Question 1
Consider the circuit shown in the figure with input $V(t)$ in volts. The sinusoidal steady state current $I(t)$ flowing through the circuit is shown graphically (where $t$ is in seconds). The circuit element $Z$ can be ________.

 A a capacitor of 1 F B an inductor of 1 H C a capacitor of $\sqrt{3}$ F D an inductor of $\sqrt{3}$ H
GATE EC 2022   Network Theory
Question 1 Explanation:
$i(t)=\frac{V(t)}{1+Z}$ where, $V(t)= \sin t$
$\because$ Given i(t) is lagging (from plot)
\begin{aligned} I_{max}&=\frac{V_{max}}{Z}=\frac{1}{Z}\\ \Rightarrow Z&=1/\sqrt{2}\\ Z&=(1+j\omega L)\Rightarrow L=1\\ as\; \omega &=1(\sin \omega t)=\sin t \end{aligned}
 Question 2
For the circuit shown, the locus of the impedance $Z(j\omega )$ is plotted as $\omega$ increases from zero to infinity. The values of $R_1$ and $R_2$ are:

 A $R_1=2k\Omega ,R_2=3k\Omega$ B $R_1=5k\Omega ,R_2=2k\Omega$ C $R_1=5k\Omega ,R_2=2.5k\Omega$ D $R_1=2k\Omega ,R_2=5k\Omega$
GATE EC 2022   Network Theory
Question 2 Explanation:
\begin{aligned} Z(j\omega )&=R_1+\frac{R_2\cdot 1/j\omega c}{R_2+ 1/j\omega c}\\ &=R_1+\frac{R_2}{\frac{1}{2}+jR_2\omega c}\\ Z(j\omega )_{\omega =0}&=R_1+R_2\\ \Rightarrow R_1+R_2&=5k\Omega \\ Z|j\omega |_{\omega \rightarrow \infty }&=R_1\\ \Rightarrow R_1&=2k\Omega \\ R_2&=3k\Omega \end{aligned}

 Question 3
In the circuit shown, if $v(t)=2sin(1000t)$ volts, $R=1k\Omega$ and $R=C=1\mu F$, then the steady-state current $i(t)$, in milliamperes (mA), is
 A sin(1000t) + cos(1000t) B 2sin(1000t) + 2cos(1000t) C 3sin(1000t) + cos(1000t) D sin(1000t) + 3cos(1000t)
GATE EC 2019   Network Theory
Question 3 Explanation:

\begin{aligned} \text{Here,}X_{C}&=\frac{1}{\omega C}=\frac{1}{10^{3} \times 10^{-6}}=\frac{1}{10^{-3}}\\ x_{C} &=10^{3} \Omega \\ R &=10^{3} \Omega \quad \text { (Given) } \\ v(t) &=2 \sin 1000 t \vee=2 \angle 0^{\circ} \mathrm{V} \end{aligned}
Redrawing the given network, we get,

As the bridge is balanced, it can be redrawn as

\begin{aligned} \therefore \quad Y_{e q} &=Y_{1}+Y_{2} \\ &=\frac{3}{2} R+\frac{1}{-2 j X_{C}} \\ &=\frac{3}{2} \times 10^{-3}+j \frac{1}{2} \times 10^{-3} \\ \therefore \quad i(t) &=(t) \times Y_{\mathrm{eq}}=2 \angle 0^{\circ}\left[\frac{3}{2}+j \frac{1}{2}\right] \mathrm{m} \mathrm{A} \\ &=(3+j 1) \mathrm{mA} \\ &=3 \sin (1000 t)+\cos (1000 t) \mathrm{mA} \end{aligned}
 Question 4
For the circuit given in the figure, the voltage $V_{C}$ (in volts) across the capacitor is
 A $1.25\sqrt{2}sin(5t-0.25\pi )$ B $1.25\sqrt{2}sin(5t-0.125\pi )$ C $2.5\sqrt{2}sin(5t-0.25\pi )$ D $2.5\sqrt{2}sin(5t-0.125\pi )$
GATE EC 2018   Network Theory
Question 4 Explanation:
$\frac{1}{\omega C}=\frac{1}{5 \times 10^{-6}}=200 \mathrm{k} \Omega$

\begin{aligned} V_{C} &=\frac{5 \angle 0^{\circ}}{200-j 200} \times(-j 200) \vee \\ &=\frac{5 \angle 0^{\circ} \times 1 \angle-90^{\circ}}{\sqrt{2} \angle-45^{\circ}} \mathrm{V} \\ &=\frac{5}{\sqrt{2}} \angle-45^{\circ} \mathrm{V}=2.5 \sqrt{2} \sin \left(5 t-\frac{\pi}{4}\right) \mathrm{V} \\ &=2.5 \sqrt{2} \sin (5 t-0.25 \pi) \mathrm{V} \end{aligned}
 Question 5
In the circuit shown, V is a sinusoidal voltage source. The current I is in phase with voltage V.
The ratio $\frac{amplitude \; of \; voltage \; across \; the \; capacitor} {amplitude \; of \; voltage \; across \; the \; resistor}$ is
 A 0.1 B 0.2 C 0.3 D 0.4
GATE EC 2017-SET-2   Network Theory
Question 5 Explanation:

Given that, V and J have same phase. So, the circuit is in resonance.
At resonance,
\begin{aligned} &V_{C}=Q V_{R}\\ &\text { So, } \frac{\text { Amplitude of } V_{C}}{\text { Amplitude of } V_{R}}=Q=\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{1}{5} \sqrt{\frac{5}{5}}=0.2 \end{aligned}

There are 5 questions to complete.