# Electric Circuits

 Question 1
In the circuit shown below, the magnitude of the voltage $V_1$ in volts, across the $8k\Omega$ resistor is ______________. (round off to nearest integer) A 100 B 120 C 150 D 175
GATE EE 2022      Basics
Question 1 Explanation:
Apply kVL :
$75-(2k)I-0.5V_1=0$ ...(1)
From circuit:
$V_1=(8k)I$
$I=\frac{V_1}{8k}$ ...(2)
From equation (1) & (2)
$75-(2k) \times \frac{V}{8k}-0.5V_1=0$
$\Rightarrow V_1=100V$
 Question 2
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      Transients and Steady State Response
Question 2 Explanation:
Concept: At steady state, capacitor behaves as open circuit. Using voltage division,
$V=\frac{2}{2+2} \times 10=5V$
 Question 3
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      Steady state AC Analysis
Question 3 Explanation:
$Q=\frac{\omega _o}{BW}=\frac{f_o}{BW}=\frac{150 \times 10^3}{600}=250$
 Question 4
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      Steady state AC Analysis
Question 4 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 5
In the circuit shown below, a three-phase star-connected unbalanced load is connected to a balanced three-phase supply of $100\sqrt{3}$ with phase sequence $ABC$. The star connected load has $Z_A=10\Omega$ and $Z_B=20\angle 60^{\circ}$. The value of $Z_C$ in $\Omega$, for which the voltage difference across the nodes $n$and $n'$ is zero, is A $20\angle -30^{\circ}$ B $20\angle 30^{\circ}$ C $20\angle -60^{\circ}$ D $20\angle 60^{\circ}$
GATE EE 2022      Three-Phase Circuits
Question 5 Explanation:
Given: n & n' are at same potential, therefore,
$I_{nn'}=0$
$I_{A}+I_{B}+I_{C}=0$
$\frac{E_A}{Z_A}+\frac{E_B}{Z_B}+\frac{E_C}{Z_C}=0$
where ,
$E_A=100\angle 0^{\circ}V \text{ and }Z_A=10\Omega$
$E_B=100\angle -120^{\circ}V \text{ and }Z_B=20\angle 60^{\circ}$
$E_C=100\angle 120^{\circ}V \text{ and }Z_C=?$
$\therefore \frac{100\angle 0^{\circ}}{10}+\frac{100\angle -120^{\circ}}{20\angle 60^{\circ} }+\frac{100\angle 120^{\circ}}{Z_C}=0$
$\Rightarrow Z_C=20\angle -60^{\circ} \Omega$
 Question 6
The transfer function of a real system, $H(s)$, is given as:
$H(s)=\frac{As+B}{s^2+Cs+D}$
where A, B, C and D are positive constants. This system cannot operate as
 A low pass filter. B high pass filter C band pass filter. D an integrator.
GATE EE 2022      Magnetically Coupled Circuits, Network Topology and Filters
Question 6 Explanation:
Put $s=0, H(0)=\frac{A \times 0+B}{0+C \times 0+D}=\frac{B}{D}$
So, the system pass low frequency component. Put $s=\infty , H(\infty )=0$
For high pass filter, high frequency component should be non zero. Hence this system cannot be operated as high pass filter.
 Question 7
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      Transients and Steady State Response
Question 7 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 8
An air-core radio-frequency transformer as shown has a primary winding and a secondary winding. The mutual inductance M between the windings of the transformer is ____________ $\mu H$.(Round off to 2 decimal places.) A 12.14 B 68.26 C 51.1 D 78.4
GATE EE 2021      Magnetically Coupled Circuits, Network Topology and Filters
Question 8 Explanation: \begin{aligned} I_{1}&=\frac{5}{22}(p-p) \\ V_{0}&=j \omega M I_{1}=7.3=\left(2 \pi \times 10 \times 10^{3}\right) \times M \times\left(\frac{5}{22}\right) \\ M&=51.10 \mu \mathrm{H} \end{aligned}
 Question 9
In the given circuit, for maximum power to be delivered to $R_{L}$, its value should be ______________ $\Omega$.
(Round off to 2 decimal places.) A 2.14 B 3.32 C 1.42 D 4.12
GATE EE 2021      Network Theorems
Question 9 Explanation: $=\frac{-j}{\omega C}=\frac{-j}{1000 \times 0.5 \times 10^{-3}}=-j 2 \Omega$ $Z_{\text {in }}=2 \| j 2=\frac{j 4}{2+j 2}=\frac{j 2}{1+j 1}$
For maximum power transfer,
$R_{L}=\left|Z_{\mathrm{TH}}\right|=\frac{2}{\sqrt{2}}=\sqrt{2}=1.414 \Omega$
 Question 10
A three-phase balanced voltage is applied to the load shown. The phase sequence is $\text{RYB}$. The ratio $\frac{\left | I_{B} \right |}{|I_{R}|}$ is ____________. A 1 B 2 C 3 D 4
GATE EE 2021      Three-Phase Circuits
Question 10 Explanation: \begin{aligned} I_{R} &=\frac{V_{R B}}{-j 10}=\frac{V_{L} \angle-60^{\circ}}{-j 10}=\frac{V_{L}}{10} \angle 30^{\circ} \\ I_{Y} &=\frac{V_{Y B}}{j 10}=\frac{V_{L} \angle-120^{\circ}}{-j 10}=\frac{V_{L}}{10} \angle 150^{\circ} \\ \text { and }\qquad \qquad I_{B}&=-\left(I_{R}+I_{Y}\right)=\frac{V_{L}}{10} \angle-90^{\circ} \\ \therefore \qquad \qquad \left|\frac{I_{B}}{I_{R}}\right|&=1 \end{aligned}

There are 10 questions to complete.