Previous Year Questions- Transients and Steady State Response - 1 | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download

Q1: The circuit shown in the figure with the switch S open, is in steady state. After the switch S is closed, the time constant of the circuit in seconds is     (2024)
(a) 1.25
(b) 0
(c) 1
(d) 1.5
Ans: (a)
Sol: 
Q2: The circuit shown in the figure is initially in the steady state with the switch K in open condition and  in closed condition. The switch K is closed and  is opened simultaneously at the instant t = t₁, where t₁ > 0. The minimum value of t₁ in milliseconds, such that there is no transient in the voltage across the 100μF capacitor, is ___ (Round off to 2 decimal places).      (2023)
(a) 0.87
(b) 1.57
(c) 1.88
(d) 2.26
Ans: (b)
Sol: Switch K is open and  is closed.
Redraw the circuit :
From circuit, using current division,
Case (ii) :
Switch K is closed and  is open.
Current source and 10Ω resistor becomes short circuited.
Redraw the circuit :
From circuit,
We have,
For transient free voltage,

Q3: In the circuit shown below, the switch S is closed at t = 0. The magnitude of the steady state voltage, in volts, across the 6Ω resistor is _________. (round off to two decimal places).       (2022)
(a) 5
(b) 8.25
(c) 12.55
(d) 3.35
Ans: (a)
Sol: Concept: At steady state, capacitor behaves as open circuit. 
Using voltage division,

Q4: A 100 Hz square wave, switching between 0 V and 5 V, is applied to a CR high-pass filter circuit as shown. The output voltage waveform across the resistor is 6.2 V peak-to-peak. If the resistance R is 820 Ω, then the value C is ______________ μF. (Round off to 2 decimal places.)      (2021)
(a) 18.5
(b) 12.46
(c) 10.06
(d) 15.48
Ans: (b)
Sol: ∴⇒ ∴ From (ii),
⇒ Using equation (iii)
⇒
⇒ as and ⇒
∴ C = 12.46 μF

Q5: In the circuit, switch 'S' is in the closed position for a very long time. If the switch is opened at time t = 0, then in i_L(t) amperes, for t ≥ 0 is      (2021)
(a) 8e^-10t
(b) 10
(c) 8 + 2e^-10t
(d) 10(1 - e^-2t)
Ans: (c)
Sol: At t = ∞

Q6: The initial charge in the 1 F capacitor present in the circuit shown is zero. The energy in joules transferred from the DC source until steady state condition is reached equals ______. (Give the answer up to one decimal place.)      (SET-2 (2017))
(a) 100
(b) 200
(c) 50
(d) 400
Ans: (a)
Sol: Consider the following circuit diagram,
After minimizing circuit elements we can have the following circuit,
Energy supplied by the source,

Q7: The switch in the figure below was closed for a long time. It is opened at t = 0. The current in the inductor of 2 H for t ≥ 0, is     (SET-1 (2017))
(a) 2.5 e^-4t
(b) 5e^-4t
(c) 2.5e^-0.25t
(d) 5e^-0.25t
Ans: (a)
Sol: From the given circuit, consider the following circuit diagram,
After rearrangement

Q8: In the circuit shown below, the initial capacitor voltage is 4 V. Switch S₁ is closed at t = 0. The charge (in μC) lost by the capacitor from t = 25 μs to t = 100 μs is ____________.        (SET-2(2016))
(a) 5
(b) 6
(c) 7
(d) 8
Ans: (c)
Sol: Change lost by capacitor from t = 25 μs to 100 μs is

Q9: In the circuit shown, switch S₂ has been closed for a long time. At time t = 0 switch  S₁ is closed. At  t = 0⁺, the rate of change of current through the inductor, in amperes per second, is _____.      (SET-1 (2016))
(a) 1
(b) 2
(c) 3
(d) 4
Ans: (b)
Sol: KCL at node A,

Q10: A series RL circuit is excited at t = 0 by closing a switch as shown in the figure. Assuming zero initial conditions, the value of  is       (SET-2 (2015))
(a) V/L
(b) (-V)/R
(c) 0
(d) (-RV)/(L²)
Ans: (d)
Sol: Initially (t = 0⁻) the inductor would be uncharged.
So, I(0⁺) = 0  
The KVL in th loop will be
Now, lets differentiate the above equation

Q11: The switch SW shown in the circuit is kept at position '1' for a long duration. At t = 0+, the switch is moved to position '2'. Assuming ∣V_o2∣ > ∣V_o1∣, the voltage v_c(t) across the capacitor is      (SET-2 (2014))
(a)
(b)
(c)
(d)
Ans: (d)

Q12: A combination of 1 μF capacitor with an initial voltage v_c(0) = −2V in series with a 100 Ω resistor is connected to a 20 mA ideal dc current source by operating both switches at t = 0s as shown. Which of the following graphs shown in the options approximates the voltage  vs across the current source over the next few seconds ?      (SET-1  (2014))
(a) (b) (c) (d) Ans: (c)
Sol: Given C =1μF, V_c(0) = −2V,  R= 100Ω, I = 20mA. Circuit fot the given condition at time t > 0 is shown below:
Applying KVL, we have, Putting values of R, C and I, we get,
Which is equation of a straight line passing through origin. Hence option (C) is correct.

Q13: In the following figure, C₁ and C₂ are ideal capacitors. C₁ has been charged to 12 V before the ideal switch S is closed at t = 0. The current i(t) for all t is     (2012)
(a) zero
(b) a step function
(c) an exponentially decaying function
(d) an impulse function
Ans: (d)
Sol: Circuit is s-domain
By applying KVL,
∴ Current i(t) is an implulse function.

Q14: The L-C circuit shown in the figure has an inductance L = 1mH and a capacitance C =  μF.
The initial current through the inductor is zero, while the initial capacitor voltage is 100 V. The switch is closed at t = 0. The current i through the circuit      (2010)
(a) $5𝑐𝑜𝑠(5\times 103𝑡)𝐴$5cos(5 × 10³t)A
(b) 10sin(10⁴t)A  
(c) $10𝑐𝑜𝑠(5\times 103𝑡)𝐴$10cos(5 × 10³t)A
(d) 10sin(10⁴t)A
Ans: (d)
Sol: Initial current through the inductor is zeroand capacitor voltage is charged upto voltage V_c(0⁻) = 100V
As current through inductor and voltage across capacitor can not change abruptly
So, after closing the switch,
The circuit id s-domain
taking inverse laplace transform

Q15: The switch in the circuit has been closed for a long time. It is opened at t = 0. At t = 0⁺ , the current through the 1 μF capacitor is       (2010)
(a) 0A
(b) 1A
(c) 1.25 A
(d) 5A
Ans: (b)
Sol: As the switch has been closed for a long time, the circuit is in steady state. At steadystate, capacitor is open circuit,
Using KVL,
As the voltage across capacitorcan not change abruptly,
Current through capacitor at t = 0⁺