# Transients and Steady State Response

 Question 1
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 1 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 2
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 2 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 3
In the circuit shown below, the initial capacitor voltage is 4 V. Switch $S_1$ is closed at t=0. The charge (in $\mu$C) 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 3 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 4
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 4 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 5
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 5 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 6
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 7
A combination of 1 $\mu$F 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 7 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.
 Question 8
In the following figure, $C_1 \; and \; C_2$ are ideal capacitors. $C_1$ has been charged to 12 V before the ideal switch S is closed at t=0. The current i(t) for all t is A zero B a step function C an exponentially decaying function D an impulse function
GATE EE 2012   Electric Circuits
Question 8 Explanation: Circuit is s-domain By applying KVL,
$\frac{12}{s}+\frac{I(s)}{s}\left ( \frac{1}{C_1} +\frac{1}{C_2}\right )=0$
$I(s)=-\frac{12C_1C_2}{C_1+C_2}=k(constant)$
$\Rightarrow \;\; i(t)=k\delta (t)$
$\therefore$ Current $i(t)$ is an implulse function.
 Question 9
The L-C circuit shown in the figure has an inductance L=1mH and a capacitance C=$\mu$F. 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
 A $5 cos(5\times 10^{3}t)A$ B $10 sin(10^{4}t)A$ C $10 cos(5\times 10^{3}t)A$ D $10 sin(10^{4}t)A$
GATE EE 2010   Electric Circuits
Question 9 Explanation:
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,
$i_L(0^+)=i_L(0^-)=0$
and $V_c(0^+)=V_c(0^-)=100V$
The circuit id s-domain $I(s)=\frac{100/s}{\left ( sL+\frac{1}{sC} \right )}$
$\;\;=\frac{100}{L}\left ( \frac{1}{s^2+\frac{1}{LC}} \right )$
$\;\;=100\sqrt{\frac{C}{L}}\left ( \frac{1/\sqrt{LC}}{s^2+(1/\sqrt{LC})^2} \right )$
taking inverse laplace transform
$i(t)=L^{-1}[I(s)]$
$\;\;=100\sqrt{\frac{C}{L}} \sin \frac{1}{\sqrt{LC}}t$
$\;\;=100 \times \sqrt{\frac{10 \times 10^3}{1 \times 10^{-3}}}\times$ $\sin\left ( \frac{1}{\sqrt{1 \times 10^{-3} \times 10 \times 10^{-6}}} \right )$
$i(t)=10\sin (10^4 t)A$
 Question 10
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 $\mu$F capacitor is A 0A B 1A C 1.25A D 5A
GATE EE 2010   Electric Circuits
Question 10 Explanation:
As the switch has been closed for a long time, the circuit is in steady state. At steadystate, capacitor is open circuit, Using KVL,
$5-I-4I=0 I=1A$
$V_c(0^-)=4 \times 1=4V$
As the voltage across capacitorcan not change abruptly,
So, $V_c(0^+)=V_c(0^-)=4V$
Circuit at $t=0^+$ Current through capacitor at $t=0^+$
$I_c(0^+)=\frac{4}{4}=1A$
There are 10 questions to complete. 