Basics

Question 1
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 1 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 2
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 2 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 3
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 3 Explanation: 
Using KCL at node , we get
I+0.4I=14
I=10A
Now, power supplied,
P=25 \times 10=250W
Question 4
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 4 Explanation: 
Consider the following circuit diagram,

After rearrangement we get

Now, R_{AB}=1+\frac{6}{5}+0.8=3\Omega
Question 5
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 5 Explanation: 


So, V_{out}=(5 \times 2 )+10=20V
Question 6
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 6 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 7
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 7 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 8
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 8 Explanation: 


(2I)^2 \times 5 =10
\Rightarrow \; I^2=\frac{10}{5 \times 5}=0.5
So, I^2 \times 4 =2 cal/sec.
Question 9
R_{A} \; and \; R_{B} are the input resistances of circuits as shown below. The circuits extend infinitely in the direction shown. Which one of the following statements is TRUE?
A
R_{A} = R_{B}
B
R_{A} = R_{B} =0
C
R_{A} \lt R_{B}
D
R_{B} = R_{A}/(1+R_{A})
GATE EE 2016-SET-1   Electric Circuits
Question 9 Explanation: 
If the equavalent resistance of first figure is R_A then from the second figure, we can see that R_B=R_A||1\Omega.
R_B=\frac{R_A}{R_A+1}
Question 10
The current i (in Ampere) in the 2\Omega resistor of the given network is ______.
A
0
B
1
C
2
D
4
GATE EE 2015-SET-2   Electric Circuits
Question 10 Explanation: 
Redrawing the circuit,

Bridge is balance, so current flowing through 2\Omega resistor is 0A.
There are 10 questions to complete.