Question 1
The value of parameters of the circuit shown in the figure are

$\mathrm{R}_{1}=2 \Omega, \mathrm{R}_{2}=2 \Omega, \mathrm{R}_{3}=3 \Omega, \mathrm{L}=10 \mathrm{mH}, \mathrm{C}=100 \mu \mathrm{F}$

For time $t \lt 0$, 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 $\left(\mathrm{V}_{\mathrm{L}}\right)$ at $t=0^{+}$in Volts is _____ (Round off to 1 decimal place). A 6.5 B 8 C 12.5 D 16
GATE EE 2023   Electric Circuits
Question 1 Explanation:
Case (i) $\mathbf{t} \lt 1$

At steady state, capacitor behaves as open circuit and inductor behaves as short circuit.
Redraw the circuit: Using current division,
\begin{aligned} \mathrm{i}_{\mathrm{L}}\left(0^{-}\right) & =\frac{3}{3+2} \times 10=6 \mathrm{~A} \\ \therefore \quad \mathrm{V}_{\mathrm{C}}\left(0^{-}\right) & =5 \times 2=12 \mathrm{~V} \end{aligned}

Case (ii) $\mathbf{t} \gt 0$ :
Switch is opened.
$\because \quad \mathrm{V}_{\mathrm{C}}\left(0^{-}\right)=\mathrm{V}_{\mathrm{C}}\left(0^{+}\right)=12 \mathrm{~V}$
and $\quad \mathrm{i}_{\mathrm{L}}\left(0^{-}\right)=\mathrm{i}_{\mathrm{L}}\left(0^{+}\right)=6 \mathrm{~A}$

Redraw the circuit $\mathrm{t}=0^{+}$: Apply KVL in loop,
$\begin{array}{rlrl} 12+8-12-\mathrm{V}_{\mathrm{L}} & =0 \\ \Rightarrow & \mathrm{V}_{\mathrm{L}} & =8 \mathrm{~V} \end{array}$
 Question 2
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)
 A 250 B 100 C 150 D 450
GATE EE 2022   Electric Circuits
Question 2 Explanation:
$Q=\frac{\omega _o}{BW}=\frac{f_o}{BW}=\frac{150 \times 10^3}{600}=250$

 Question 3
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)
 A 12 B 24 C 48 D 96
GATE EE 2022   Electric Circuits
Question 3 Explanation:
We have, overall Q-factor of given circuit is,
$Q=\frac{Q_LQ_C}{Q_L+Q_C}=\frac{60 \times 240}{60+240}=48$
 Question 4
A 0.1$\mu F$ capacitor charged to 100 V is discharged through a 1 $k\Omega$ resistor. The time in ms (round off to two decimal places) required for the voltage across the capacitor to drop to 1V is ______
 A 0.25 B 0.65 C 0.45 D 0.85
GATE EE 2019   Electric Circuits
Question 4 Explanation: \begin{aligned} v_c(t)&=V_0e^{-t/\tau } \\ V_0&=100V \\ \tau &=RC=(10^3)(10^{-7}) \\ &=10^{-4}sec \\ \therefore \;v_c(t)&=100e^{-10^4 t }V \end{aligned}
Let the time required by the voltage across the capacitor to drop to 1 V is $t_1,$
\begin{aligned} \therefore \; v_c(t_1)&=100e^{-10^4t_1} \\ \text{But, } v_c(t_1)&=0 \\ \text{So, }0&=100e^{-10^4t_1} \\ t_1&=0.46msec \end{aligned}
 Question 5
The voltage across the circuit in the figure, and the current through it, are given by the following expressions:
$v(t) = 5 - 10 cos(\omega t + 60^{\circ}) V$
$i(t) = 5 + X cos(\omega t) A$
where $\omega =100 \pi$ 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). A 5 B 10 C 15 D 20
GATE EE 2018   Electric Circuits
Question 5 Explanation:
Given that,
$v(t)=5-10 \cos (\omega t+60^{\circ})$
$i(t)=5+C \cos (\omega t-0^{\circ})$
$P_{req}=0$
$0=5 \times 5 +\frac{1}{2}[(-10)(X)\cos(60^{\circ})]$
$-25=\frac{1}{2}[(-10)(X)\cos(60^{\circ})]$
$X=10$

There are 5 questions to complete.