Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download

Q1: The value of parameters of the circuit shown in the figure are
For time t < 0, the circuit is at steady state with the switch 'K' in closed condition. If the switch is opened at t = 0, the value of the voltage across the inductor (V_L) at t = 0⁺in Volts is _____ (Round off to 1 decimal place).       (2023)
(a) 6.5
(b) 8
(c) 12.5
(d) 16
Ans: (b)
Sol: Case (i) t < 1
At steady state, capacitor behaves as open circuit and inductor behaves as short circuit.
Redraw the circuit:
Using current division,
Case (ii) t > 0 :
Switch is opened.
Redraw the circuit t = 0⁺:
Apply KVL in loop, 
Q2: The network shown below has a resonant frequency of 150 kHz and a bandwidth of 600 Hz. The Q-factor of the network is __________. (round off to nearest integer)       (2022)
(a) 250
(b) 100
(c) 150
(d) 450
Ans: (a)
Sol: 
Q3: An inductor having a Q-factor of 60 is connected in series with a capacitor having a Q-factor of 240. The overall Q-factor of the circuit is ________. (round off to nearest integer)      (2022)
(a) 12
(b) 24
(c) 48
(d) 96
Ans: (c)
Sol: We have, overall Q-factor of given circuit is,

Q4: A 0.1 μF capacitor charged to 100 V is discharged through a 1 kΩ resistor. The time in ms (round off to two decimal places) required for the voltage across the capacitor to drop to 1V is ______       (2019)
(a) 0.25
(b) 0.65
(c) 0.45
(d) 0.85
Ans: (c)
Sol: Let the time required by the voltage across the capacitor to drop to 1 V is t₁,

Q5: The voltage across the circuit in the figure, and the current through it, are given by the following expressions:
where ω = 100π radian/s. If the average power delivered to the circuit is zero, then the value of X (in Ampere) is _____ (up to 2 decimal places).      (2018)
(a) 5
(b) 10
(c) 15
(d) 20
Ans: (b)
Sol: Given that,

Q6: In the figure, the voltages are
v₁(t) = 100cos(ωt),
v₂(t) = 100cos(ωt + π/18) and
v₃(t) = 100cos(ωt + π/36).
The circuit is in sinusoidal steady state, and R<<ωL. P₁, P₂ and P₃ are the average power outputs. Which one of the following statements is true?       (2018)

(a) $𝑃1=𝑃2=𝑃3=0$P₁ = P₂ = P₃ = 0
(b) $𝑃1<0,𝑃2>0,𝑃3>0$P₁ < 0, P₂ > 0, P₃ > 0
(c) $𝑃1<0,𝑃2>0,𝑃3<0$P₁ < 0, P₂ > 0, P₃ < 0  
(d) P₁ > 0, P₂ < 0, P₃ > 0
Ans: (c)
Sol: V₂ leads V₁ and V₃,
So, V₂ is a source, V₁ and V₃ are absorbing.  
Hence, P₂ > 0 and P₁, P₃ < 0

Q7: The voltage (V) and current (A) across a load are as follows.
v(t) = 100sin(ωt),
i(t) = 10sin(ωt − 60°) + 2sin(3ωt) + 5sin(5ωt).  
The average power consumed by the load, in W, is___________.        (SET-2 (2016))
(a) 100
(b) 150
(c) 200
(d) 250
Ans: (d)
Sol: The average power consumed by the load = 
Q8: A resistance and a coil are connected in series and supplied from a single phase, 100 V, 50 Hz ac source as shown in the figure below. The rms values of plausible voltages across the resistance (V_R) and coil (V_C) respectively, in volts, are      (SET-2  (2016))
(a) 65, 35
(b) 50, 50
(c) 60, 90
(d) 60, 80
Ans: (d)
Sol: It is a series RL circuit and

Q9: In the circuit shown below, the supply voltage is 10sin (1000t) volts. The peak value of the steady state current through the 1Ω  resistor, in amperes, is ______.     (SET-1 (2016))
(a) 0.5
(b) 1
(c) 2
(d) 3
Ans: (b)
Sol: If we observe the parallel LC combination, we get that at ω = 1000 rad/sec the parallel LC is at resonance, thus it is open circuited.
The circuit given in question can be redrawn as
So, So, peak value is 1 Amp.

Q10: A symmetrical square wave of 50% duty cycle has amplitude of ±15 V and time period of 0.4 π ms. This square wave is applied across a series RLC circuit with R = 5 Ω, L = 10 mH, and C = 4 μF. The amplitude of the 5000 rad/s component of the capacitor voltage (in Volt) is ______.       (SET-2 (2015))
(a) 100
(b) 140
(c) 192
(d) 212
Ans: (c)

Q11: The total power dissipated in the circuit, shown in the figure, is 1 kW.
The voltmeter, across the load, reads 200 V. The value of X_L is _____.      (SET-2 (2014)
(a) 8.5
(b) 17.3
(c) 22.4
(d) 28.6
Ans: (b)
Sol: 

Given, total power dissipated in the circuit = 1kW =1000 Watt
Also, voltage drop across R,
Voltage drop across load,
∴ voltage drop across inductor,

Q12: In the circuit shown, the three voltmeter readings are V₁ = 220 V , V₂ = 122 V , V₃ = 136 V.
If  R_L = 5Ω, the approximate power consumption in the load is      (2012)
(a) 700 W
(b) 750 W
(c) 800 W
(d) 850 W
Ans: (b)
Sol: Given,
Power consumed by load,

Q13: In the circuit shown, the three voltmeter readings are V₁ = 220 V , V₂ = 122 V , V₃ = 136 V.
The power factor of the load is       (2012)
(a) 0.45
(b) 0.5
(c) 0.55
(d) 0.6
Ans: (a)
Sol: 
Q14: The average power delivered to an impedance (4 − j3)Ω by a current 5cos(100 πt+100)A is      (2012)
(a) 44.2 W
(b) 50 W
(c) 62.5 W
(d) 125 W
Ans: (b)
Sol: The average power is
Alternate Method:

Q15: An RLC circuit with relevant data is given below.
The current I_C in the figure above is      (2011)
(a) -j2A
(b)
(c)
(d) +j2A
Ans: (d)
Sol: Using KCL,

Q16: An RLC circuit with relevant data is given below.
The power dissipated in the resistor R is       (2011)
(a) 0.5 W
(b) 1 W
(c)
(d) 2W
Ans: (b)
Sol: Power supplied by the source = V_sI_s cosϕ
Where, ϕ= angle between V_s and I_s = π/4
Inductor and capacitor do not consume power. Therefore, power dissipated in R = Power supplied by the source
 P_R = V_SI_S cosϕ  

Q17: The voltage applied to a circuit is 100√2𝑐𝑜𝑠(100𝜋𝑡) volts and the circuit draws a current of 10√2sin(100πt + π/4) amperes. Taking the voltage as the reference phasor, the phasor representation of the current in amperes is     (2011)
(a)
(b)
(c)
(d)
Ans: (b)
Sol: Voltage represented in phasor form:

Q18: The r.m.s value of the current i(t) in the circuit shown below is      (2011)
(a) 1/2 A
(b) 1/√2 A
(c) 1 A
(d) √2 A
Ans: (b)
Sol: Impedance of the branch containing inductor and capacitor
So, this branch is short circuit and the whole current flow through it

Q19: An energy meter connected to an immersion heater (resistive) operating on an AC 230 V, 50 Hz, AC single phase source reads 2.3 units (kWh) in 1 hour. The heater is removed from the supply and now connected to a 400 V peak to peak square wave source of 150 Hz. The power in kW dissipated by the heater will be      (2006)
(a) 3.478
(b) 1.739
(c) 1.54
(d) 0.87
Ans: (b)
Sol: Assuming resistance of the heater = R
(i) When heater connected to 230 V, 50 Hz source, energy consumed by the heater = 2.3 units or 2.3 kWh in 1 hour
Power consumed by the heater
rms value of the input voltage
(ii) When heater connected to 400 V (peak to peak) square wave source of 150Hz

Q20: The RL circuit of the figure is fed from a constant magnitude, variable frequency sinusoidal voltage source V_IN. At 100 Hz, the R and L elements each have a voltage drop μ_RMS. If the frequency of the source is changed to 50 Hz, then new voltage drop across R is      (2005)
(a)
(b)
(c)
(d)
Ans: (c)
Sol:  At f = 100Hz
∣V_R∣ = ∣V_L∣
as R and L are series connected, current through R and L is same, so

Q21: The RMS value of the voltage u(t)= 3 + 4 cos (3t) is      (2005)
(a) √17 V
(b) 5V
(c) 7V
(d) (3+2√2)V  
Ans: (a)
Sol: R.M.S. value of d.c voltage R.M.S. value of a.c. voltage
Therefore, R.M.S. value of the voltage

Q22: In figure, the admittance values of the elements in Siemens are Y_R = 0.5 + j0, Y_L = 0 - j1.5, Y_C = 0 + j0.3 respectively. The value of I as a phasor when the voltage E across the elements is 10∠0^∘ V is      (2004)
(a) 1.5 + j0.5
(b) 5 - j18
(c) 0.5 + j1.8
(d) 5 - j12
Ans: (d)
Sol: