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).


4.2 | |
6 | |
8 | |
10.6 |
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}

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)


100 | |
120 | |
150 | |
175 |
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
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.


15 | |
50 | |
5 | |
20 |
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 ____


0.8 | |
1.1 | |
1.4 | |
1.9 |
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


j12 \Omega | |
-j12 \Omega | |
j13 \Omega | |
13 \Omega |
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.