Electric Circuits - PracticePaper

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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

\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

\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 ____________