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 nand 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.