# Basics

 Question 1
For the circuit shown in the figure, $\mathrm{V}_{1}=8 \mathrm{~V},$ DC and $I_{1}=8 A$, DC. The voltage $V_{a b}$ in Volts is ___ (Round off to 1 decimal place).

 A 4.2 B 6 C 8 D 10.6
GATE EE 2023   Electric Circuits
Question 1 Explanation:
Reraw the circuit:

Now, using voltage division,
$\mathrm{V}_{\mathrm{ab}}=\frac{1.5 \times 8}{1.5+0.5}=6 \mathrm{~V}$
 Question 2
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 2 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 3
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 3 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 4
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 4 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 5
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 5 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,
$Z_{eq}=j12$

There are 5 questions to complete.