Question 1
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 1 Explanation:
$Q=\frac{\omega _o}{BW}=\frac{f_o}{BW}=\frac{150 \times 10^3}{600}=250$
 Question 2
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 2 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 3
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 3 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 4
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 4 Explanation:
Given that,
$v(t)=5-10 \cos (\omega t+60^{\circ})$
$i(t)=5+C \cos (\omega t-0^{\circ})$
$P_{req}=0$
$0=5 \times 5 +\frac{1}{2}[(-10)(X)\cos(60^{\circ})]$
$-25=\frac{1}{2}[(-10)(X)\cos(60^{\circ})]$
$X=10$
 Question 5
In the figure, the voltages are
$v_{1}(t)=100cos(\omega t)$,
$v_{2} (t) = 100cos(\omega t + \pi /18)$ and
$v_{3}(t) = 100cos(\omega t + \pi /36)$.
The circuit is in sinusoidal steady state, and $R \lt \lt \omega L$. $P_{1}$,$P_{2}$ and $P_{3}$ are the average power outputs. Which one of the following statements is true? A $P_{1}=P_{2}=P_{3}=0$ B $P_{1} \lt 0,P_{2} \gt 0,P_{3} \gt 0$ C $P_{1} \lt 0,P_{2} \gt 0,P_{3} \lt 0$ D $P_{1} \gt 0,P_{2} \lt 0,P_{3} \gt 0$
GATE EE 2018   Electric Circuits
Question 5 Explanation: $V_2:\frac{\pi}{18}=\frac{180^{\circ}}{18}=10^{\circ}$
$V_3:\frac{\pi}{36}=\frac{180^{\circ}}{36}=5^{\circ}$
$V_2$ leads $V_1$ and $V_3$,
So, $V_2$is a source, $V_1$ and $V_3$ are absorbing.
Hence, $P_2 \gt 0$ and $P_1,P_3 \lt 0$
 Question 6
The voltage (V) and current (A) across a load are as follows.
$v(t) = 100 sin(\omega t)$,
$i(t) = 10 sin(\omega t - 60^{\circ}) + 2 sin(3\omega t) + 5 sin(5\omega t)$.
The average power consumed by the load, in W, is___________.
 A 100 B 150 C 200 D 250
GATE EE 2016-SET-2   Electric Circuits
Question 6 Explanation:
The average power consumed by the load =
$P=V_1I_1 \cos \phi$
$\;\;=\frac{100}{\sqrt{2}}\frac{10}{\sqrt{2}} \cos 60^{\circ}=250W$
 Question 7
A resistance and a coil are connected in series and supplied from a single phase, 100 V, 50 Hz ac source as shown in the figure below. The rms values of plausible voltages across the resistance ($V_{R}$) and coil ($V_{C}$) respectively, in volts, are A 65, 35 B 50, 50 C 60, 90 D 60, 80
GATE EE 2016-SET-2   Electric Circuits
Question 7 Explanation:
As per GATE Official answer key MTA (Marks to All)
 Question 8
In the circuit shown below, the supply voltage is $10sin(1000t)$ volts. The peak value of the steady state current through the 1$\Omega$ resistor, in amperes, is ______. A 0.5 B 1 C 2 D 3
GATE EE 2016-SET-1   Electric Circuits
Question 8 Explanation:
If we observe the parallel LC combination, we get that at $\omega$=1000 rad/sec the parallel LC is at resonance, thus it is open circuited.
The circuit given in question can be redrawn as So, $I=\frac{10 \sin 1000t}{10}=\sin 100t$
So, peak value is 1 Amp.
 Question 9
A symmetrical square wave of 50% duty cycle has amplitude of $\pm$15 V and time period of 0.4$\pi$ ms. This square wave is applied across a series RLC circuit with R=5 $\Omega$, L=10 mH, and C=4 $\mu$F. The amplitude of the 5000 rad/s component of the capacitor voltage (in Volt) is ______. A 100 B 140 C 192 D 212
GATE EE 2015-SET-2   Electric Circuits
 Question 10
The total power dissipated in the circuit, shown in the figure, is 1 kW. The voltmeter, across the load, reads 200 V. The value of $X_L$ is _____.
 A 8.5 B 17.3 C 22.4 D 28.6
GATE EE 2014-SET-2   Electric Circuits
Question 10 Explanation: Given, total power dissipated in the circuit = 1kW =1000 Watt
$\therefore \;\; 2^2 \times 1 +10^2 \times R=1000$
$R=\frac{998}{100}=9.98\Omega$
Also, voltage drop across R,
$V_R=IR$
$=10 \times 9.98=99.8 volt$
$V=200 volt =\sqrt{V_R^2+V_{X_L}^2}$
$\therefore$ voltage drop across inductor,
$V_{X_L}=\sqrt{V^2-V_R^2}$
$\;\;=\sqrt{(200)^2-(99.8)^2}$
$\;\;=173.32 volt$
Now, $V_{X_L}=IX_L$
$X_L=\frac{V_{X_L}}{I}$
$X_L=\frac{173.32}{10}=17.332\Omega$