Video Lecture
Theory For Making Notes
Mean Value of AC
Mean value of a time varying current can be found using the following equation
Iav=∫dt∫Idt the limits of integrations depends upon the time interval in which mean value is required.
Usually the mean value of ac is calculated for two time intervals
(1) Over half time period (half cycle).
(2) Over full time period (full cycle).
The average value of ac over half cycle can be calculated as follows:
Iav = T/20∫T/2Idt=T20∫T/2Idt … (1)
= T20∫T/2I0sinωt(dt)
= T2I0[−ωcosωt]0T/2 = ωT2I0[−cosω2T−(−cos0)]
= T2π.T2I0[−cosT2π.2T+cos0]
= πI0[−cos(T.22π.T)−(−1)]
= πI0[−cosπ+1]
(ii) Similarly for calculating mean value of alternating current over full cycle we have to replace the limit of integration in equation (1) by 0 to T. Solving integration in the same way we find that the mean value of ac over full cycle is zero. Also looking at graph, the positive current pattern is equal and opposite to negative pattern, hence mean value of ac over full cycle is zero.
RMS value
The rms value of any quantity is “square root of the arithmetic mean of squares of all those values whose rms value is required”. RMS value is also called effective values.
Thus rms value of 1, 2, 3, 4 is
(12+22+32+42)/4=215
RMS value of AC
In case of AC, the rms value of an AC comes out to be that value of DC which will produce the same heating effect on passing through the same resistance over the same period of time. Hence rms value of ac is also called Equivalent DC (Ieq)
If the magnitude (I) of AC at any instant of time is given by
I = I0sinωt,
Then heat (H) produced by this AC when passed through a resistance (R) during one time period (T) will given by
H=0∫TdH=0∫TI2Rdt … (i)
And during this one time period T, the heat produced by a DC (Irms or Ieq) will be
H =Irms2RT … (ii)
Hence, equation (i) and (ii), when equated will give us the value of the equivalent DC or rms value of ac (Ieq).
Irms2RT = 0∫TI2Rdt=0∫T(I0sinωt)2Rdt
= I02R0∫Tsin2ωtdt
= I02R0∫T(21−cos2ωt)dt [because cos(2 ω t)= 1 — 2 sin2 ω t)
= 2I02R⎣⎡0∫Tdt−0∫Tcos2ωtdt⎦⎤
= 2I02R[T−{2ωsin2ωt}0T]
=2I02RT
[Because when t = T, sin2 ω t = sin 2 ω T = sin(2. (2 π /T) × T)= sin 4π= 0; and also at t= 0, sin 2ωt= sin 0 = 0].
hence Irms2RT=2I02RT or Irms2=2I02
or Irms=2I0 Similarly Erms=2E0
Illustration
The electric current in a circuit is given i=ioτtfor some time. Calculate the rms current for the period t = 0 to t =τ.
Solution
Irms=0∫τdt0∫τI2.dt=τ0∫ττ2I02t2.dt
=τ3I02[3t3]0τ=τ3I02.3τ3−0 = 3I0 Ans.
Simple AC Circuits
AC circuit containing only a Resistor
Figure shows a circuit containing an AC source E = Eosinωt and a resistor of a resistance R.
Such a circuit is also known as a purely resistive circuit.

If the current at time t is I(t), Kirchhoff’s loop law gives I(t).R= Eosinωt
or, I(t)=RE0sinωt
I(t)= Iosinωt
where I0=RE0
Phasor Diagram
A diagram representing alternating voltage and current as rotating vectors with angular velocity ω in the counter clockwise direction.
In these diagram the instantaneous value of a quantity that varies sinusoidally with time is represented by the projection onto a vertical axis (if it is a sine function) or onto a horizontal axis (if it is a cosine function) of a vector with a length equal to the amplitude (I0) of the quantity. The vector rotates counter clockwise with constant angular velocity ω.
As in case of a pure resistor current and voltages are in same phase hence on phasor diagram the peak value or rms value of voltage and current can be shown by two collinear vectors as given below

Graphically the current and emf can be shown by two waves which reach zero and peak value at the same time.

Average Power
The mean value of power over one complete cycle can be found as given below:
Pav=T0∫TP.dt=T0∫TE.I.dt
Putting source E = Eosinωt and I= Iosinωt in the above equation and solving we get
Pavg = Erms . Irms
Circuit containing only an inductor
Figure shows an inductor connected to an AC source. Such a circuit also known as a purely inductive circuit.

The induced emf across the inductor is −LdtdI so from Kirchhoff’s loop law,
E0sinωt−LdtdI=0 or, dtdI=LE0sinωt integrating the equation I=−ωLE0cosωt or, I=ωLE0sin(ωt−π/2)
|
or, I=I0sin(ωt−π/2)
where as E = Eosinωt … (3)
where I0=ωLE0
The term ωL plays the role of effective resistance in this circuit and is denoted by XL. It is called the inductive reactance of the inductor. The graph between XL and ν is shown below

It is zero for direct current (as ω = 0) and increases as the frequency is increased.
We see from equation number (2) and (3) that the phase of the current is 2π less than that of the emf. The current lags behind the emf by 2π (radians).
Figure shows plots of the current through an inductor and of the emf as time passes

The above relation of phase angle between current and emf can be represented on a phasor diagram by drawing peak current and peak emf as two mutually perpendicular vectors as shown below:

Average power over full cycle:
As Pav=T0∫TP.dt=T0∫TE.Idt
Substituting E = Eosinωt and I = Io sin ( ω t — π /2)in the above equation and solving we get
Pav = 0
Practice Questions (Basic Level)
Q.1
The reactance of a 25 µF capacitor at the AC frequency of 4000 Hz is
(a) π5Ω
(b) π5Ω
(c) 10Ω
(d) 10Ω
Ans. (a)
Q.2
What will be the approximate resistance offered by a capacitor of 10 µF and frequency 100 Hz ?
(a) 160 Ω (b) 1600 Ω (c) 16 Ω (d) None of the above
Ans. (a)
Q.3
An AC voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3Ω, the phase difference between the applied voltage and the current in the circuit is
(a) π / 4 (b) π / 2 (c) zero (d) π / 6
Ans. (a)
Q.4
The power factor of an R–L circuit is 21. If the frequency of AC is doubled, what will be the power factor ?
(a) 31
(b) 51
(c) 71
(d) 111
Ans. (b)
Q.5
An inductive coil has a resistance of 100 Ω. When an AC signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by 45º. The inductance of the coil is
(a) 10π1
(b) 20π1
(c) 40π1
(d) 60π1
Ans. (b)
Q.6
Two coils have a mutual inductance 0.005 H. The current changes in the first coil according to equation I=I0sinωt, where I0=10A and ω=100πrad/s. The maximum value of emf in the second coil is (in volt).
(a) 2π
(b) 5π
(c) π
(d) 4π
Ans. (b)
Q.7
In an AC circuit, V and I are given by V = 150 sin (150 t) volt and I=150sin(150t+3π) The power dissipated in the circuit is
(a) 106 W (b) 150 W (c) 5625 W (d) zero
Ans. (c)
Q.8
In an AC circuit an alternating voltage e=2002sin100tvolt is connected to a capacitor of capacity 1µF. The rms value of the current in the circuit is
(a) 100 mA (b) 200 mA (c) 20 mA (d) 10 mA
Ans. (c)
Q.9
An AC voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3Ω, the phase difference between the applied voltage and the current in the circuit is
(a) π/4 (b) π/2 (c) zero (d) π/6
Ans. (a)
Q.10
In the case of an inductor
(a) voltage lags the current by 2π
(b) voltage leads the current by 2π
(c) voltage leads the current by 3π
(d) voltage leads the current by 4π
Ans. (b)
Q.11
The inductive time constant in an electrical circuit is
(a) LR
(b) RL
(c) RL
(d) LR
Ans. (b)
Q.12
In an AC circuit, V and i are given by V = 100 sin (100 t) volt and i=100sin(100t+3π)amp The power dissipated in the circuit is
(a) 104 W (b) 2.5 kW (c) 5 kW (d) 5 W
Ans. (b)
Q.13
The reactance of 25 µF capacitor at the AC frequency of 4000 Hz is
(a) π5Ω
(b) π5Ω
(c) 10 Ω
(d) 10Ω
Ans. (a)
Q.14
The frequency for which a 5 µF capacitor has a reactance of 10001Ω is given by
(a) π100MHz
(b) π1000Hz
(c) 10001Hz
(d) 1000 Hz
Ans. (a)
Q.15
The capacity of a pure capacitor is 1 F. In DC circuits, its effective resistance will be
(a) zero (b) infinite (c) 1 Ω (d) 1/2 Ω
Ans. (b)
Q.16
The rms voltage of the wave form shown is
(a) 10 V
(b) 7 V
(c) 6.37 V
(d) None of these

Ans. (a)
Q.17
The output sinusoidal current versus time curve of a rectifier is shown in the figure. The average value of output current in this case is
(a) 0
(b) 2I0
(c) π2I0
(d) I0

Ans. (c)
18.
The capacitor reactance of a capacitor in d.c. circuit in ohms is
(A) ωC1
(B) ωC
(C) zero
(D) ∞
Ans (A)
19.
The reactance of a inductor at 50 Hz is 10 Ω. What will be its reactance at 200 Hz ?
(A) 10 Ω
(B) 40 Ω
(C) 2.5 Ω
(D) 20 Ω
Ans (B)
20.
An inductive circuit contains a resistance of 10 Ω and inductance of 2H. If an a.c. voltage of 120 V and frequency 60 Hz is applied to this circuit, the current in circuit would be nearly
(A) 0.32 A
(B) 0.16 A
(C) 0.48 A
(D) 0.80 A
Ans (B)
21.
What is the average value of a.c. over a half cycle
(A) zero
(B) πI0
(C) π2I0
(D) none of these
Ans (C)
22.
In an AC circuit having R and L in serries
(A) voltage leads current
(B) current leads voltage
(C) voltage and current are increase
(D) voltage and current are in opposite phase
Ans (A)
23.
Average power in a pure inductive circuit is
(A) irms´ vrms
(B) 2 irms´ vrms
(C) 1
(D) zero
Ans (D)
24.
The root mean square value of voltage (V) in an AC circuit is
(A) 0.637Vmax
(B) 0.707 Vmax
(C) 2Vmax
(D) 2Vmax
Ans (B)
25.
In an AC series circuit, instantaneous voltage is maximum while instantaneous current is zero. The source is connected to
(A) pure inductor only
(B) pure resistance only
(C) pure capacitor only
(D) (A) or (C)
Ans (D)
26.
In the circuit shown, the voltage in C & L are

(A) out of phase byπ
(B) out of phase by π/2
(C) out of phase by π/4
(D) In Phase
Ans (A)
27.
If the mean value of an alternating voltage is π282 volt, its rms value is
(A) 28 volt
(B) 14 volt
(C) 142 volt
(D)zero
Ans (B)
28.
When an alternating voltage source is connected to a pure inductor, the difference between phase of instantaneous current through the inductor and the phase of instantaneous potential difference across the inductor at t = T/4 sec. is
(A)zero
(B) π/2 rad.
(C) –π/2 rad.
(D) π/4 rad.
Ans (C)
Practice Questions (JEE Main Level)
1.
An alternating current is given by i = i1 cos ωt + i2 sin ωt. The rms current is given by
(A) i1 + i2
(B) (i1+i2)2
(C) 2i1+i2
(D) none of the above
Ans (D)
2.
The current in a series RL circuit decays asI=I0e−t/τ. Obtain the rms current in the interval0≤t≤τ.
(a) 2I0
(b) eI0
(c) 2eI0
(d) eI02e2−1
Ans (d)
3.
An alternating voltage is given by e=e1sinωt+e2cosωt. Then, the root mean square value of voltage is given by
(a) e12+e22
(b) e1+e2
(c) 2e1e2
(d) 2e12+e22
Ans (d)
Practice Questions (JEE Advance Level)
1.
In an inductor of self-inductance L = 2mH, current changes with time according to relation I=t2e−t.At what time e.m.f. is zero?
(a) 4 s
(b) 3 s
(c) 2 s
(d) 1 s
Ans (c)