# Network Theorems

 Question 1
For the circuit shown, if $\mathrm{i}=\sin 1000 t$, the instantaneous value of the Thevenin's equivalent voltage (in Volts) across the terminals a-b at time $\mathrm{t}=5 \mathrm{~ms}$ is ___ (Round off to 2 decimal places).

 A -12.25 B 12.25 C 11.97 D -11.97
GATE EE 2023   Electric Circuits
Question 1 Explanation:
By source transformation, the circuit become

Apply KVL in loop,
\begin{aligned} & (10+j 1)+4 i_{x}-(10+j 10+10-j 10) i_{x}=0 \\ & 10+\mathrm{j} 10+4 \mathrm{i}_{x}-20 \mathrm{i}_{x}=0 \\ & \Rightarrow \quad \mathrm{i}_{\mathrm{x}}=0.884 \angle 45^{\circ} \mathrm{A} \\ & \text { Now, } \quad \mathrm{V}_{\text {Th }}=\mathrm{i}_{x}(10-\mathrm{j10}) \\ & =0.884 \angle 45^{\circ} \times 14.142 \angle-45^{\circ} \\ & =12.5 \angle 0^{\circ} \mathrm{V} \\ & \therefore \quad \mathrm{V}_{\text {Th }}=12.5 \sin 1000 \mathrm{t} \\ & \text { At } \quad \mathrm{t}=5 \mathrm{msec} \\ & \mathrm{V}_{\text {Th }}=12.5 \times \sin \left(\frac{5 \times 180^{\circ}}{\pi}\right) \\ & =-11.986 \mathrm{~V} \end{aligned}
 Question 2
In the given circuit, for maximum power to be delivered to $R_{L}$, its value should be ______________ $\Omega$.
(Round off to 2 decimal places.)

 A 2.14 B 3.32 C 1.42 D 4.12
GATE EE 2021   Electric Circuits
Question 2 Explanation:

$=\frac{-j}{\omega C}=\frac{-j}{1000 \times 0.5 \times 10^{-3}}=-j 2 \Omega$

$Z_{\text {in }}=2 \| j 2=\frac{j 4}{2+j 2}=\frac{j 2}{1+j 1}$
For maximum power transfer,
$R_{L}=\left|Z_{\mathrm{TH}}\right|=\frac{2}{\sqrt{2}}=\sqrt{2}=1.414 \Omega$

 Question 3
A signal generator having a source resistance of $50\:\Omega$ is set to generate a $\text{1 kHz}$ sinewave. Open circuit terminal voltage is $\text{10 V}$ peak-to-peak. Connecting a capacitor across the terminals reduces the voltage to $\text{8 V}$ peak-to-peak. The value of this capacitor is _____________$\mu F$. (Round off to 2 decimal places.)
 A 4.25 B 6.32 C 1.25 D 2.38
GATE EE 2021   Electric Circuits
Question 3 Explanation:
\begin{aligned} \frac{V_{0}(j \omega)}{V_{i}(j \omega)} &=\frac{1}{1+j \omega R C} \\ \frac{8}{10} \angle \theta &=\frac{1}{1+j \omega R C} \\ 1+j \omega R C &=1.25 \angle \theta \\ \omega R C &=\sqrt{1.25^{2}-1^{2}} \\ \omega R C &=0.75 \\ C &=\frac{0.75}{2000 \pi \times 50}=2.38 \mu\mathrm{F} \end{aligned}
 Question 4
In the given circuit, for voltage $V_y$ to be zero, the value of $\beta$ should be _________. (Round off to 2 decimal places).

 A -4.58 B -3.25 C 4.65 D 3.45
GATE EE 2021   Electric Circuits
Question 4 Explanation:
\begin{aligned} \frac{V_{x}-6}{1}+\frac{V_{x}}{4}+\frac{V_{x}-V_{y}}{2} & =0 \\ 4 V_{x}-24+V_{x}+2 V_{x}-2 V_{y} & =0 \\ 7 V_{x}-2 V_{y} & =24 \\ \text { If } V_{y}=0 \qquad\qquad 7 V_{x} & =24 \\ \Rightarrow\qquad\qquad V_{x} & =\frac{24}{7} \\ \Rightarrow\qquad\qquad \frac{V_{y}-V_{x}}{2}+\frac{V_{y}-\beta V_{x}}{3} & =2 \\ 3 V_{y}-3 V_{x}+2 V_{y}-2 \beta V_{x} & =12 \\ 5 V_{y}-(3+2 \beta) V_{x} & =12 \\ V_{y} & =0\\ -(3+2 \beta) \frac{24}{7} &=12 \\ (3+2 \beta) &=\frac{-7}{2} \\ 2 \beta &=\frac{-7}{2}-3=\frac{-7-6}{2}=\frac{-13}{2} \\ \beta &=\frac{-13}{4}=-3.25 \end{aligned}
 Question 5
For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals 'ab' is

 A $\text{10 V}$ in series with $12\:\Omega$ B $\text{65 V}$ in series with $15\:\Omega$ C $\text{50 V}$ in series with $2\:\Omega$ D $\text{35 V}$ in series with $2\:\Omega$
GATE EE 2021   Electric Circuits
Question 5 Explanation:
Given circuit can be resolved as shown below,

$V_{T H}=15+50=65 \mathrm{~V}$

\begin{aligned} V_{x} &=2+3+10=15 \mathrm{~V} \\ R_{\mathrm{TH}} &=\frac{V_{x}}{1}=15 \Omega \end{aligned}

There are 5 questions to complete.