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Formula Sheet
Alternating Current
SECOND
EDITION
23.
A L T E R N A T I N G
C U R R E N T
1. INTRODUCTION
A majority of electrical power in the world is generated, distributed, and consumed in the form of 50-Hzor60-Hz
sinusoidal alternating current (AC) and voltage. It is used for household andindustrial applications.
AC has several advantages over DC. The major advantage of AC is the fact that it can be transformedinto any form,
whereas direct current (DC) cannot. A transformer permits voltage to be stepped up or down for the purpose of
transmission. Transmission of high voltage (in terms of KV) implies that less current is required to produce the same
amount of power. Less current permits thinner wires to be used for transmission.
In this chapter, we will introduce a sinusoidal signal and its basic mathematic equation. We will discuss and analyse
circuits where currents i(t) and voltages v(t) vary with time. The phasor analysis techniques will be used to analyse
electronic circuits under sinusoidal steady-state operating conditions. The chapter will conclude with single-phase
power.
2. SINUSOIDAL WAVEFORMS
AC, unlike DC, flows first in one direction, then in the
opposite direction. The most common AC waveform is a
sine (or sinusoidal) waveform.
In discussing AC signal, it is necessary to express the
current and voltage in termsof maximum or peak values,
peak-to-peakvalues, effective values, average values, or
instantaneous values. Each of these values has a different
meaning and is used to describe a different amount of
current or voltage. V(t)=V 0 sin wt. Where Vo is the peak
voltage, ω = 2 π f is the angular frequency expressed in
radian per second (rad/s), f is the frequencyexpressed in
Hertz (Hz), t is time expressed in second (s).
2.1 Instantaneous Value
The instantaneous value of an AC signal is the value of voltage or current at one particular instant. The value may
be zero, if the particular instant is the time in the cycle at which the polarity of the voltage is changing. It may also
be the same as the peak value, if the selected instant is the time in the cycle at which the voltage or current stops
increasing and starts decreasing. There are actually an infinitenumber of instantaneous values between zero and
the peak value.
1 cycle
RMS value
Peak-to-Peak
Peak value
Figure 23.1: Sinusoidal Waveform.
Physics | 23.
PLANCESS CONCEPTS
(b) RMS value of current = rms
I =2.5 A so, (^) ( ) ( )
2 2 2 rms av
I = I =6.25 A
m rms
I
( c ) I ;
2
So, current amplitude Im = 2 Irms^ =^ 2 2.5A( )=3.5 A
m rms
V
( d) V 220V ;
2
So, supply voltage amplitude Vm^ =^ 2 V( (^) rms^ ) =^ 2 220V( )=311 V.
2.3 Effective Value (RMS Value)
This is the value of AC signal that will have the same effect on a resistance as a comparable value of direct voltage
or current will have on the same resistance. It is possible to compute the effective value of a sine wave of current to
a good degree of accuracy by taking equally spacedinstantaneous values of current along the curve and extracting
the square root of the average of the sum of the squared values. For this reason, effective value is sometimes called
RMS value.
Root mean square value of a function, from t 1 to t 2
is defined as =
∫
t 2 2
t 1 rms 2 1
f dt
f
t t
The magnitude of Irms is given by
( )
T T 2 2 2 (^0 ) (^2 0 0 ) rms (^) T T
0 0
I dt I sin dt I I 2 dt dt
ωτ
0 eff rms 0
I
I I 0.707 I
Where I 0 is the peak value of the current. Similarly Ve f f^ or
0 rms 0
V
V 0.707 E
A
RMS value is actually more important because in the context of power transmission,the loss in energy
due to a resistor plays an important role. And the power is given by (^) i R,^2 where R is the resistance.
Yashwanth Sandupatla (JEE 2012, AIR 821)
Illustration 2: Find the RMS value of current I = Im sin ω tfrom (i) t=0 to t=
π
ω
(ii)t=
to t 2 2
π π
ω ω
(JEE MAIN)
Sol: In AC circuit over time interval 0 ≤^ t^ ≤^ Tthe RMS value of current is given by
( )
T T 2 2 2 0 0 0 0 rms (^) T T
0 0
I dt I sin dt
I
I
dt dt^2
∫ ∫
∫ ∫
where
T
π
ω
23.4 | Alternating Current
PLANCESS CONCEPTS
(i)
2 2 m (^2) (^0) m m rms
I sin ( t)dt
I I
I
π ω
∫
(ii)
3 2 2 2 m
2 2 m m rms
I sin ( t)dt
I I
I A
π π
π
∫
The RMS value of one cycle or half cycle (either a positive or negative cycle) is same.
GV Abhinav (JEE 2012, AIR 329)
2.4 Difference between Sine and Cosine Representation of AC Signal
The sine and cosine are essentially the same function, but with a 90^0 phase difference. For example, sin (^) ωt =cos
( )
0
ω −t 90. Multiples of 360^0 may be added to or subtracted from the argument of any sinusoidal function, without
changing the value of the function. To realize this, let us consider
( )
0 1 P
V = V cos 10t + (^20) ( )
0 0 P
= V sin 10t + 90 + (^20) … (i)
( )
0 = VP1 sin 10t + (^110) Leads (^) ( )
0 V 2 = VP2 sin 10t − (^40) … (ii)
by 150 0. It is also correct to say that v 1 lags v 2 by 210^0 , since v 1 may be written as
( )
0 V 1 = VP1 sin 10t − (^250) V … (iii)
q
V (^) Psin( wt+ ) q V sinp wt
v
wt
-V (^) P
V (^) P
Figure 23.2: Representation of voltage as sine and cosine function
23.6 | Alternating Current
PLANCESS CONCEPTS
m m
V
I
R
rms rms
V
I
R
=
=
2 rms rms rms
V
V I cos R
φ =
4.2 Purely Capacitive
Writing KVL along the circuit shown in Fig. 23.
S
q
V 0
c
− = And current in the circuit is
( ) ( m )
m m
dq d cv^ d cv^ sin^ t^ V cv sin t cos t. dt dt dt 1 / C
I
ω = = = ω ω = ω ω
m m C
V
cos t I cos t X
= ω = ω
where (^) C
where X
C
and is called capacitive reactance. Its unit is Ohm ( Ω ).
From the graph of current versus time and voltage versus time,
it is clear that current attains its peak value at a time
T
before the
time at which voltage attains its peak value. Corresponding to
T
phase difference.
2 T
t
T 4 2
i cleadsvcby π / 2diagrammatically (phasediagram) represented as
Since
0
=
rms rms
V I cos φ = 0
The current leads the voltage by π / 2in a capacitive circuit
=0 doesn’t mean it is zero in any period less than the time period. In actuality, first the capacitor
gets charged up, gaining energy during the first half cycle, and loses it for the next half cycle.So overall,
power becomes zero.Same goes for the inductor in a different fashion (magnetic field plays a role there).
Yashwanth Sandupatla (JEE 2012, AIR 821)
4.3 Pure Inductive Circuit
Writing KVL along circuit, s
di
V L 0
dt
m
di
L V sin t;
dt
m
Ldi = V sin ωt dt ∫ ∫
i
m
V
cos t C
L
; =0 ; C=0;
∴i
m
V
cos t
L
m L
Vm I X
= From the graph of current versus time and voltage
versus time, it is clear that voltage attains its peak value at a time
T
before the time
at which current attains its peak value. Corresponding to
T
, the phase difference
2 T 2
t
T 4 T 2
i
V = Vs m sin wt
C
~
Figure 23.4: AC voltage
applied to capacitive load
Figure 23.6: AC voltage
applied to inductive load
Figure 23.
V
t
T
i
t
V m
I m
i
V = Vs m sin wt
~
L
Physics | 23. V
t
T
i
t
Figure 23.7: Variation of current and voltage with respect to time
Diagrammatically (See Fig. 23.7) it is represented
Vm
I (^) m
as
iL lags behind V L by π / 2since
0
φ = 90 ,
=Vrms I rms cosφ =0. The current lags voltage by π / 2in a purely inductive
circuit.
Induced
voltage
Applied
voltage
Figure 23.8: AC voltage applied to purely inductive circuit
5. IMPEDANCE
We have already seen that the inductive reactance L
X = ω Land capacitance reactance
C
X = 1 / ωL play the role of
an effective resistance in apurely inductive and capacitive circuit respectively. In the series RLC circuit, the effective
resistance is the impedance,defined as = + (^) ( − (^) )
2
L C
Z R X X …(iv)
The relationship between Z, X (^) L, and XC can be represented by the diagram shown in Fig. 23.9.
Following is a diagrammatic representation of the relationship between Z, X L
and X C
Figure 23.9: Impedance
Triangle
R
f
Z X (^) L - XC The impedance has SI unit of Ω. In terms of Z the current may be rewritten as I(t)
= (^) ( )
V 0
sin t
Z
ω − φ …(v)
Notice that the impedance Z also depends on the angular frequency ω ,as do X
L
and X C
Using the above equations for phase φ and Z, we may readily recover the limit for
simple circuit (with only one element).
**Physics | 23.**
= (^) ( )
1/ 2 62 2 3.14 40 0.
+ × × ×
(a) RMS value of the current: I rms
rms
V 220
Z 6.
= = A
(b) The potential difference across the resistance is given by: V R
=I
rms ×
R=33.83 × 6= 202.83 V
(c) Potential difference across the inductance is given by:
VL = Irms × ( ωL ) = 33.83 × (^) ( 2 × 3.14 × 0.01)= 96.83 V
(d) Phase angle 1
L
tan R
− ^ ω φ = (^)
; so, φ = tan-1^ (0.4189=22.46)
Now time lag=
T 0.0623 s.
Illustration 5: A
H inductor and a 12 Ω resistance are connected in a series to a 225 V, 50 Hz ac source.
Calculate the current in the circuit and the phase angle between the current and the source voltage. (JEE MAIN)
Sol: Theimpedance of LR circuit is ( )
2 2
Z = R + ωL. The RMS value of the current is^
rms rms
V
I
Z
=. In LR circuit, the
current lags the applied voltage by phase angleφ obtained as
1 L
tan R
− ^ ω φ = (^)
Here XL= ωL =
fL 2 50 9
π = π × × = Ω
So,
2 2 2 2 L
Z = R + X = 12 + 9 = 15 Ω
(a) l=
V 225
15A
Z 15
= = and (b)
1 XL^19
tan tan
R 12
− −
^
1 0 tan 3 / 4 37
− = =
o
i.e., the current will lag the applied voltage by
0
37 in phase.
Illustration 6: A chokecoil is needed to operate an arc lamp at 160 V (RMS) and 50 Hz. The arc lamp has an
effective resistance of 5Ω when running of 10 A (RMS). Calculate the inductance of the choke coil. If the same arc
lamp is to be operated on 160V (dc), what additional resistance is required? Compare the power losses in both
cases.
(JEE ADVANCED)
Sol: The choke coil is a LR circuit having large inductanceand small
resistance. The potential difference across the resistor and inductor is added
vectorially:V
2 =VR
2 +V (^) L
2 .
As for the lamp, VR=IR = 10 × 5 = 50V, so when it is connected to 160 V ac source
though a choke in series,V 2 =VR^2 +V (^) L^2 , VL=
2 2
160 − 50 = 152 V
And as, VL=IXL=I ωL = 2 fLIπ L= =
π × π × ×
V L 152
2 fI 2 50 10
2
10 H
−
×
Now the lamp is to be operated at 160 V dc; instead of choke, if additional
resistance r is I put in a series with it, V = I(R+r), i.e. 160 = 10(5+r) i.e. r = 11Ω In
case of AC, as choke has no resistance, power loss in the choke will be zero, while
~
R
V R
Ark lamp
L
V = V sin 0 wt
VL
Figure 23.
23.10 | Alternating Current
the bulb will consume P=I
2 R=
2 × 5=500 W. However, in case of DC,as resistance r is to be used instead of choke,
the power loss in the resistance r will be PL=10^2 X^ 11=1100 W
While the bulb will still consume 500 W, i.e., when the lamp is run on resistance r instead of choke, more than
double the power consumed by the lamp is wasted by the resistance r.
6.2 RC Circuits
I S
f
V S
V R
V C
(b)
~V^ S
R
C
V R
V C
(a)
Figure 23.12: (a) AC voltage applied to RC circuit (b) Phasor diagram of voltage drops across R and C
If V s
, V
R
and V C
are RMS voltages across a source, resistance and capacitor respectively
2 2 2 2 S R C S C
V = V + V = I = R +X
Impedance of circuit, Z=
S 2 2 2 C (^2 ) S
V 1
R X R
I C
V
S
leads I S
by
1 X^ C 1 1
tan tan
R CR
− −
^
The current leads the applied voltage by angel φ.
Illustration 7: An ac source of angular frequency ω is fed across a resister R and a capacitor C in series. The current
registered is I. If now, the frequency of source is changed to w/3 (but maintaining the same voltage), the current in
the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency w.
(JEE MAIN)
Sol: The impedance of RC circuit is:
2
Z R
C
. The RMS current is
rms rms
V
I
Z
According to the given problem, I=
( )
1/ 2 2
V V
Z
R 1 / C
… (i)
And for frequency of
( )
1/ 2 2
I V
R 3 / C
… (ii)
Substituting the value of I from equation (i) in (ii),
2 2 2 2 2 2
4 R R
C C
+^ =^ +
ω^ ω
i.e..
2 2 2
R
C 5
23.12 | Alternating Current
PLANCESS CONCEPTS
( j)
= Vrms I (^) rms
cos 200 2 2
φ = × × W
(k)
I
rms
2 R=200 W
(l)
Physics | 23. PLANCESS CONCEPTS
V (^) S ~
R
L
VR
C
VL
VC
(a) (b)
IS
f
V -V
L
C VS
VR
VL
V C
Figure 23.15: (a) AC voltage applied to LCR circuit. (b) Phasor diagram of voltage drops across L, C and R
Where cos φ is called the power factor of the LCR circuit.
6.4.1 Resonance in RLC Circuits
At a particular angular frequency ωο^ of the source, when XL^ =^ XC or ωο^ L=
0
ω C
, the impedance of the circuit
becomes minimum and equal to R and therefore, the current will be maximum. The circuit is then said to be
in resonance. The resonance angular frequency ωο^ andfrequency V 0 given by
0 0
2 2 LC
ω = ν = π
The variation of RMS current with the frequency of the applied voltage is shown
in the Fig. 23.16. If the applied voltage consists of a number of frequency
components, the current will be large for the components having frequency V 0
The Q factor of an LCR series circuit is given by Q=
0 L
R
. A direct current of a
flows uniformly throughout the cross-section of theconductor. An alternating
current on the other hand, flows mainly along the surface of the conductor.
This effect is known as the skin effect. The reason is that when ac flows through
aconductor, the flux change in the inner part of the conductor is higher.
The idea of resonance is used in TV channelsfor clarity: a particular frequency is assignedto a channel
and when this frequency is received by the receiver,the current corresponding to this frequency becomes
maximum. This helps in maximum possible separation of channels, thus increasing their individual clarity.
It is also used by intelligence agencies to intercept the signals of anti-social elements. They generally use
frequency of a very high order.
Nivvedan (JEE 2009, AIR 113)
V
0
I rms
V
Figure 23.
Physics | 23.
(a) Potential difference across resistance: V R
=iR = (^) 5 × 16 =80 V
Potential difference across inductance: VL=i × ω( L^ )=^5 ×^24 =120 V
Potential difference across capacitor: Vc = ×i (^) ( 1 / ωC (^) ) = 5 × 12 =60 V
( ) ( )
2 2 1 2 2 (b) Z R L 16 12 20 C
= ^ + − ω = + = Ω (^) ω
(c) The voltage of ac supply is given by: V = IZ = 5 × 20 =100 V
(d)
1 L^ (^ 1 /^ c)
tan
R
−
tan tan 0.75 36 46 16
− ^ − −
Illustration 11: An oscillating voltage drives an alternating current through a resistor, an
inductor, and a capacitor that are all connected in series. Calculate the RMS voltage across
each another by multiplying the reactance or resistance of each element by the RMS current.
To calculate the RMS current, divide the RMS voltage by the impedance. (JEE ADVANCED)
Sol: In series LCR circuit, the impedance is ( )
2 2
Z = R + XC − XL where XC and XL are the
capacitive reactance and inductive reactance respectively. The phase angle between voltage
and current is given by
1 XL^ XC
tan R
−
φ = (^)
. Find the current in the series circuit, and multiply the resistance or
reactance of each element with the currrent to find the voltage drop across it.
- Calculate XC ;
( )
C
C (^2) 60.0Hz 0.1 F
X
ω (^) π μ
=17.68 k Ω
- Calculate XL^ ; XL^ = ωL^ =^2 π^ ( 60.0Hz^ )( 25mH)^ =^ 9.42π Ω
- Calculate the impedance:
( ) ( ) ( )
2 2 2 2 L C
Z = R + X − X = 9.9k Ω + 0.00942k Ω − 17.68kΩ (^) = 20.25 KΩ
- Divide the voltage by the impedance:
rms rms
V 115 V
I 5.7 mA Z 20.25 k
- Multiply the current by the resistance: rms.R rms
V = I R = 5.68 mA(9.9k Ω =) 56 V
6. Multiply the current by the inductive reactance: Vrms.L^ =^ Irms^ XL^ =^ 5.68 mA( 9.42k^ Ω^ )=^ 54 V
- Multiply the current by the capacitive reactance:
rms.C rms C
V = I X = 5.68 m A(17.68k Ω ) = 100V= 0.10 KV
L
V (^) rms
~
R
l (^) rms C
Figure 23.
23.16 | Alternating Current
6.5 Parallel RCL Circuits
Consider the parallel RLC circuit illustrated in Fig. 23.19.
The voltage source is V (t) =V 0 sin ωt (^).
Unlike the series RLC circuit, the instantaneous voltage acrossall three circuit
elements R, L, and C are the same, and each voltage is in phase with the current
through the resistor. However, the current through each element will be different.
In analysing this circuit, we make use of the results derived before. The current
in the resistor is
0 R R
V(t) V
I (t) sin t I sin t
R R
= = = ω = ω … (i)
Where R
I =
0
V /R. The voltage across the inductor is
L L 0
dI
V (t) V(t) V sin t L
dt
= = ω = … (ii)
which gives
t 0 0 0 L L 0 L
V V V
I (t) sin t'dt' cos t sin t I sin t
L L X 2 2
∫
… (iii)
where L0 0 L L
I = V / X andX = ωL is the inductive reactance.
Similarly, the voltage across the capacitor is V C
(t)=V 0
sin (^) ωt =Q(t)/c, which implies
0 C 0 C C
dQ V I (t) CV cos t sin t I sin t dt X 2 2
(^) π (^) π = = ω ω = (^) ω + (^) = (^) ω +
… (iv)
where C0 0 C C
I = V / X and X = 1 / ωL is the capacitive reactance.
Using Kirchhoff’s junction rule, the total current is simply the sum of all three currents.
R L c
I(t) = I (t) + I (t) + I (t)=
R0 L0 C
I sin t I sin t I sin t 2 2
π π ω + (^) ω − (^) + (^) ω +
… (v)
The current can be represented with the phasor diagram shown in Fig. 23.
I CO
I 0
f
I CO
ILO
I LO
I RO
V 0
Figure 23.20: Phase difference between current and voltage
From the phasor diagram, we see that. I 0 = IR0 + IL0 +IC
… (vi)
And the maximum amplitude of the total current, I 0 , can be obtained as
V(t)
~ R L C
Figure 23.19 Parallel LRC circuit
23.18 | Alternating Current
Illustration 12: The image shows an inductor (L=0.22 mH) in series with a 15 Ω
resistor. These elements are in parallel with a second 15 Ω resistor. An AC generator
powers the circuit with an RMS voltage of 65V.
In the limit of high frequency, the inductor behaves like a very large resistor. In such a
case, nearly all of the current flows through the branch with the lone resistor. Calculate
the current by dividing the RMS voltage by the single resistor.
In the limit of low frequency, the reactance of the inductor approaches zero.
In such a case, the current flows through each resistor equally. Calculate the
equivalent resistor and divide the voltage by the equivalentresistance to determine the current. (JEE ADVANCED)
Sol: For very high source frequency, the reactance of the inductor becomes practically infinite so that the current
doesn’t flow through the inductor. Thus, the inductor acts as an open circuit. For very low source frequency, the
reactance of the inductor becomes practically zero, and theinductor behaves as a short circuit.
- Calculate the current at high frequency:
rms rms
V 65 V
I
R 15
4.3 A
- Calculate the equivalent resistance at low frequency:
1
eq
1 1 R 15
R 7.
R R 2 2
− Ω = (^) + (^) = = = Ω
Divide the voltage by the equivalent resistance:
rms rms eq
V 65V
I
R 7.
= = =8.7 A
Illustration 13: For the circuit shown in Fig. 23.23, current in inductance is 0.8 A while its capacitance is 0.6A. What
is the current drawn from the source? (JEE ADVANCED)
~
C
I L
I I^ C
Figure 23.
Sol: For LC circuit, total current in the circuit is I = I 0 sin( ω + φt (^) )= IL + IC. The current in the inductor lags the applied
voltage by phase difference of
while in capacitor, the current leads applied voltage by
In parallel ac circuit,
V = V sin 0 ωt is applied across both the inductorand capacitor, current in inductor lags the applied voltage while
current in capacitor leads the applied voltage.
So, I (^) L=
C
V
sin t 0.8cos t X 2
π ω −^ = −^ ω
C C
V
I sin t 0.6cos t X 2
π = (^) ω + (^) = ω
So, the current drawn from the source, L C
I = I + I = −0.2cos ωt , i.e.
0
I =0.2 A
7. MORE ON POWER FACTOR
(a) The factor cosφ present in the relation for average power of an ac circuit is called power factor.
So, cosφ =
ac avg
rms rms V
P P
E I p
=. Thus, ratio of average power and virtual power in the circuit is equal to power factor.
L R
~
R
Figure 23.
Physics | 23.
(b) Power factor is also equal to the ratio of the resistance and the impedance of the
ac circuit.
Thus, cosφ =
R
Z
(c) Power factor depends upon the nature of the components used in the circuit.(d) If
a pure resistor is connected in the ac circuit then,
φ = 0,cos φ = 1 ;
2 0 0 0 av
E I E
p Erms 2 2R
= = = Irms
Thus, the power loss is maximum and electrical energy is converted in the form of heat.
(e) If a pure inductor or capacitor are connected in the ac circuit, then
φ ≠ 90
o , cosφ = 0 ∴ Pav =0 (minimum)
Thus is no loss of power.
(f) If a resistor and an inductor or a capacitor are connected in an ac circuit, then φ ≠ 0 or φ ≠ 90 o. Thus φ is in
between 0 & 90
o .
(g) If the components L, C and R are connected in series in a circuit, then
( ω^ −^ ω ) φ = = φ =
X L^ 1 /^ C R
tan and cos R R Z ( )
1/ 2 2
Pow
R R
; cos
Z
R L 1
er factor
/ C
(h) Power factor is a unit less quantity.
(i) If there is only an inductance coil in the circuit, there will be no loss of power, and energy will be stored in the
magnetic field.
( j) If a capacitor is only connected in the circuit, there will also be no loss of power, and energy will be stored in the
electrostatic field.
(k) In reality, an inductor and capacitor do have some resistance. So, there is always some loss of power.
(l) In the state of resonance, the power factor is one.
8. WATTLESS CURRENT
(a) The component of current whose contribution to the average power is nil, is called wattless current.
(b) The average wattle of power iszero because the average of the second
component of instantaneous power for a full cycle will be
E sin 0 ωt l sin (^) ( 0 φ (^) ) sin (^) ( ω − πt / 2 (^) )= 0
(c) The component of current associated with this part is called Wattless
current. Thus the current
( l sin 0 φ)^ sin^ ( ω − πt^ / 2)is a wattless current whose amplitude is^ l sin 0 φ^.
(c) If RMS value of current in the circuit is Irms , then the RMS value of a
wattless current will be Irms , sinφ. A wattless current lags or leads the e.m.f.
by an angle π / 2. RMS value of wattless current:
Z
wC
f
0 R
Figure 23.
Irms sinf
I (^) rms cosf
f
I (^) rms
E
Z
X
Figure 23.
