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ALTERNATING CURRENT
Important Points:

1. The alternating current (AC) is generally expressed as

I = I0 sin ( ωt + φ ) Where i0 = peak value of alternating current.

2. emf of an alternating current source is generally given by E = E0 sin ( ωt + φ )

Where e0 = BAN ω is known as peak value of alternating emf

3. Average or Mean Value:

a) It is the steady current (DC) which when passes through a circuit for half time period of AC
sends the same charge as done by the AC in the same time.

b) The average value of AC for the complete cycle is zero.

c) The average value of AC for the half cycle of AC is given by.

2I0
I avg = 0.636 I 0
π

2E0
Similarly, Eavg is also for half cycle of AC
π

4. RMS Value OR Virtual Value OR Effective Value:

The RMS value of AC is the steady current (DC) which when flowing through a given
resistance for a given time, produces the same amount of heat as produced by the AC when
flowing through the same resistance for the same time.

I0 E0
Iv = = 0.707 I 0 And Ev = = 0.707 E0
2 2

RMS value of AC for half cycle is also same as above

5. AC through A Pure Resistor

E = E0 sin ωt And I = I 0 sin ωt Where E0 and I0 are the peak values of

Voltage and current of AC respectively

Since the emf and current raise or fall simultaneously they are in phase with each other.

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6. AC through A Pure Inductor:

E0  π
E = E0 sin ωt And I= sin  ω t − 
Lω  2

The term Lω has the units of resistance and it is called as resistance of the inductor or
inductive reactance (XL)

E0  π  π
I= sin  ωt −  ⇒ I = I 0 sin  ωt − 
XL  2  2

Where I0 is the peak value of current. Hence the current is lagging behind emf or emf is
π
leading current by a phase difference of .
2

7. AC through A Pure Capacitor:

E0
E = E0 sin ωt And I= cos ωt
1/ cω

1
The term is called capacitive reactance where it has the dimensions of resistance. It is also
cω
called resistance of capacitor.
E0  π  π
I= sin  ωt +  ⇒ I = I 0 sin  ωt +  Where I0 is peak value of AC
Xc  2  2

π
Hence current leads emf or emf is lagging behind current by
2

8. AC through LR Circuit:

Let VL and VR be the instantaneous voltages across the inductor and resistor respectively.

VL = IX L = ILω And VR = IR

Where XL is the inductive reactance when AC flow through a pure resistor,

∴ E = I R 2 + ( Lω ) 2

Lω
Z = R 2 + ( Lω ) 2 And tan φ =
R

This is the phase angle by which the emf leads the current in L – R circuit. z is called
impedance of LR circuit.

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9. AC through C – R Circuit:

I
Vc = I × X c = And VR = IR Where XC = capacitive reactance
Cω

2
 I 
E = V + V ⇒ E = ( IR) + 
R
2
C
2

2

 Cω 

 1  VC 1
E = I  R2 +  And Tanφ = =
 (Cω ) 2 VR Cω R
 

This is the phase angle by which the emf lags behind the current in C – R circuit
2
 1 
Here Z = R +   is called the impedance of C – R circuit.
2

 Cω 

10. AC through L – C – R Circuit:

1
a) VL = LX L = ILω;VC = IX C = and VR = IR
Cω

2
 1 
b) E = I R 2 +  Lω − 
 Cω 

c) Z = R 2 +  Lω −
1 
 is called the impedance of L – C – R circuit.
 Cω 

 1 
 Lω − 
V −V Cω 
d) Tanφ = L C =  This is the phase angle by which the emf leads the current.
VR R
1
e) If Lω = , then Tanφ = 0 or φ = 0 and hence emf and current are in phase. The circuit in
Cω
this condition behaves like a pure resistor circuit. This condition is called resonance condition.

1 1
f) At resonance, Lω = or ω 2 =
Cω LC

1 1
ω = 2π n = ⇒n=
LC 2π LC

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11. Advantages of AC Over DC:

a) The generation of AC is more economic than DC

b) AC voltages can be easily stepped up or stepped down using transformers.

c) AC can be transmitted to longer distances with less loss of energy.

d) AC can be easily converted into DC by using rectifiers.

12. Disadvantages:

a) AC is more fatal and dangerous than DC.

b) AC always flows on the outer layer of the conductor (skin effect) and hence AC requires
stranded wires.

c) AC cannot be used in electrolysis like electroplating etc.

13. Transformer:
This works on the principle of mutual inductance between two circuits linked by common
magnetic flux.

Step up transformers converts low voltage high current into high voltage low current.

Step down transformer: Converts high voltage low current into low voltage high current.

E.g.: In bed lamps, we will use this type of transformer.

14. Transformer works only with ac (Alternating Current)
ε s Ns ip
15. Transformation ratio = =
ε p Np is

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Very Short Answer Questions

1. A transformer converts 200 V ac into 2000 V ac. Calculate the number of turns in the
secondary if the primary has 10 turns?
A. VP = 200V VS = 2000V
nP = 10 nS = ?
VP nP
=
VS nS
VS 2000
nS = .nP = X 10 = 100
VP 200

2. What type of transformer is used in a 6V bed lamp?

A. Step down transformer.

3. What is the phenomenon involved in the working of a transformer?

A. A transformer works on the principle of mutual induction.

4. What is transformer ratio?

vS Number of turns in secondary ( N S )

A. =
vP Number of turns in primary ( N P )

5. Write the expression for the reactance of i) an inductor and ii) a capacitor?
A. Inductive reactance X L = ω L
1
Capacitive reactance X C =
ωC

6. What is the phase difference between AC emf and current in the following Pure resistor,
pure inductor and pure capacitor?
A. Phase difference between ac emf and current
i) In pure resistor: zero
ii) In pure Inductor: Voltage leads current by 900
iii) In pure capacitor: Current leads voltage by 900

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7. Define power factor. On which factors does power factor depend?

A. The average power dissipated depends not only on the voltage and current but also on the
cosine of the phase angle φ between them. The quantity cos φ is called power factor.
p = VI cos φ
Power factor depends on nature of elements (Resistor, Inductor, Capacitor) in the circuit.

8. What is meant by wattless component of current?

A. If the voltage and current differ in phase by π / 2 , then Power factor, cos φ = cos 900 = 0 .

In this case, the current has no power. Such a current is, therefore, called wattless current.
Since this current does not perform any work, this current may also be called idle current. Such
a current flows only in purely inductive or in purely capacitive circuits.

9. When does a LCR circuit have minimum impedance?

1
A. When X L = X C or Lω = . Then tan φ = 0 or φ = 0° tan φ = 0 or φ = 0° . Thus, there is no phase
Cω
difference between current and potential difference. Therefore, the given LCR circuit is
equivalent to a pure resistive circuit. The impedance of such LCR circuit is given by Z = R.
which is minimum.

10. What is the phase difference between voltage and current when the power factor in LCR
series circuit is unity?
A. The phase difference between voltage and current is zero.
Power factor = cos φ = 1
φ =0
φ is phase difference between Voltage and current

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Short Answer Questions

1. Obtain an expression for the current through an inductor when an AC emf is applied.

A. A.C. through a Pure Inductance:

Consider a pure inductor of inductance L (no resistance) connected to a source of emf ε . The
instantaneous emf is given by v = vm sin ωt

di
Let I be the current through the circuit and be the rate of change of current in the circuit at
dt
any instant. The net emf in the circuit is given by

di
v−L =0
dt

Since there is no resistance in the circuit, there is no P.D. in the circuit.

di v vm
= = sin ωt
dt L L

Integrating the above equation,

di vm
∫ dt dt = L ∫ sin ωt dt
vm
∴i = ( − cos ωt ) + Constant
Lω

Since the current is oscillatory, time independent constant does not exist.

 π
∴ i = im sin  ω t − 
 2

The term Lω has the units of resistance and it is called as inductive reactance (XL)

vm  π  π
i= sin  ωt −  ⇒ i = im sin  ωt −  Here is the peak value of current. Hence the current is
XL  2  2
π
lagging behind emf or emf is leading current by a phase difference of
2

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2. Obtain an expression for the current in a capacitor when an Ac emf is applied?

A. A.C. through a pure capacitor:

Consider a capacitor of capacity C connected to a source of alternating emf ε . The

instantaneous emf is given by.

v = vm sin ωt

The net emf in the circuit is given by,

q q
v− = 0 (Or) v =
C C

q
Or = vm sin ωt
C

dq d ( vmC sin ωt )
∴i = = = Cω vm cos ωt
dt dt

vm v  π  π
(or) i = cos ωt Or i = m sin  ωt +  ⇒ i = im sin  ωt + 
1 XL  2  2
cω

1
The term is called capacitive reactance and it has the dimensions of resistance. Here im is
Cω
π
peak value of AC. Hence current leads emf or emf is lagging behind current by .
2

3. State the principle on which a transformer works. Describe the working of a transformer
with necessary theory?

A. Transformer:

A transformer converts high voltage low currents into low voltage high currents and vice-versa.
Transformer works only for AC.

Principle:

A transformer works on the principle of mutual inductance between two coils linked by a
common magnetic flux.

Construction:

A transformer consists of two mutually coupled insulated coils of wire wound on a continuous
iron core. One of the coils is called primary coil and the other is called secondary coil. The
primary is connected to an AC emf. And secondary to a load. Due to this alternating flux
linkage, an emf. Is induced in the secondary due to mutual induction.
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P N1 N2 S P N1 N2 S

(a) (b) (c)

Soft iron core Laminated core

Working:

Let N P and N S be the number of turns in the primary and secondary coils respectively. The
induced emf’s produced in primary and secondary coils are given by

 dφ   dφ 
ε p = − N p   and ε s = − N s   ,
 dt   dt 

εs N s v N
Hence = Or s = s
εp Np vp N p

Where vP and vS are the primary and secondary voltages.

ip vs N s
If the efficiency of the transformer is 100 % , then vs is = v p i p or = =
is vp N p

Ns
is called transformer ratio. If N s > N p , then it is called a step-up transformer. If N s < N p ,
Np
then it is called a step-down transformer.

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Long Answer Questions

1. Obtain an expression for impedance and current in series LCR circuit. Deduce an
expression for the frequency of an LCR series resonating circuit?
A. Phasor Diagram Solution:
Let an alternating emf V= V0 sin ω t be applied to a circuit containing a resistor of resistance
R, a capacitor of capacitance C and an inductor of inductance L connected in series as shown
in the fig.

Let i = i0 sin (ωt + φ ) be ac current in each element at any time where φ is the phase difference
between voltage of source and the current in the circuit. Let VL ,VR ,VC and V represent the
voltage across the inductor, resistor, capacitor and the source respectively which are shown in
the phasor.
The voltage equation for the circuit is VL + VR + VC = V and the amplitudes of
VL = i0 X L , VR = i0 R, VC = i0 X e and V = V0
From the diagram (B)
V02 = VR2 + (VC − VL )
2

V02 = ( i0 R ) + ( i0 X C − i0 X L ) = i0  R 2 + ( X C − X L ) 
2 2 2

V0
Then i0 =
R2 + ( X C − X L )
2

V0
Or i0 = where Z = R 2 + ( X C − X L ) impedance of the circuit
2

Z
VC − VL i0 X C − i0 X L
And also tan φ = =
VR i0 R
XC − X L
i.e. tan φ = where φ is the phase angle between VR and V
R
Resonance: At a resonant frequency, the total reactance of the circuit is zero and the
impedance will be minimum.

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From the expression Z = R 2 + ( X C − X L )

2

The impedance of the circuit Z = R , (∵ X L = X C )

1 1 1
∴ω L = ⇒ ω2 = ⇒ω =
ωC LC LC
1 1
But ω = 2π f where f is the resonant frequency then 2π f = f =
LC 2π LC
At resonant frequency of LCR series circuit, the impedance is minimum equal to R and the
current in the circuit is maximum.

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PROBLEMS

1. An ideal inductor (no internal resistance for the coil) of 20 mH is connected in series with
an AC ammeter to an AC source whose emf is given by e = 20 2 sin ( 200t + π / 3 ) V ,
where t is in seconds. Find the reading of the ammeter?
 π
Sol: e = 20 2 sin  200t + 
3 
Comparing with e = eo sin ( ωt + φ )

ω = 200 rads −1 ; eo = 20 2 v.

L = 20 mH = 20 × 10−3 H
eo e 20 2
io = = o =
xL ωL 200 × 20 × 10−3

io = 5 2 A

io 5 2
irms = = = 5A
2 2

2. The instantaneous current and instantaneous voltage across a series circuit containing
resistance and inductance are given by i = 2 sin ( 100t − π / 4 ) A and v = 40 sin (100t) V.
Calculate the resistance?
 π
Sol: i = 2 sin  100t −  A
 4 
Comparing with i = io sin ( ωt − φ )

io = 2, ω = 100 rads −1
Comparing v = 40 sin (100t) with
V = Vo sin ω t
vo = 40v, ω = 100 rads −1
vo 40
Z= = = 20 2 Ω
io 2

Z = R 2 + ( Lω ) = 20 2
2

R 2 + ( Lω ) = 800
2

Lω
But Tanφ =
R
Lω
Tan π / 4 = ⇒ Lω = R
R
∴ R 2 + R 2 = 800 ⇒ R 2 = 400
R = 20Ω

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3. In an AC circuit a condenser, a resistor and a pure inductor are connected in series across
an alternator (AC generator). If the voltage across them is 20 V, 35 V and 20V
respectively, find the voltage supplied by the alternator?
Sol: VC = 20v, VR = 35v, VL = 20v

V = V22 + (VL − VC ) = ( 35 ) + ( 20 − 20 ) = 35V

2 2 2

4. An Ac circuit contains resistance R, an inductance L and a capacitance C connected in

series across an alternator of constant voltage and variable frequency. At resonant
frequency, it is found that the inductive reactance, the capacitive reactance and the
resistance are equal and the current in the circuit is i0. Find the current in the circuit at a
frequency twice that of the resonant frequency?
A. R = X L = XC
V0 V0
i0 = = (at resonance XL = XC)
Z R
XC
When frequency is doubled X L' = 2 X L ⇒ X C' =
2
2
 R 13R
Z ' = R 2 + ( X L' − X C' ) 2 = R 2 +  2 R −  =
 2 2
V0 V0 2V0 2i
i' = = = = 0
Z' 13R 13R 13
2

5. A series resonant circuit L1, R1 and C1. The resonant frequency is F. Another series
resonant circuit contains L2, R2 and C2. The resonant frequency is also F. If these two
circuits are connected in series, calculate the resonant frequency?
1 ω
A. f = = and
2π L1C1 2π
1 ω
f = =
2π L2C2 2π
1 1
R1 = L1ω = and R2 = L2ω =
C1ω C2ω
R1 R2
L1 = , L2 =
ω ω
R1 + R2
In series combination effective inductance (L) = L1 + L2 =
ω
1 1
And = R1ω , = R2ω
C1 C2

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C1C2 1
In series combination effective capacitance (C) = =
C1 + C2 ( R1 + R2 ) ω
1
∴ LC =
ω2
1 ω
Resonating frequency f ' = = = f
2π Lc 2π

6. In a series LCR circuit R=200 W and the voltage and the frequency of the mains supply
is 200V and 50 Hz respectively. On taking out the capacitance from the circuit the
current lags behind the voltage by 450. On taking out the inductor from the circuit the
current leads the voltage by 450. Calculate the power dissipated in the LCR circuit?
A. R = 200Ω
Vrms = 200V

Z R 2 + X L2 R 2 + X C2
tan φ = = =
R R R
2
Vrms 200 × 200
∴ XL = XR =P = = = 200W
R 200

7. The primary of transformer with primary to secondary turns ratio of 1: 2, is connected to

an alternator of voltage 200 V. A current of 4 A is flowing though the primary coil.
Assuming that the transformer has no losses, find the secondary voltage and current are
respectively?
NP 1
Sol: = , VP = 200V .
NS 2

N S VS 2 V
= ⇒ = S ⇒ VS = 400V
N P VP 1 200

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