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Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download

Q16: In the figure shown, all elements used are ideal. For time t<0, S1 remained closed and S2 open. At t = 0, S1 is opened and S2 is closed. If the voltage VC2 across the capacitor C2 at t = 0 is zero, the voltage across the capacitor combination at t = 0+ will be      (2009)Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)

(a) 1 V
(b) 2 V
(c) 1.5 V
(d) 3 V
Ans: 
(a)
Sol: t = 0, S1 is closed,  S2 is open
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)C1 gets charged upto 3 V
Charge stored in  C1
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Voltage across C2 is zero at t = 0−, so no charge is stored in C2.
At t > 0, S1 is open and S2 is closed.
Charge stored (Q0) initially in Cgets redistributed between C1 and  C2.
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Let charge stored in  C1 = Q1
Charge stored in  C2 = Q1
According to conservation of charge  
Q1 + Q2 = Q0 = 3...(i)
Voltage across C1 =Voltage across C2
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Solving equation (i) and (ii) we get
Q= 1C and  Q= 2C  
Voltage across combination = Q1/C1 = 1/1 = 1.

Q17: The current i(t) sketched in the figure flows through a initially uncharged 0.3 nF capacitor.
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)The capacitor charged upto 5 μs, as per the current profile given in the figure, is connected across an inductor of 0.6 mH. Then the value of voltage across the capacitor after 1 μs will approximately be      (2008)
(a) 18.8 V
(b) 23.5 V
(c) -23.5 V
(d) -30.6 V
Ans:
(d)
Sol: Capacitor charge upto 5μs, so total charge stored in capacitor = Q = 13 nC.
Voltage across the capacitor before connecting to inductor
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Voltage across the capacitor at time t
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q18: The current i(t) sketched in the figure flows through a initially uncharged 0.3 nF capacitor.
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)The charge stored in the capacitor at t = 5 μs, will be     (2008)
(a) 8 nC
(b) 10 nC
(c) 13 nC
(d) 16 nC
Ans: 
(c)
Sol: Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Charged stored in the capcitor =Area under i-t curve
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q19: The time constant for the given circuit will be      (2008)
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 1/9 s
(b) 1/4 s
(c) 4s
(d) 9 s
Ans:
(c)
Sol: For finding time constant, we neglect current source as a open circuit.
Therefore, circuit becomes
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Therefore, Time constant = ReqCeq
= 6 x (2/3) = 4 sec

Q20: In the circuit shown in figure. Switch SW1 is initially closed and SW2 is open. The inductor L carries a current of 10 A and the capacitor charged to 10 V with polarities as indicated.  SW2 is closed at t = 0  and SW1 is opened at t = 0. The current through C and the voltage across L at t = 0+ is       (2007)
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 55 A, 4.5V
(b) 5.5A, 45V
(c) 45A, 5.5V
(d) 4.5A, 55V
Ans: 
(d)
Sol: Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)By KCL,
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q21: In the figure, transformer T1 has two secondaries, all three windings having the same number of turns and with polarities as indicated. One secondary is shorted by a 10 Ω resistor R, and the other by a 15 μF capacitor. The switch SW is opened (t = 0) when the capacitor is charged to 5 V with the left plate as positive. At t = 0+ the voltage VP and current IR are      (2007)
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) -25 V, 0.0 A
(b) very large voltage, very large current
(c) 5.0 V, 0.5 A
(d) -5.0 V, -0.5 A
Ans:
(d)
Sol: Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)All the three windings has same number of turns, so magnitude of induced emf's in all the three windings will be same i.e.
∣VP∣ = ∣Vs∣ = ∣VT
Polarity of the winding is decided on the basis of dot-convention.
As capacitor is carged to 5V with left plate as positive
So, T1 is positive w.r.t. T2
VT = VT1 − VT2 = 5V
As T2 has negative polarity. So P1 has negative polarity.  
Therefore , VP = VP1−VP2 = −5V
Similarly, S1 has negative polarity
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q22: An ideal capacitor is charged to a voltage V0 and connected at t = 0 across an ideal inductor L. (The circuit now consists of a capacitor and inductor alone). If we let Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)the voltage across the capacitor at time t > 0 is given by      (2006)
(a) V0
(b) Vcos (ω0t)
(c) 𝑉0sin(𝜔0𝑡)V0 sin (ω0t)
(d) Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)

Ans: (b)
Sol: Voltage across capacitor will discharge through inductor upto voltage across the capcitor becomes zero. During thos period , electrostatic energy stored in capacitor is transferred into magnetic energy which is stored in inductor.
Now, inductor will start charging capacitor, magnetic energy in inductor is converted into electostatic energy in capacitor.
Expression for Vc(t) can be obtained in s-domain. As capacitor is charged initially to voltage V0, then representation of capacitor in s-domain
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)As current through the inductor is zero at t = 0, then
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)The circuit at t > 0
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Voltage across the capacitor
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q23:  In the circuit shown in the figure, the current source I = 1A, the voltage source V = 5 V, R1 = R2 = R3 = 1Ω, L1 = L2 = L3 = 1H,  C= C= 1F
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)The currents (in A) through R3 and through the voltage source V respectively will be      (2006)
(a) 1, 4
(b) 5, 1
(c) 5, 2
(d) 5, 4
Ans:
(d)
Sol: In steady state, inductor behaves as short-circuit and capacitor behaves as open-circuit
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Voltage across, R= V = 5V
Current through, Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Apply KCL,
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Current through voltage source Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)

Q24:  A coil of inductance 10 H and resistance 40 Ω is connected as shown in the figure. After the switch S has been in contact ith point 1 for a very long time, it is moved to point 2 at, t = 0.  
For the value of R = 40ΩΩ, the time taken for 95% of the stored energy to be dissipated is close to       (2005)
(a) 0.10 sec
(b) 0.15 sec
(c) 0.50 sec
(d) 1.0 sec
Ans:
(b)
Sol: The circuit (in s-domain)
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Initial stored energy in inductor
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Remaining energy in inductor
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q25: A coil of inductance 10 H and resistance 40 Ω is connected as shown in the figure. After the switch S has been in contact ith point 1 for a very long time, it is moved to point 2 at, t = 0.
If, at t = 0+, the voltage across the coil is 120 V, the value of resistance R is      (2005)
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 0 Ω
(b) 20 Ω
(c) 40 Ω
(d) 60 Ω
Ans:
(c)
Sol: Before moving the switch at t = 0
The circuit is in steady state and inductor behaves as short-circuit.
The circuit at t = 0
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)After moving the switch at 𝑡 = 0+ 
Current through inductor can not change abruptly.  
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q26: The circuit shown in the figure is in steady state, when the switch is closed at t = 0. Assuming that the inductance is ideal, the current through the inductor at t = 0+ equals     (2005)
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 0 A
(b) 0.5 A
(c) 1 A
(d) 2 A
Ans:
(c)
Sol: Before closing the switch, at t = 0, the circuit is in steady state. So, inductor behaves as short-circuit
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)After closing the switch, at 𝑡=0+ t = 0+
Current through inductor can not change abruptly.
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q27: In the figure given, the initial capacitor voltage is zero. The switch is closed at t = 0. The final steady-state voltage across the capacitor is     (2005)
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 20 V
(b) 10 V
(c) 5 V
(d) 0 V
Ans:
(b)
Sol: At (t → 0+). The capacitor act as short-circuit. At (t → ∞), the capacitor will become open circuit.
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Therefore, voltage across capacitor = Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)

Q28: In figure, the capacitor initially has a charge of 10 Coulomb. The current in the circuit one second after the switch S is closed will be    (2004)
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 14.7 A
(b) 18.5 A
(c) 40.0 A
(d) 50.0 A
Ans:
(a)
Sol: Using KVL,
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q29: In the circuit shown in figure, the switch S is closed at time (t = 0). The voltage across the inductance at t = 0+, is     (2003)
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 2 V
(b) 4 V
(c) -6 V
(d) 8 V
Ans: 
(b)
Sol: Before closing the switch, the circuit was not energized, therefore, current through inductor and voltage across capacitor are zero.
After closing the switch, at t = 0+ inductor acts as open-circuit and capacitor acts as shortc ircuit. Equivalent circuit at t = 0+
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q30: Consider the circuit shown in figure. If the frequency of the source is 50 Hz, then a value of t0 which results in a transient free response is      (2002)
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 0 ms
(b) 1.78 ms
(c) 2.71 ms
(d) 2.91 ms
Ans:
(b)
Sol: For transient free response,
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q31: An 11 V pulse of 10 μs duration is applied to the circuit shown in figure. Assuming that the capacitor is completely discharged prior to applying the pulse, the peak value of the capacitor voltage is     (2002)
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 11 V
(b) 5.5 V
(c) 6.32 V
(d) 0.96 V
Ans:
(c)
Sol: Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)and
∵ pulse of duration 10μs is applies. Hence, capacitor charges till 10μs and then starts discharging, so Vc will be maximum at t = 10μs.

Q32: A unit step voltage is applied at t = 0 to a series RL circuit with zero initial conditions.     (2001)
(a) It is possible for the current to be oscillatory.
(b) The voltage across the resistor at t = 0+ is zero.
(c) The energy stored in the inductor in the steady state is zero.
(d) The resistor current eventually falls to zero.
Ans: 
(b)
Sol: At t = 0+ inductor works as open circuit. Hence, complete source voltage drops across it and consequently, current through the resistor R is zero. Hence, voltage across the resistor at t = 0is zero. And further with time it rises accroding to VR(t) = (1−e−Rt/L)u(t).  
Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)

The document Previous Year Questions- Transients and Steady State Response - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Network Theory (Electric Circuits).
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FAQs on Previous Year Questions- Transients and Steady State Response - 2 - Network Theory (Electric Circuits) - Electrical Engineering (EE)

1. What is the difference between transients and steady state response?
Ans. Transients refer to the temporary behavior of a system before it reaches a stable state, while steady state response is the behavior of a system after it has settled into a stable state.
2. How do transients affect the performance of a system?
Ans. Transients can impact the stability and reliability of a system by causing temporary fluctuations in its response before reaching a steady state. It is important to analyze and control transients to ensure optimal system performance.
3. What factors can influence the duration of transients in a system?
Ans. The duration of transients in a system can be influenced by factors such as the system's initial conditions, input signals, damping ratio, and natural frequency. These factors can affect how quickly a system reaches a steady state response.
4. How can transients be minimized in a system?
Ans. Transients can be minimized by designing the system with appropriate damping and control mechanisms, optimizing the system's parameters, and using feedback control strategies to regulate the system's response.
5. Why is it important to analyze both transients and steady state response in system dynamics?
Ans. Analyzing both transients and steady state response is crucial for understanding the overall behavior and performance of a system. Transients provide insights into the system's initial behavior and response to disturbances, while steady state response helps evaluate the system's long-term stability and performance.
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