# 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.