc:
Xc = 1 / 2πfC
Xc = 1 / 2×π×160/π×0.000004
Xc = 1 / 2×160×0.000004
Xc = 1 / 0.00128
Xc = 781.25 ohms.
3. Impedance, Z
Z = √( R² + ( XL – xc )² )
R = 100 ohms, Xc = 781.25 ohms, XL = 960 ohms
Z = √( 100² + ( 960 – 781.25)² )
Z = √(100² + 178.75²)
Z = √(10000 + 31951.56)
Z = √ 41951.56
Z = 208.82 ohms
4. Current, I:
V = I × Z, V = 100 volt, Z = 208.82 ohms
Let us substitute for the variables, therefore,
100 = I × 208.82. Make I the subject, then we have
I = 100 / 208.82
I = 0.488 ampere
5. Power:
Power = I² × R = I² × Z
Power = 0.488² × 208.82
Power = 48.82 / 2 = 24.41 watt.
Example2:
From the circuit diagram above, calculate the :
I. Resistance of the resistor
II. Inductance of the inductor
III. Capacitance of the capacitor
IV. Voltage across the capacitor
V. Impedance
Solution:
From the diagram,
V = 240V, f = 50Hz, I =10A, VL = 50V, Vr = 140V, Vc =?,
Note that since the components are connected in series, the same value of current flows through each of them.
I. Vr = Ir × R
Vr = 140v, Ir = 10 A, R =?
140 = 10 × R, make R the subject, Then,
R = 140 / 10 = 14 ohms.
II. Inductive reactance XL:
VL = IL × XL. VL = 50v, IL = 10 A, then,
50 = 10 × XL. Make XL the subject of formula, then
XL = 50 / 10 , XL = 5 ohms
Now , we will calculate the inductance L of the inductor thus:
XL = 2πfL, f = 50Hz, XL = 5 ohms,
XL = 2 × 22/7 × L. Let us substitute for the values
5 = 2 × 22/7 × 50 × L. Let us make L the subject, then we will get
Therefore,
L = 5 × 7 / 2 × 22 × 50 , L = 35 / 2200, then,
L = 0.0159 Henry
III. Capacitive reactance Xc:
Vc = Ic × Xc. Ic = 10 A, Vc = ?, Xc = ?.
We can not use this formula to calculate Xc because the value of Vc is not given in the question.
We must first of all calculate Vc before we can calculate C. Therefore, we will use the voltage formula relation which is ,
Vr.m.s = Vr + VL + Vc ( the reason is ,it is the 240V that is shared across each of the three components)
Vr.m.s = 240V, Vr = 140V, VL = 50V, Vc = ?
Therefore,
240 = 140 + 50 + Vc. Make Vc the subject of the formula,
then,
Vc = 240 – 50 – 140 = 50V
Vc = 50V.
Now , we can calculate the value of Xc using the formula,
Vc = Ic × Xc
50 = 10 × Xc. Make Xc the subject of the formula, then we get,
Xc = 50 / 10 . Xc = 5 ohms.
Now that we have calculated Xc, we will now calculate the capacitance of the capacitor using the formula:
Xc = 1 / 2πfC. F = 50Hz, Xc = 5 ohms. We substitute for the values,
5 = 1 / 2 × 22/7 × 50 × C. We make the subject of the formula, then,
C = 5 × 7 / 2 × 22 × 50.
C = 35 / 2200. Then C = 0.0159 microfarad.
IV. Voltage across the capacitor Vc.
Note ! We have already calculated Vc in the working above because we needed the value of Vc to calculate the value of C.
V. Impedance of the circuit, Z:
Z = √( R² + ( XL – Xc )² ).
R = 140 ohms, XL = 5 ohms, Xc = 5 ohms, then
Z = √( 140² + ( 5 – 5)² )
Z = √(140² + 0² )
Z = √ 140² = √ 19600
Z = 140 ohms.
Parallel R-C Circuit:
In a parallel R-C Circuit, resistor is connected in parallel with a capacitor.
The current that flow through the resistor and the capacitor are different but the voltage across the resistor and the capacitor are the same.
Circuit diagram:
Impedance of R-C parallel Circuit:
It is the resistance/ opposition that R-C offer to the flow of ac current. It is measured in ohm.
Formula for Calculating impedance in R-C Parallel Circuit:
R is the resistance of the resistor, Xc is the capacitive reactance of the capacitor, and Z is the total resistance of the circuit called impedance.
1 / Z = 1 / R + 1 / Xc . Recall that Xc = 1 / 2πfC.
Let us make the subject of the by fraction concept, then we will get,
L.C.M = R × Xc, then, 1 / Z = (Xc + R ) / ( R × Xc ), then ,
Z = ( R × Xc ) / ( Xc + R ),
therefore,
Impedance, Z = ( R × Xc ) / ( Xc + R ).
From ohm's law, V = I × R. Where R is the impedance of the circuit.
Therefore,
V = I × R, R = Z, then, V = I × Z , and Z = ( R × Xc ) / ( R + Xc ). then,
V = I × ( R × Xc ) / ( R + Xc ). , or , I = V / ( ( R × Xc ) / ( R + Xc ) )
Worked Example 1:
A 2 ohms resistor is connected in parallel with a 5 microfarad capacitor across 240 voltage source of frequency 50 hertz. Draw the circuit diagram and calculate the circuit impedance, the circuit current and the in each components.
Solution:
From the question, R = 2 ohms, C = 5 microfarad, V = 240 volt, f = 50 Hz
I. Circuit diagram.
II. Impedance Z:
Z = ( R × Xc ) / ( Xc + R ). We must first calculate Xc using the formula,
Xc = 1 / 2πfc. , then , Xc = 1 / 2 × 22/7 × 50 × 0.000005
Xc = 1 / 0.001571. Xc = 636.54 ohms.
Now that we have calculated Xc, we can now calculate the impedance Z using the formula we stated earlier.
Therefore,
Z = ( Xc × R ) / ( R + Xc )
Z = ( 636.54 × 2 ) / ( 2 + 636.54 ), Z = 1.994 ohms.
III. Circuit current:
V = I × Z. Then , 240 = I × 1.994. Make I the subject of the formula. I = 240 / 1.994. I = 120.36 amperes.
IV. Current that flow in each components:
Note that since the components are connected in parallel, the voltage across them are the same. Therefore,
Current that flow through resistor R is:
Vr = Ir × R. We substitute for the variables. Then,
240 = Ir × 2. Ir = 240 / 2. Ir = 120 amperes.
Current that flow through the capacitor C is :
Vc = Ic × Xc. We substitute for the variables. Then,
240 = Ic × 636.54. We make Ic the subject of the formula. Then,
Ic = 240 / 636.54. Ic = 0.38 ampere.
Worked Example 2:
From the circuit diagram, the current that flow in the circuit, the Capacitor and the capacitive rectance capacitor and impedanc
A.C. means alternating current. It is a current that flow in the positive and negative direction periodically.
Waveform of Alternating current:
The waveform of alternating current is sinusoidal waveform.
Diagram of a.c waveform:
Alternating Current Circuit:
Equation of A.C current:
I = Io Sin@
In the production of A.C voltage or current, the coil is in a magnetic field and sweep through angle @.
Therefore,
I = Io Sin @
W = @/t. If we make @ the subject, we will have, @ = w*t. Understand?
If we substitute for @ in the equation, I = Io Sin @
Then the equation will become,
I = Io sin wt.
The next thing is:
When the coil sweep through one revolution or one circle, angle @ = 360°. We will change 360° to radian
Therefore,
@ = 360° = 2π radian. Therefore,
W = @/t = 360°/t = 2π/t. Therefore, w = 2π/t.
At this point, remember the formula that connect t and f,
t = 1/f. You will now substitute for t in the formula,
w = 2π/t thus:
W = 2π/t = 2π ÷ 1/f = 2π x f/1 = 2πf,
therefore
W = 2πf, where f is frequency. It is measured in hertz.
At this point, we will substitute for w in I = Io Sin wt.
Therefore,
I = Io sinwt = Io sin 2πf * t. That is, I = Io sin 2πft
Therefore,
the general equation for a.c current is
I = Io sin 2πft, where
f is the frequency. it is measured in hertz
t is the time. It is measured in second
Io is the peak current. It is measured in ampere
I is the instantaneous current, measured in ampere.
W is angular velocity measured in radian per second.
Equation of A.C current:
Every thing that we did above is applicable to a.c voltage. Therefore the general equation for a.c voltage is the same as that of a.c current. The only difference is that letter V is used instead of Letter I.
Therefore,
V = Vo sin @ , V = sin wt, V = Vo sin 2πft. That is all.
Application of general equation of a.c. current/voltage to solve problems
An a.c. voltage is represented by the equation, V = 4Sin900πt.
Calculate the: I) the peak voltage. II) the frequency. III) the angular velocity.
Solution:
You have to compare each parts of the give equation with the general equation thus:
General equation is I = Io Sin 2πft
The given equation is V = 4 Sin 900πt
I) Peak voltage is Vo = 4volt
II) To find frequency, you compare by saying,
Sin 2πft = Sin 900πt
Sin will cancel Sin, π will cancel π and t will also cancel t.
Therefore we have,
2f = 900, make f the subject of formula, then
f = 900/2 = 450 hertz
III) To find angular velocity, w, you will use
W = 2πf, therefore,
W = 2 x π x f = 2 x π x 450 , w = 900π radian per second.
Or, w = 2 x 22/7 x 450 = 2828.57 rad/sec
Peak Current, Io / peak Voltage, Vo
It is the maximum value of the a.c current or a.c Voltage.
Root Mean Square (r.m.s) of current/ voltage
It is the steady part or d.c part of the alternating current that will produce the same quantity of heat if it flow through the same resistance in the same amount of time.
Relationship between peak current (Io) and I(r.m.s)
Let us derive the formula that connect Io and I(r.m.s).
Firstly,
The rate at which heat is produced is directly proportional to the square of the current.
The I(r.m.s)² is the square of the average of Io².
Therefore,
I(r.m.s)² = average of Io²
I(r.m.s)² = Io²/2. We will square both sides . then we will have,
I(r.m.s)² = Io²/2, if we make I(r.m.s) the subject, we will get,
I(r.m.s) = square root of Io²/2 = √(Io²/2) = √Io²/√2 = Io/√2
Therefore,
I(r.m.s) = Io/√2. Or I(r.m.s) = 0.7070*Io, Io = √2*I(r.m.s)
Io is the peak current while I(r.m.s) is the root mean square of the current.
Relationship between peak voltage(Vo) and V(r.m.s):
Every thing we did above is also applicable to voltage. Therefore, the formula is the for voltage. The only difference is that letter V is used instead of letter I.
Therefore,
V(r.m.s) = Vo/√2. Or V(r.m.s) = 0.7070*Vo Or, Vo = √2*V(r.m.s)
Where Go is the peak voltage, V(r.m.s) is the root mean square of the voltage.
Note:
Why is it that when a voltmeter or ammeter is connected across a.c source, the voltmeter or ammeter give a steady reading?
Answer:
When a voltmeter or ammeter is connected across a.c source, the voltmeter or ammeter give a steady reading because the instrument is only measuring the steady or d.c Part of the a.c current or voltage and not the a.c part.
It is the d.c part of the a.c current/voltage that is powering the electrical devices, not the a.c part. If not so, the devices would power on and off repeatedly for as long as it is powered by ac current/voltage.
Alternating Current Circuit
These are circuit which are powered by alternating current or alternating voltage.
Types of a.c circuits
Resistive a.c circuit:
Circuit diagram:
A resistive circuit is an a.c. circuit that contains only resistor.
Resistance of a R- circuit:
Since the circuit contains only resistor, the circuit therefore obey ohm’s law. Therefore,
V = I*R ( ohm’s law)
Remember that for a.c, voltage and current,
V = Vo Sin wt, and I = Io Sin wt . we will put (substitute) this into the ohm’s law that we stated above.
Therefore, the formula will become,
V = I*R
Vo Sin wt = (Io Sin wt)* R.
We will make R the subject, then we will get,
R = (Vo Sin wt)/(Io Sin wt). ( Sin wt will cancel sin wt, then we will get)
R = Vo/Io (for peak voltage and current)
R = V(r.m.s)/I(r.m.s) ( for r.m.s voltage and current)
Phase Relationship between Current and Voltage in Resistive Circuit
In a resistive circuit, the current is in phase with the voltage. This means that current and voltage start their journey from the same point at the same time, reach their maximum and minimum points at the same time.
Diagram that illustrate
Vector or Phasor diagram of R- Circuit
Worked Examples:
Capacitive a.c Circuit
Diagram of C- circuit:
A capacitive circuit is an a.c circuit that contains only capacitor as the active components.
Phase Relationship between Current and Voltage in Capacitive Circuit
In a capacitive circuit, current leads the voltage by 90°or π/2 radian or ¼ circle. Or voltage lag behind current by 90° or π/2 radian or ¼ circle. The reason is that voltage always wait behind to charge the capacitor before it will start to flow in the circuit while current start to flow immediately in the circuit.
Current and voltage are out of phase. This means that current and voltage start their journey from different points at different time and reach their maximum and minimum points at different time.
Diagram to show phase relationship:
Vector or phasor diagram of C- Circuit:
Phase Difference between voltage and current in c- circuit
This is the amount in degree by which the current leads the voltage in a capacitive circuit. The phase difference is 90° or π/2 or ¼ circle.
Therefore,
V = Vo Sin wt , and I = Io Sin ( wt + π/2 ) .
Note that π/2 is added because current leads the voltage in capacitive circuit.
Capacitive Reactance ( Xc) of a capacitor in a capacitive circuit
The resistance of a capacitor in a capacitive circuit is called capacitive reactance. The symbol is Xc. Its unit is ohm.
Definition of capacitive reactance Xc:
Capacitive reactance is the resistance that a capacitor offers to the flow of an a.c. current.
Formula for calculating capacitive reactance Xc
Xc = 1/wC.
Remember that w = 2πf ( I had derived this formula above)
Therefore,
Xc = 1/2πfC .
where f is the frequency of current or voltage, measured in hertz. C is the capacitance of the capacitor, measured in microfarad.
At this point, we will put ( substitute for ) Xc in the formula of ohm’s law. Therefore,
From ohm’s law, V = I*R. Where R is represent capacitive reactance Xc.
Therefore,
V = I*R = I * 1/2πfC = I/2πfC
Worked Examples:
A 2microfarad capacitor is connected across 150V(r.m.s.), 60 Hz a.c source. Calculate: I) the r.m.s value if the current ii) the peak value.
Solution:
From the question, V(r.m.s) = 150Volt, f = 60Hz,
C = 2microfarad = 2x10~6Farad
Note that:
we are to calculate the r.m.s value of the current. i.e I(r.m.s)
The only formula that we can use to calculate the r.m.s value of current is Xc = V(r.m.s)/I(r.m.s)
Xc is not given in the question. You will notice that C and f are given in the question. So we must first of all calculate Xc and then use the calculated value of Xc to calculate I(r.m.s).
Therefore, we use the formula:
I. Xc = 1/2πfC, to calculate Xc
Xc = 1/ 2*3.142*60*0.000002 = 1/ 0.00075408
Xc = 1326.11924 ohms
At this point, we use the calculated value of Xc to calculate the current. Therefore,
Xc = V(r.m.s) / I(r.m.s) , make I(r.m.s) the subject of the formula,
I(r.m.s) = V(r.m.s) / Xc = 150 / 1326.11924 = 0.113112 Amps
II. The peak value of the current.
To calculate the peak value of the current, we use the formula:
I(r.m.s) = Io/√2 . Let us substitute
0.113112 = Io/√2, 0.113112 * √2 = Io
Io = 0.113112 * 1.4142 = 0.1600 Amps.
A.C Circuit that contain only Inductor L:
Circuit diagram of inductive circuit
An inductive circuit is an a.c. circuit that contains only inductor. The resistance of the inductor is called inductance. It is measured in Henry. The inductance of an inductor is represented by letter L.
Phase Relationship between Current and Voltage in an Inductive Circuit
In an inductive circuit, the voltage leads the current by 09° or π/2 radian or ¼ circle. Or we say the current lag behind the voltage by 90° , or π/2 radian or ¼ circle. The reason is that current always wait behind to build a magnetic field around the inductor before it flow in the circuit while voltage start to flow in the circuit immediately.
Diagram to show phase relationship between V and I in inductive- circuit
Phase Difference between voltage and current in inductive - circuit
This is the amount in degree by which the voltage leads the current or the amount in degree by which current lag behind voltage. The phase difference is 90° or π/2 radian or ¼ circle.
Now, since current lag behind voltage, the formula of a.c current will become,
I = Io Sin ( wt – π/2 ) while V = Vo Sin wt. ( no change).
Inductive Reactance XL of inductive circuit
This is the resistance that an inductor offer to the flow of a.c current. Its unit is ohm. Its symbol is L.
Formula for calculating inductive reactance XL
XL = wL.
Remember that w = 2πf. Therefore,
XL = 2πfL.
Where L is the inductance of the coil, f is frequency of current or voltage. Π is 22/7.
At this point, we will put ( substitute for ) XL in the formula of ohm’s law. Therefore,
From ohm’s law, V = I*R. Where R is represent inductive reactance XL.
Therefore,
V = I*R = I * 2πfL = I*2πfL. Therefore,
V = I*2πfL
Worked Examples
Find the impedance ( inductive reactance) across an inductor of 0.2H when an a.c voltage of 60Hz is applied across it, if the voltage is given by V = 150 Sin 120πt, calculate the r.m.s and the peak value of the current.
Solution:
Note: impedance is another name for inductive reactance.
From the question,
L = 0.2 henry, and V = 150 Sin 120πt.
If you compare the equation with the general equation for a.c. voltage or current
( V = Vo Sin wt), you will agree that,
Vo = 150 volt,
If you look at the question, you will notice that frequency is not given. Ok! Then you have to calculate the frequency using the method I used in worked example above under general equation of a.c current or voltage.
Therefore, general equation is V = Vo Sin 2πft while in the question,
v = 150 Sin 120πt, then Vo = 150 V, compare the sin parts.
Sin 2πft = Sin 120πt. The sin,π and t will cancel one another.
Then,
2f = 120.
If we make f the subject of formula, then f = 120/2 = 60Hz. Let us bring all the data together:
Vo = 150V, f = 60 Hz and L = 0.2 Henry
i) Impedance Z: XL = 2πfL = 2*3.142*60*0.2
Impedance Z( XL) = 75.408 ohms
ii) We must first calculate Io before we can calculate I(r.m.s) because the value Vo is given in the question. Also, we can not calculate R.m.s value of current now because the value of V(r.m.s) is not given in the question. Also, note that you can’t use Vo and V(r.m.s) in the same formula. You can only use Vo and Io together in the same formula or V(r.m.s) and I(r.m.s).
We will use the formula, Vo = Io * XL,
Vo = 150 V , XL = 75.408 ohms, Io = ?
therefore, if we substitute the values in the formula, then we will get:
150 = Io * 75.408, then, Io = 150/ 75.408 , Io = 1.989 Amps.
iii) Now that we have calculated peak current(Io), we can now calculate r.m.s current by using the formula I(r.m.s) = Io/√2.
Therefore,
Io = 1.989 Amps., then , I(r.m.s) = 1.989/1.4142 = 1.406 Amps.
Therefore, r.m.s current = 1.406 amps.
Series R-C circuit:
A series R-C circuit is an AC circuit that contains a resistor and a capacitor connected in series. The circuit connection is as shown below:
Circuit diagram:
Impedance of R-C Circuit , Z:
It is the opposition that the two components offer to the flow of an AC current. Its unit is ohm.
Vector or Phasor diagram of R-C circuit:
Formula for calculating the impedance of R-C circuit
From the vector/ phasor diagram, we will obtain the formula for calculating the impedance of R-C circuit. Thus:
Vector diagram here:
Using Pythagoras theorem,
Impedance, Z² = R² + Xc², Z = √(R² + Xc²). Recall that Xc = 1/2πfc. Then, Z = √(R² + (1/2πfc)²). Where Xc is the capacitive reactance. R is the resistance. c is the capacitance of the capacitor. F is the frequency of the ac voltage or current.
At this point, we will bring in the formula into ohm's law.
Therefore, V = I * R, where R represent Z. i.e R = Z. Then
V = I *R ; V = I * Z;
V = I*√(R² + (1/2πfc)²)
Formula for calculating phase angel in R- C circuit:
Using shocahtoa concept, tan @ = opposite / adjacent, tan @ = Xc / R. If we make A the subject of the formula, then we will have,
@ = tan~ ¹(Xc/R). Where R is the resistance of the resistor measured in ohms. Xc is the capacitive reactance of the capacitor measured in ohms. A is the phase angle by which current leads the voltage.
Worked Examples:
Series R-L circuit:
Is an ac circuit that contains a resistor and an inductor connected in series. The circuit is as shown below.
Series R-L circuit diagram:
Impedance of R- L series circuit(Z):
This is the opposition/ resistance that a resistor and an inductoroffer to the flow of an ac current or in an ac circuit.
Vector / Phasor diagram of R-L series circuit:
Vector diagram here:
From the vector / phasor above,
Tan @ = XL / R. If we make A he subject, then we will get,
@ = tan~¹ (XL / R). Recall that XL = 2πfL. Therefore,
@ = tan~¹(2πfL/R). R is the resistance measured in ohm, f is the frequency of the ac voltage / current measured in hertz, L is the inductance of the inductor measured in henry. @ is the phase angle between current and voltage.
Formula of impedance (Z) in R-L series circuit:
From the vector / phasor diagram above, we will obtain the formula for calculating the impedance of R-L series circuit thus:
Z² = R² + XL² ; Z = √(R² + XL²). Recall that XL = 2πfL, then we will get,
Z = √( R² + (2πfL)² ).
At this point, we will bring in the formula into ohm's law .
From ohm’s law ,
V = I * R. Where R represent impedance, Z. I.e R = Z. Therefore,
V = I * Z; V = I * √(R²+XL²) ; V = I * √(R² + (2πfL)²).
Note: you can equally use the voltage vector diagram to get the impedance.
Worked Examples:
Series R-L-C circuit:
This circuit contains resistor, inductor and capacitor connected in series as shown in the diagram below.
Impedance of R-L-C series circuit(Z):
This is the resistance that resistor, inductor and a capacitor offer to the flow of ac current in an a. Circuit.
Vector or phasor diagram of R-L-C series circuit:
Phase angle of R-L-C series circuit:
From the vector diagram,
If we use voltage vector diagram, Tan @ = (VL – VC) / Vr
Also, if we use reactance vector diagram, we will get,
Tan @ = ( XL – XC ) / R. If we makeanglew @ the subject, we will get,
@ = Tan~1 ( XL – XC ) / R. Where XL is the inductive reactance of the inductor, XC is the capacitive reactance of the capacitor and R is the resistance of the resistor. @ is the phase angle of lead or lag between V and I.
Recall that XL = 2π*f*L and XC = 1 /2π*f*C. have it in mind that these formulae can be substituted for in the formula above. Then we will get, @ = tan~1 ( 2πfL + 1/2πfC )
Formula for calculating impedance in R-L-C series circuit:
From the vector diagram above, if we use the voltage vector diagram and we will use pythagoras theorem, then, we will get ,
(Vr.m.s)² = Vr² + ( VL - VC)²; Recall that V = I*R, then Vr.m.s = I*Z, VL = I*XL, VC = I*Xc and Vr = I*R. We will now substitute for all of them in the above formula . then we will get,
Vr.m.s² =Vr² + ( VL – VC)² ;
(I*Z)² = ( I*R)² + ( I*XL – I*XC)².if we factorise I², then we will get,
I²Z² = I² ( R² + ( XL – XC )² ). I² at the right hand side cancels I² at the left hand side. Then we will get,
Z² = R² + ( XL – XC )².if we make Z the subject of the formula. Then we will get,
Impedance, Z = √( R² + ( XL – XC )² )
At this point, we will substitute for Z in ohm's law, V = I * R, where R = Z,
Then, V = I * Z ,
V = I * √( R² + ( XL – XC )² ). If we make I the subject of the formula, then we will get,
I = V / ( √( R² + ( XL – XC )² )
Worked Example:
From the circuit below, calculate the capacitive, Xc, inductive reactance XL, impedance, the current and power.
Circuit diagram:
Solution:
R = 100 ohms, L = 3 henry, C = 4 microfarad, V = 100 volt, f = 160/π hertz.
1. Inductive reactance, XL
XL = 2πfL , f = 160/π Hz, L = 3 H
XL = 2×π×160/π×3
XL = 2×160×3 ( the π cancels each other)
XL = 960 ohms.
2. Capacitive reactance, Xc:
Xc = 1 / 2πfC
Xc = 1 / 2×π×160/π×0.000004
Xc = 1 / 2×160×0.000004
Xc = 1 / 0.00128
Xc = 781.25 ohms.
3. Impedance, Z
Z = √( R² + ( XL – xc )² )
R = 100 ohms, Xc = 781.25 ohms, XL = 960 ohms
Z = √( 100² + ( 960 – 781.25)² )
Z = √(100² + 178.75²)
Z = √(10000 + 31951.56)
Z = √ 41951.56
Z = 208.82 ohms
4. Current, I:
V = I × Z, V = 100 volt, Z = 208.82 ohms
Let us substitute for the variables, therefore,
100 = I × 208.82. Make I the subject, then we have
I = 100 / 208.82
I = 0.488 ampere
5. Power:
Power = I² × R = I² × Z
Power = 0.488² × 208.82
Power = 48.82 / 2 = 24.41 watt.
Example2:
From the circuit diagram above, calculate the :
I. Resistance of the resistor
II. Inductance of the inductor
III. Capacitance of the capacitor
IV. Voltage across the capacitor
V. Impedance
Solution:
From the diagram,
V = 240V, f = 50Hz, I =10A, VL = 50V, Vr = 140V, Vc =?,
Note that since the components are connected in series, the same value of current flows through each of them.
I. Vr = Ir × R
Vr = 140v, Ir = 10 A, R =?
140 = 10 × R, make R the subject, Then,
R = 140 / 10 = 14 ohms.
II. Inductive reactance XL:
VL = IL × XL. VL = 50v, IL = 10 A, then,
50 = 10 × XL. Make XL the subject of formula, then
XL = 50 / 10 , XL = 5 ohms
Now , we will calculate the inductance L of the inductor thus:
XL = 2πfL, f = 50Hz, XL = 5 ohms,
XL = 2 × 22/7 × L. Let us substitute for the values
5 = 2 × 22/7 × 50 × L. Let us make L the subject, then we will get
Therefore,
L = 5 × 7 / 2 × 22 × 50 , L = 35 / 2200, then,
L = 0.0159 Henry
III. Capacitive reactance Xc:
Vc = Ic × Xc. Ic = 10 A, Vc = ?, Xc = ?.
We can not use this formula to calculate Xc because the value of Vc is not given in the question.
We must first of all calculate Vc before we can calculate C. Therefore, we will use the voltage formula relation which is ,
Vr.m.s = Vr + VL + Vc ( the reason is ,it is the 240V that is shared across each of the three components)
Vr.m.s = 240V, Vr = 140V, VL = 50V, Vc = ?
Therefore,
240 = 140 + 50 + Vc. Make Vc the subject of the formula,
then,
Vc = 240 – 50 – 140 = 50V
Vc = 50V.
Now , we can calculate the value of Xc using the formula,
Vc = Ic × Xc
50 = 10 × Xc. Make Xc the subject of the formula, then we get,
Xc = 50 / 10 . Xc = 5 ohms.
Now that we have calculated Xc, we will now calculate the capacitance of the capacitor using the formula:
Xc = 1 / 2πfC. F = 50Hz, Xc = 5 ohms. We substitute for the values,
5 = 1 / 2 × 22/7 × 50 × C. We make the subject of the formula, then,
C = 5 × 7 / 2 × 22 × 50.
C = 35 / 2200. Then C = 0.0159 microfarad.
IV. Voltage across the capacitor Vc.
Note ! We have already calculated Vc in the working above because we needed the value of Vc to calculate the value of C.
V. Impedance of the circuit, Z:
Z = √( R² + ( XL – Xc )² ).
R = 140 ohms, XL = 5 ohms, Xc = 5 ohms, then
Z = √( 140² + ( 5 – 5)² )
Z = √(140² + 0² )
Z = √ 140² = √ 19600
Z = 140 ohms.
Parallel R-C Circuit:
In a parallel R-C Circuit, resistor is connected in parallel with a capacitor.
The current that flow through the resistor and the capacitor are different but the voltage across the resistor and the capacitor are the same.
Circuit diagram:
Impedance of R-C parallel Circuit:
It is the resistance/ opposition that R-C offer to the flow of ac current. It is measured in ohm.
Formula for Calculating impedance in R-C Parallel Circuit:
R is the resistance of the resistor, Xc is the capacitive reactance of the capacitor, and Z is the total resistance of the circuit called impedance.
1 / Z = 1 / R + 1 / Xc . Recall that Xc = 1 / 2πfC.
Let us make the subject of the by fraction concept, then we will get,
L.C.M = R × Xc, then, 1 / Z = (Xc + R ) / ( R × Xc ), then ,
Z = ( R × Xc ) / ( Xc + R ),
therefore,
Impedance, Z = ( R × Xc ) / ( Xc + R ).
From ohm's law, V = I × R. Where R is the impedance of the circuit.
Therefore,
V = I × R, R = Z, then, V = I × Z , and Z = ( R × Xc ) / ( R + Xc ). then,
V = I × ( R × Xc ) / ( R + Xc ). , or , I = V / ( ( R × Xc ) / ( R + Xc ) )
Worked Example 1:
A 2 ohms resistor is connected in parallel with a 5 microfarad capacitor across 240 voltage source of frequency 50 hertz. Draw the circuit diagram and calculate the circuit impedance, the circuit current and the in each components.
Solution:
From the question, R = 2 ohms, C = 5 microfarad, V = 240 volt, f = 50 Hz
I. Circuit diagram.
II. Impedance Z:
Z = ( R × Xc ) / ( Xc + R ). We must first calculate Xc using the formula,
Xc = 1 / 2πfc. , then , Xc = 1 / 2 × 22/7 × 50 × 0.000005
Xc = 1 / 0.001571. Xc = 636.54 ohms.
Now that we have calculated Xc, we can now calculate the impedance Z using the formula we stated earlier.
Therefore,
Z = ( Xc × R ) / ( R + Xc )
Z = ( 636.54 × 2 ) / ( 2 + 636.54 ), Z = 1.994 ohms.
III. Circuit current:
V = I × Z. Then , 240 = I × 1.994. Make I the subject of the formula. I = 240 / 1.994. I = 120.36 amperes.
IV. Current that flow in each components:
Note that since the components are connected in parallel, the voltage across them are the same. Therefore,
Current that flow through resistor R is:
Vr = Ir × R. We substitute for the variables. Then,
240 = Ir × 2. Ir = 240 / 2. Ir = 120 amperes.
Current that flow through the capacitor C is :
Vc = Ic × Xc. We substitute for the variables. Then,
240 = Ic × 636.54. We make Ic the subject of the formula. Then,
Ic = 240 / 636.54. Ic = 0.38 ampere.
Worked Example 2:
From the circuit diagram, the current that flow in the circuit, the Capacitor and the capacitive rectance capacitor and impedanc
Very nice resource,what about RLC in parallel
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