Basics

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   Electric Circuits
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 given circuit, the value of capacitor C that makes current I=0 is _________ \mu F.

A
15
B
50
C
5
D
20
GATE EE 2021   Electric Circuits
Question 2 Explanation: 




\begin{aligned} Z_{L} &=(j 5) \|\left(j 5-j X_{c}\right) \\ \frac{(j 5)\left(j 5-j X_{c}\right)}{j 5+j 5-j X_{c}} &=\infty \\ j 5+j-j X_{c} &=0 \\ \Rightarrow\qquad \qquad j X_{c} &=j 10 \\ \Rightarrow\qquad \qquad X_{c} &=10 \Omega \\ X_{c} &=\frac{1}{\omega C}=10 \\ \Rightarrow\qquad \qquad C&=\frac{1}{10 \times 5 \times 10^{3}} \\ C &=\frac{1}{5 \times 10^{4}} \times \frac{10^{2}}{10^{2}}=\frac{10^{2}}{5 \times 10^{6}}=20 \mu \mathrm{F} \end{aligned}
Question 3
The current I flowing in the circuit shown below in amperes (round off to one decimal place) is ____
A
0.8
B
1.1
C
1.4
D
1.9
GATE EE 2019   Electric Circuits
Question 3 Explanation: 


Applying nodal at node x,
-I-2+\frac{V_x-5I}{3}=0
-3I-6+V_x-5I=0
\Rightarrow \; 8I=V_x-6\;\;...(i)
As, I=\frac{20-V_x}{2}
\Rightarrow \; V_x=20-2I\;\;...(ii)
Substituting (ii) in (i),
8I=20-2I-6
10I=14
I=1.4A
Question 4
The equivalent impedance Z_{eq} for the infinite ladder circuit shown in the figure is
A
j12 \Omega
B
-j12 \Omega
C
j13 \Omega
D
13 \Omega
GATE EE 2018   Electric Circuits
Question 4 Explanation: 




Z_1=j9
Z_2=j5-j1=j4
Z_{eq}=Z_1+\frac{Z_2Z_{eq}}{Z_2+Z_{eq}}
By solving above equation,
Z_{eq}=j12
Question 5
The power supplied by the 25 V source in the figure shown below is ________W.
A
25
B
250
C
2.5
D
100
GATE EE 2017-SET-1   Electric Circuits
Question 5 Explanation: 
Using KCL at node , we get
I+0.4I=14
I=10A
Now, power supplied,
P=25 \times 10=250W
Question 6
The equivalent resistance between the terminals A and B is ______ \Omega.
A
2.2
B
1.2
C
1
D
3
GATE EE 2017-SET-1   Electric Circuits
Question 6 Explanation: 
Consider the following circuit diagram,

After rearrangement we get

Now, R_{AB}=1+\frac{6}{5}+0.8=3\Omega
Question 7
In the circuit shown below, the voltage and current sources are ideal. The voltage (V_{out}) across the current source, in volts, is
A
0
B
5
C
10
D
20
GATE EE 2016-SET-2   Electric Circuits
Question 7 Explanation: 


So, V_{out}=(5 \times 2 )+10=20V
Question 8
In the circuit shown below, the node voltage In the circuit shown below, the node voltage V_{A} is ___________ V.
A
11.42
B
5.55
C
7.25
D
15.25
GATE EE 2016-SET-1   Electric Circuits
Question 8 Explanation: 


Applying KCL at node A, we get
\frac{V_A}{5}+\frac{V_A-10}{10}+\frac{V_A+10I_1}{5}=5
So, 2V_A+V_A-10+2V_A+20I_1=5
5V_A+20I_1=60
Since, I_1=\frac{V_A-10}{10}
So, 5V_A+2V_A-20=60
7V_A=80
V_A=11.42 volt
Question 9
In the given circuit, the current supplied by the battery, in ampere, is _______.
A
0.1
B
0.5
C
1
D
1.5
GATE EE 2016-SET-1   Electric Circuits
Question 9 Explanation: 


Applying KCL at node A,
-I_1+I_2+I_2=0
2I_2=I_1\;\;...(i)
and applying KVL in loop ABCD,
1-I_1-I_2-I_2=0
I_1+2I_2=1\;\;...(ii)
From equation (i) and (ii),
\Rightarrow \; 2I_2+2I_2=1
\Rightarrow \; 4I_2=1
\Rightarrow \; I_2=\frac{1}{4}A
\Rightarrow \; I_1=2 \times \frac{1}{4}=\frac{1}{2}A
Question 10
In the portion of a circuit shown, if the heat generated in 5 \Omega resistance is 10 calories per second, then heat generated by the 4 \Omega resistance, in calories per second, is _______.
A
1
B
2
C
3
D
4
GATE EE 2016-SET-1   Electric Circuits
Question 10 Explanation: 


(2I)^2 \times 5 =10
\Rightarrow \; I^2=\frac{10}{5 \times 5}=0.5
So, I^2 \times 4 =2 cal/sec.
There are 10 questions to complete.