Transients and Steady State Response

Question 1
In the circuit shown below, the switch S is closed at t=0. The magnitude of the steady state voltage, in volts, across the 6 \Omega resistor is _________. (round off to two decimal places).

A
5
B
8.25
C
12.55
D
3.35
GATE EE 2022   Electric Circuits
Question 1 Explanation: 
Concept: At steady state, capacitor behaves as open circuit.

Using voltage division,
V=\frac{2}{2+2} \times 10=5V
Question 2
A \text{100 Hz} square wave, switching between \text{0 V} and \text{5 V}, is applied to a \text{CR} high-pass filter circuit as shown. The output voltage waveform across the resistor is \text{6.2 V} peak-to-peak. If the resistance R is \text{820 $\Omega$}, then the value C is ______________\mu F. (Round off to 2 decimal places.)

A
18.5
B
12.46
C
10.06
D
15.48
GATE EE 2021   Electric Circuits
Question 2 Explanation: 






\begin{aligned} v_{0}&=v_{i}-v_{c}\\ \text{For }1^{\text {st }}\text{ half cycle}, \quad v_{0}&=5-v_{c} \\ \text{For }2^{\text {nd }}\text{ half cycle}, \quad v_{0}&=-v_{c}\\ v_{p-p} &=\left(5-V_{c \;\min}\right)-\left(-V_{c}\; \max \right) \\ 6.2 &=5+V_{c \;\max }-V_{c \;\min} \\ \Rightarrow \quad V_{c \max }-V_{c \;\min } &=1.2 \ldots(\alpha) \end{aligned}
For first half cycle i.e. 0 \lt t \lt \frac{T}{2}
\begin{aligned} v_{c}\left(0^{+}\right) &=v_{c}(0)=v_{c}\left(0^{-}\right)=v_{c} \min \\ v_{c}(\infty) &=5 \mathrm{~V} \\ \therefore \qquad\qquad v_{c}(t) &=v_{c}(\infty)+\left[v_{c}\left(0^{+}\right)-v_{c}(\infty)\right] e^{-t / \tau} \\ v_{c}(t) &=5+\left[V_{c m i n}-5\right] e^{-t / 2 \tau}=V_{c m a x} \\ \Rightarrow \qquad\qquad V_{c} \max &=5\left[1-e^{-T / 2 \tau}\right]+V_{c m i n} e^{-T / 2 \tau} \end{aligned}
For \frac{T}{2} \lt t \lt T


\begin{aligned} v_{c}(t) &=v_{c}\left(\frac{T}{2}\right) e^{-t(t-T / 2) \tau} \\ \therefore\qquad v_{c}(t) &=V_{c m a x} e^{-(t-T / 2) \tau} \\ \text{at } t =T,\qquad \qquad v_{c} &=V_{c} \mathrm{~min} \\ \Rightarrow \qquad \qquad V_{C \mathrm{~min}} &=V_{C \max } e^{-T / 2 \tau}\\ \text { As } \quad V_{c \text { max }}-V_{c \text { min }}&=1.2 \qquad \qquad [\text { From }(\alpha)]\\ \therefore \quad V_{\text {cmax }}-V_{c \max } e^{-T / 2 t} &=1.2 \\ V_{c} \max &=\frac{1.2}{1-e^{-T / 2 \tau}} \\ \Rightarrow \qquad\qquad V_{c} \max &=\frac{1.2}{1-e^{-T / 2 \tau}}=5\left[1-e^{-T / 2 \tau}\right]+V_{c \min } e^{-2 \tau} \end{aligned}
From (ii),
\begin{aligned} V_{c \max } &=5\left[1-e^{-T / 2 \tau}\right]+\left(V_{c m a x} e^{-T / 2 \tau}\right) e^{-T / 2 \tau} \\ V_{c \max } &=5\left[1-e^{-T / 2 \tau}\right]+V_{c \max } e^{-T / \tau} \\ \Rightarrow \qquad \qquad V_{c \max }\left[1-e^{-T / \tau}\right] &=5\left[1-e^{-T / 2 \tau}\right] \\ V_{c} \max &=\frac{5\left[1-e^{-T / 2 \tau}\right]}{\left[1+e^{-T / 2 \tau}\right]\left[1-e^{-T / 2 \tau}\right]} \end{aligned}
Using equation (iii)
\begin{aligned} \frac{1.2}{1-e^{-T / 2 \tau}} &=\frac{5}{1+e^{-T / 2 \tau}} \\ \Rightarrow \qquad \qquad 1.2+1.2 e^{-T / 2 \tau} &=5-5 e^{-\pi / 2 t} \\ \Rightarrow \qquad \qquad 6.2 e^{-T / 2 \tau} &=3.8 \\ e^{-T / 2 \tau} &=\frac{3.8}{6.2}=0.6129 \\ \frac{T}{2 \tau} &=0.4895\\ \text { as }\qquad \qquad T&=\frac{1}{f}=\frac{1}{100} \mathrm{sec}\\ \text { and }\qquad \qquad \tau & =R C=820 \mathrm{C} \\ \Rightarrow \qquad \qquad \frac{1}{(100)(2)(820) C} & =0.4895 \\ \therefore \qquad \qquad C & =12.46 \mu \mathrm{F} \end{aligned}
Question 3
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\geq0 is

A
8e^{-10t}
B
10
C
8+2e^{-10t}
D
10\left ( 1-e^{-2t} \right )
GATE EE 2021   Electric Circuits
Question 3 Explanation: 
At t = 0^-


i_{L}\left(0^{-}\right)=\frac{10}{1}=10 \mathrm{~A}
For t > 0


At t = \infty


i(\infty)=\frac{40}{5}=8 \mathrm{~A}
R_{\text{eq}}:


\begin{aligned} R_{\mathrm{eq}} &=5 \Omega \\ \tau &=\frac{L}{R_{\mathrm{eq}}}=\frac{0.5}{5}=0.1 \mathrm{sec} \\ i(t) &=8+[10-8] e^{-t / 0.1} \\ &=8+2 e^{-10 t} \mathrm{~A} \end{aligned}
Question 4
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.)
A
100
B
200
C
50
D
400
GATE EE 2017-SET-2   Electric Circuits
Question 4 Explanation: 
Consider the following circuit diagram,

After minimizing circuit elements we can have the following circuit,

Here, \tau =RC=5 sec.
Now current,
i(t)=\frac{V}{R}e^{\frac{t}{\tau }}
\;\;=\frac{10}{5}e^{-t/5}=2e^{-0.2t}
Energy supplied by the source,
E=\int_{0}^{\infty }10 \times 2e^{-0.2t}dt
\;\;=100J
Question 5
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 \geq 0, is
A
2.5e^{-4t}
B
5e^{-4t}
C
2.5e^{-0.25t}
D
5e^{-0.25t}
GATE EE 2017-SET-1   Electric Circuits
Question 5 Explanation: 
From the given circuit, consider the following circuit diagram,

After rearrangement

For t \geq 0
I_0=i(0^-)=2.5A
We can write,
i(t)=I_0e^{-\frac{Rt}{L}}
i(t)=2.5e^{-4t}A
Question 6
In the circuit shown below, the initial capacitor voltage is 4 V. Switch S_1 is closed at t=0. The charge (in \muC) lost by the capacitor from t=25\mu s to t=100\mu s is ____________.
A
5
B
6
C
7
D
8
GATE EE 2016-SET-2   Electric Circuits
Question 6 Explanation: 


i(t)=\frac{4}{5}e^{-t/\tau }
\tau =RC
\;\;=20 \times 10^{-6}sec
Change lost by capacitor from t=25\mu s to 100\mu s is
\int_{25 \mu s}^{100\mu s}i(t)dt=6.99 \times 10^{-6}C
Question 7
In the circuit shown, switch S_2 has been closed for a long time. At time t=0 switch S_1 is closed. At t = 0^{+}, the rate of change of current through the inductor, in amperes per second, is _____.
A
1
B
2
C
3
D
4
GATE EE 2016-SET-1   Electric Circuits
Question 7 Explanation: 


KCL at node A,
\frac{V_A-3}{1}+\frac{3}{2}+\frac{V_A-3}{2}=0
2(V_A-3)+3+(V_A-3)=0
3V_A=6
V_A=2
V_A=L\frac{di(0^+)}{dt}=2
\frac{di(0^+)}{dt}=\frac{2}{L}=\frac{2}{1}=2A/sec
Question 8
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 \frac{d^{2}i}{dt^{2}}\; at \; t=0^{+} is
A
\frac{V}{L}
B
\frac{-V}{R}
C
0
D
\frac{-RV}{L^{2}}
GATE EE 2015-SET-2   Electric Circuits
Question 8 Explanation: 


Initially (t=0^-) the inductor would be uncharged.
So, I(0^+)=0
The KVL in th loop will be
V=RI+L\frac{dI}{dt}
At t=0^+
V=RI(0^+)+L\frac{dI}{dt}(0^+)
Since, I(0^+)=0
So, \frac{dI}{dt}(0^+)=\frac{V}{L}
Now, lets differentiate the above equation
So, \frac{dV}{dt}=R\frac{dI}{dt}+L\frac{d^2I}{dt^2}
\;\;0=R\frac{dI}{dt}+L\frac{d^2I}{dt^2}
At t=0^+
0=R\frac{dI}{dt}(0^+)+L\frac{d^2I}{dt^2}(0^+)
So, \frac{d^2I}{dt^2}(0^+)=-\frac{R}{L^2}\cdot V
Question 9
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} | \gt |V_{o1}|, the voltage v_{c}(t) across the capacitor is
A
v_{c}(t)=-V_{o2}(1-e^{-t/2RC})-V_{o1}
B
v_{c}(t)=V_{o2}(1-e^{-t/2RC})+V_{o1}
C
v_{c}(t)=-(V_{o2}+V_{o1})(1-e^{-t/2RC})-V_{o1}
D
v_{c}(t)=(V_{o2}-V_{o1})(1-e^{-t/2RC})+V_{o1}
GATE EE 2014-SET-2   Electric Circuits
Question 10
A combination of 1 \muF capacitor with an initial voltage v_c(0)=-2V in series with a 100 \Omega 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 v_s across the current source over the next few seconds ?

A
A
B
B
C
C
D
D
GATE EE 2014-SET-1   Electric Circuits
Question 10 Explanation: 
Given C=1\mu F, V_c(0)=-2V, R=100\Omega , I=20mA. Circuit fot the given condition at time t \gt 0 is shown below:

Applying KVL, we have,V_s(s)=\left ( \frac{-2}{s} \right )+\frac{I}{s} \left ( R+\frac{I}{Cs} \right )
\;\;=\frac{1}{s}\left [ -2+IR+\frac{I}{Cs} \right ]
\;\;=\frac{1}{s}\left [ (IR-2) +\frac{I}{Cs}\right ]
Putting values of R, C and I, we get,
V_s(s)=\frac{1}{s}\left [ (20 \times 10^{-3} \times 200-2)+\left ( \frac{20 \times 10^{-3}}{10^{-6}} \right ) \times \frac{1}{s} \right ]
\;\;=\frac{1}{s}\left [ (2-2)+20 \times 10^3 \times \frac{1}{s} \right ]
\;\;=\frac{20 \times 10^3}{s^2}
\therefore \;\; V_s(s)=\frac{20 \times 10^3}{s^2}
V_s(t)=20000 t u(t)
\therefore \;\; V_s(s)=(20000)tu(t)
Which is equation of a straight line passing through origin. Hence option (C) is correct.
There are 10 questions to complete.