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Test: Networks- 2

20 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Test: Networks- 2

Description

Attempt Test: Networks- 2 | 20 questions in 60 minutes | Mock test for GATE preparation | Free important questions MCQ to study GATE ECE (Electronics) 2023 Mock Test Series for GATE Exam | Download free PDF with solutions

QUESTION: 1

The Y parameter for total network is

Solution:

There are two networks connected in parallel. Thus there Y-parameter will be added.

*Answer can only contain numeric values

QUESTION: 2

in the circuit shown below is

Solution:

QUESTION: 3

The value of z_infor the circuit shown below is

A.
12Ω
B.
14Ω
C.
9.6Ω
D.
None of these

Solution:

Net impedance on primary side of transformer is

*Answer can only contain numeric values

QUESTION: 4

Equivalent resistance across AB is _____Ω

Solution:

Convert Δ to Y and then use series parallel combination

QUESTION: 5

In the circuit given below v,n = u(—t) + u(t) V

Then vc (t) for t > 0

Solution:

QUESTION: 6

A voltage V = 12√2 cos 5000t V is applied to the circuit shown below.

The current i(t) is

A.
0.49 cos (5000t + 39°)
B.
0.49,√2 cos (50001 + 39°)
C.
10.49 cos(5000t + 51°)
D.
0.49√2.cos(5000t + 51°)

Solution:

QUESTION: 7

In the circuit shown below, the voltage source has been connected for a long time enough for steady state condition to be reached. At time t = 0, switch S is opened. Then open circuit voltage across AB is

A.
V.e ^{- (R / L)}
B.
V + V.e^-R/L
C.
V
D.
o

Solution:

QUESTION: 8

Calculate V_x & I_x in the circuit shown below.

A.
V_x= -4V ; I_x = -2A
B.
V_x = 12V ; I_x = -2A
C.
V_x = —4V ; I_x = 2A
D.
V_x = 12V ; I_x = 2A

Solution:

Ix = -2A
Apply KCL
-4I_x + 2I_x + V_x = 0
V_x = 4I_x -2I_x = 2I_x = -4V

QUESTION: 9

Determine the value of I' in the circuit shown below.

Solution:

QUESTION: 10

Determine RMS values of voltage across capacitor (0.25F), current throughcapacitor, if voltage across capacitor is 4 + 4 cos (9t) - 5cos (4t)

A.
6V, 7.2A
B.
4.5V, 7.2A
C.
5.33V, 7.2A
D.
4.5V, 7.8A

Solution:

Vc = 4 + 4 cos(9t) - 5cos(4t)

= 1/4 {-4(9) sin(9t) + 5(4) sin (4t)}
⇒ -9 sin(9t) + 5 sin (4t)

RMS value of V_L is =

*Answer can only contain numeric values

QUESTION: 11

The value of V, in the circuit shown below

Solution:

V₁— V₂ =12V................. (1)

At node 0:

--------(2)

At supernode 1 - 2
(3)

Solving 1, 2 & 3; V_o = 4.5V; V₁ = 5V; V₂ = —7V

QUESTION: 12

For the circuit shown in figure, find V(t) when V_s = 2sin(500t) V

A.
1.25 cos (500t 141⁰)V
B.
1.25 cos (500t - 39⁰)V
C.
0.8cos(500t-141⁰)V
D.
0.8cos(625t - 39⁰)V

Solution:

*Answer can only contain numeric values

QUESTION: 13

In the circuit shown below, the switch is closed at time t=0. The steady state value of the voltage V_c is_____ V

Solution:

At steady state inductor short circuit, capacitor is open circuit. Thus voltage V_c is equal to voltage across 1 ohm resistance.

QUESTION: 14

Which of the following statements are true for the circuit shown below?

1. It is first order with steady state value of
2. It is second order with steady state value of V = iV, I = 1A
3. The network function has one pole I(s)
4. The network function has two poles I(s)

Choose the correct one

A.
2 and 3
B.
2 and 4
C.
2 alone
D.
1 alone

Solution:

Under steady state,

Where V(t) is the voltage across the capacitor

This has a pole at S = -2.

QUESTION: 15

In the circuit shown below, it is found that the input ac voltage and current are in phase. The value of coupling coefficient K and the dot polarity of the coil PQ are

A.
K = 1/4 and dot at P
B.
K = 1/2 and dot at P
C.
K = 1/4 and dot at Q
D.
K = 1/2 and dot at Q

Solution:

If input ac voltage and current are in phase then net impedance is equal to zero. Net inductive impedance = Net capacitive impedance

, Thus k = 21 and dot must be at Q so that net inductance =

QUESTION: 16

For the given circuit, switch closes at t = 0

A.
B.
zero
C.
D.
∞

Solution:

QUESTION: 17

The voltage applied to a particular circuit comprising two components connected in series is given by:

V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by

= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)A

Q. Average power supplied is

A.
OW
B.
15.67W
C.
30W
D.
60W

Solution:

Average Power P is given by

P = 0 + 5.523+ 6.099+ 4.044 =15.67W

QUESTION: 18

The voltage applied to a particular circuit comprising two components connected in series is given by:

V = (30 + 40 sin103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by

= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)A

Q. Which of the following element is not a part of the circuit?

A.
R
B.
L
C.
C
D.
Both L and C

Solution:

The expression for the voltage contains a dc component of 30V. However there is no corresponding term in the expression for current. This indicates that one of the components is a capacitor (since in a dc circuit a capacitor offers infinite impedance to a direct current). Since power is delivered to the circuit the other component is a resistor.

QUESTION: 19

Consider the circuit shown in figure