Sinusoidal Steady State

Question 1
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 1 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 2
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 2 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 3
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 3 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}
Question 4
The figure shows an RLC circuit exited by the sinusoidal voltage 100 \cos (3t) volts, where t is in seconds. The ratio \frac{amplitude \; of \; V_{2}}{amplitude \; of \; V_{1}} is _________.
A
1.6
B
2.6
C
3.6
D
4
GATE EC 2017-SET-1   Network Theory
Question 4 Explanation: 


\begin{aligned} \omega &=3 \mathrm{rad} / \mathrm{sec} ; Z_{1}=(4+j 3) \Omega \\ \text{and}\quad Z_{2} &=(5-j 12) \Omega \\ \left|V_{2}\right| &=|i|\left|z_{2}\right|=|i| \sqrt{5^{2}+12^{2}}=13|i| \\ \left|V_{1}\right| &=|i|\left|z_{1}\right|=|i| \sqrt{4^{2}+3^{2}}=5|i| \\ \frac{\left|V_{2}\right|}{\left|V_{1}\right|} &=\frac{13|i|}{||5 1}=\frac{13}{5}=2.6 \end{aligned}
Question 5
In the circuit shown, the positive angular frequency \omega (in radians per second) at which magnitude of the phase difference between the voltages V_1 \; and \; V_2 equals \pi/4 radians, is
A
1
B
2
C
3
D
4
GATE EC 2017-SET-1   Network Theory
Question 5 Explanation: 


\begin{aligned} \text{Let,}i(t)&=I_{m} \angle \theta_{i} and Z_{2}=1+j \omega \\ &=\sqrt{1+\omega^{2}} \angle \theta_{2} \\ \theta_{2} &=\tan ^{-1}\left(\frac{\omega}{1}\right) \\ V_{1} &=i(t)(1 \Omega)=I_{m} \angle \theta_{i} \\ V_{2} &=i(t) Z_{2}=I_{m} \sqrt{1+\omega^{2}} \angle \theta_{2}+\theta_{i} \end{aligned}
From the given data,
\begin{aligned} \left(\theta_{i}+\theta_{2}\right)-\left(\theta_{i}\right)&=\frac{\pi}{4}\\ \theta_{2}&=\frac{\pi}{4} \\ \tan ^{-1}\left(\frac{\omega}{1}\right)&=\frac{\pi}{4}\\ \omega&=1 \mathrm{rad} / \mathrm{sec} \end{aligned}
Question 6
the RLC circuit shown in the figure, the input voltage is given by

v_{i}(t)=2cos(200t)+4sin(500t)

The output voltage v_{0}(t) is
A
cos(200t) + 2 sin(500t)
B
2cos(200t) + 4 sin(500t)
C
sin(200t) + 2 cos(500t)
D
2sin(200t) + 4 cos(500t)
GATE EC 2016-SET-3   Network Theory
Question 6 Explanation: 


Where, V_{i}(t)=2 \cos (200 t)+4 \sin (500 t)
As different frequencies are operating, using superposition theorem, we get
for \omega=200 \mathrm{rad} / \mathrm{sec}
\begin{array}{l} x_{L}=\omega L=(200)(0.25)=50 \Omega \\ x_{C}=\frac{1}{\omega C}=\frac{1}{200 \times 100 \times 10^{-6}}=50 \Omega \end{array}


\begin{aligned} V_{0}(t) &=V_{i}(t) \\ X_{L} &=0.4 \times 500=200 \Omega \\ X_{C} &=\frac{1}{10 \times 10^{-6} \times 500}=200 \Omega \end{aligned}


\begin{aligned} \therefore \quad V_{0}(t)&=V_{i}(t) \\ \text { Therefore } \quad V_{0}(t)&=2 \cos (200 t)+4 \sin (500 t) \end{aligned}
Question 7
The figure shows an RLC circuit with a sinusoidal current source.

At resonance, the ratio |I_{L}|/|I_{R}|, i.e., the ratio of the magnitudes of the inductor current phasor and the resistor current phasor, is ________
A
0.1
B
0.3
C
0.6
D
0.9
GATE EC 2016-SET-2   Network Theory
Question 7 Explanation: 
At resonance (for parallel RLC circuit)
\begin{array}{l} I_{R}=I \\ I_{L}=Q I \angle-90^{\circ} \\ I_{C}=Q I \angle 90^{\circ} \end{array}
For parallel RLC circuit
\begin{aligned} \frac{\left|I_{L}\right|}{\left|I_{R}\right|} &=\frac{I Q}{I}=Q=R \sqrt{\frac{C}{L}} \\ &=10 \sqrt{\frac{10 \times 10^{-6}}{10 \times 10^{-3}}}=0.316 \end{aligned}
Question 8
In the circuit shown, the current I flowing through the 50 \Omega resistor will be zero if the value of capacitor C (in \muF) is ______.
A
10
B
20
C
30
D
40
GATE EC 2015-SET-3   Network Theory
Question 8 Explanation: 


\begin{aligned} Z_{\mathrm{eq}} &=\frac{j 5 \Omega \times\left(\frac{1}{j \omega C}+j 5\right)}{j 5 \Omega+j 5 \Omega+\frac{1}{j \omega C}} \\ I &=0 \text { if } Z \text { is } \infty \\ Z &=Z_{\mathrm{eq}}+50+j 5\\ \text{For z to be } \infty, Z_{e q}&=\infty \\ \Rightarrow j 5+j 5+\frac{1}{j \omega C}&=0\\ 10 &=\frac{1}{5000 \times C} \\ C &=\frac{1}{5 \times 10^{3} \times 10}=20 \mu \mathrm{F} \end{aligned}
Question 9
At very high frequencies, the peak output voltage V_{0} (in Volts) is ________.
A
0
B
0.5
C
1
D
1.5
GATE EC 2015-SET-3   Network Theory
Question 9 Explanation: 
Circuit contains balanced Wheat-stone bridge. Also at high frequencies capacitor can be considered as short circuits.
Redrawing the circuit

\begin{aligned} V_{0} &=1 \sin \omega t \times 1 k \Omega=0.5 \sin \omega t \\ \text { Peak output } &=0.5 \mathrm{V} \end{aligned}
Question 10
An LC tank circuit consists of an ideal capacitor C connected in parallel with a coil of inductance L having an internal resistance R. The resonant frequency of the tank circuit is
A
\frac{1}{2\pi \sqrt{LC}}
B
\frac{1}{2\pi \sqrt{LC}}\sqrt{1-\frac{R^{2}C}{L}}
C
\frac{1}{2\pi \sqrt{LC}}\sqrt{1-\frac{L}{R^{2}C}}
D
\frac{1}{2\pi \sqrt{LC}}\ (1-\frac{R^{2}C}{L})
GATE EC 2015-SET-2   Network Theory
Question 10 Explanation: 
\begin{aligned} Z_{e q}=& \frac{\frac{1}{j \omega C} \times(j \omega L+R)}{\frac{1}{j \omega C}+j \omega L+R} \\ Z_{e q}=& \frac{\left(\frac{R}{j \omega C}+\frac{L}{C}\right)}{R+j\left(\omega L-\frac{1}{\omega C}\right)} \times \frac{R-j\left(\omega L-\frac{1}{\omega C}\right)}{R-j\left(\omega L-\frac{1}{\omega C}\right)} \end{aligned}


Equating imaginary part to zero.
\begin{aligned} \text{Img}&=-\frac{R^{2}}{\omega C}-\frac{L}{C}\left(\omega L-\frac{1}{\omega C}\right)=0 \\ &\frac{R^{2}}{\omega C}+\frac{L}{C}\left(\omega L-\frac{1}{\omega C}\right)=0 \\ &\frac{C R^{2}+\omega^{2} L^{2} C-L}{\omega C^{2}}=0 \\ \omega^{2}&=\frac{L-R^{2} C}{L^{2} C} \\ \omega&=\frac{1}{\sqrt{L C} \sqrt{1-\frac{R^{2} C}{L}}} \\ f&=\frac{1}{2 \pi \sqrt{L C}} \sqrt{1-\frac{R^{2} C}{L}} \end{aligned}
There are 10 questions to complete.
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