Q21GATE 2014MCQ

The steady state output of the circuit shown in the figure is given by $y(t)=A(\omega )sin(\omega t+\phi (\omega ))$ . If the amplitude $|A(\omega )|$ = 0.25, then the frequency $\omega$ is

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📖 Explanation

Applying KCL at node A, we get

\begin{aligned} &=\frac{V_{A}-\sin \omega t}{R}+\frac{V_{A}}{\frac{1}{j \omega C}}+\frac{V_{A}}{\frac{2}{j \omega C}}=0 &\ldots(i)\\ &=V_{A}\left[\frac{1}{R}+j \omega C+\frac{j \omega C}{2}\right]=\frac{\sin \omega t=1 \angle 0^{\circ}}{R} \\ &=V_{A}=\frac{2}{2+3 R C \cdot j \omega} &\ldots(ii)\\ \text{Also }\quad Y&=\frac{V_{A}}{2}=\frac{1}{2+3 j \omega R C} &\ldots(iii)\\ \because|A(\omega)|&=\frac{1}{4} \\ \therefore \quad \frac{1}{4}&=\frac{1}{\sqrt{4+9 \omega^{2}(R C)^{2}}} \\ or \quad \omega&=\frac{2}{\sqrt{3} R C} \end{aligned}

Q23GATE 2011MCQ

The circuit shown below is driven by a sinusoidal input $v_{i} = v_{p} cos (t/RC)$ . The steady state output $v_{o}$

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📖 Explanation

Redrawing the circuits-domain

\begin{aligned} V_{i}(s) &=\left(R+\frac{1}{s C}\right) I(s)+\frac{R \cdot \frac{1}{s C}}{R+\frac{1}{s C}} I(s) \\ V_{i}(s) &=\frac{1+s C R}{s C} I(s)+\frac{R}{1+s C R} I(s) \\ \because \quad v_{i} &=V_{p} \cos (t / R C) \qquad \ldots(i)\\ \text{so here,} \qquad \omega&=\frac{1}{R C} \\ Now,V_{i}(s)&=\frac{(1+j \omega C R)}{j \omega C} I(s)+\frac{R}{(1+j \omega C R)} I(s) \\ \text{Put,}\quad \omega&=\frac{1}{R C} \end{aligned}

\begin{aligned} \text { So, } \quad V_{i}(s)&=\left[\frac{(1+j) R}{j}+\frac{R}{1+j}\right] I(s)\\ \frac{V_{i}(s)}{I(s)} &=\frac{3 R}{(1+j)} \qquad\ldots(ii)\\ I(s) &=\frac{V_{i}(s)}{3 R} \times (1+j) \\ \text { Now, } V_{0}(s) &=\left(R \| \frac{1}{s C}\right) I(s) \\ \Rightarrow \quad V_{0}(s) &=\frac{R \cdot \frac{1}{s C}}{R+\frac{1}{s C}} I(s) \\ \Rightarrow \quad \quad V_{0}(s)&=\frac{R}{1+S C R} \cdot \frac{V_{i}(s)}{3 R}(1+j) \\ \Rightarrow \quad V_{0}(s)&=\frac{R}{1+j} \cdot \frac{V_{i}(s)}{3 R}(1+j) \\ \Rightarrow \quad \quad V_{0}(s)&=\frac{V_{i}(s)}{3} \\ \text{In time domain,}\\ v_{0}(t)&=\frac{1}{3} v_{i}(t) \\ \Rightarrow \quad v_{0}(t)&=\frac{V_{p}}{3} \cos \left(\frac{t}{R C}\right) \end{aligned}

Q24GATE 2010MCQ

The current I in the circuit shown is

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📖 Explanation

$(L=20 \mathrm{mH}, C=50 \mu \mathrm{H})$ Nodal analysis at node A,

\begin{array}{l} \frac{V_{A}-20}{j \omega L}+\frac{V_{A}}{1}+\frac{V_{A}}{1 / j \omega C}=0 \\ V_{A}\left[\frac{1}{j 10^{3} \times 20 \times 10^{-3}}+1+j 10^{3} \times 50 \times 10^{-6}\right]\$ \\ =\frac{20}{j 10^{3} \times 20 \times 10^{-3}} \\ V_{A}\left[\frac{-j}{20}+1+\frac{j}{20}\right]=-j 1 \\ V_{A}=-j 1 V \\ I=\frac{V_{A}}{1}=-j 1 \mathrm{A} \end{array}

Q25GATE 2010MCQ

For parallel RLC circuit, which one of the following statements is NOT correct?

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📖 Explanation

Characteristic equation for a parallel RLC circuit is $s^{2}+\frac{1}{R C} s+\frac{1}{L C}=0$ where, Bandwidth= $\frac{1}{RC}$ (i) It is clear that the bandwidth of a parallel RLC circuit is independent of L and decreases if R is increased. (ii) At resonance, imaginary part of input impedance is zero. Hence, at resonance input impedance is a real quantity. (iii) In parallel RLC circuit, the admittance is minimum at resonance. Hence magnitude of input impedance attains its maximum value at resonance.

Q26GATE 2009MCQ

An AC source of RMS voltage 20 V with internal impedance $Z_{s}=(1+2j)\Omega$ feeds a load of impedance $Z_{L}=(7+4j)\Omega$ in the figure below. The reactive power consumed by the load is

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📖 Explanation

$I=\frac{20}{8+6 j}=\frac{10}{4+3 j}=2 \angle-36.87^{\circ}$ Reactive power $Q=I^{2} X_{L}=4 \times 4=16 \mathrm{VAR}$

Q27GATE 2007MCQ

In the ac network shown in the figure, the phasor voltage $V_{AB}$ (in Volts) is

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📖 Explanation

\begin{aligned} V_{A B} &=\text { Current } \times \text { impedance } \\ &=5 \angle 30^{\circ} \times (5-3 j) \|(5+3 j) \\ &=5 \angle 30^{\circ} \times \frac{(5-3 j) \times (5+3 j)}{5-3 j+5+3 j} \\ &=5 \angle 30^{\circ} \times \frac{25+9}{10} \\ \Rightarrow \quad &=5 \angle 30^{\circ} \times 3.4=17 \angle 30^{\circ} \end{aligned}

Q28GATE 2005MCQ

The condition on R, L and C such that the step response y(t) in the figure has no oscillations, is

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📖 Explanation

Transfer function

\begin{aligned} =& \frac{\frac{1}{s C}}{R+s L+\frac{1}{s C}}=\frac{1}{s^{2} L C+s C R+1} \\ \frac{Y(s)}{U(s)}=& \frac{\frac{1}{L C}}{s^{2}+\frac{R}{L} s+\frac{1}{L C}} \\ 2 \xi \omega_{n} &=\frac{R}{L} \\ \omega_{n} &=\frac{1}{\sqrt{L C}} \\ \xi &=\frac{R}{2 L} \sqrt{L C}=\frac{R}{2} \sqrt{\frac{C}{L}} \end{aligned}

For no oscillations, $\xi \geq 1$ $\frac{R}{2} \sqrt{\frac{C}{L}} \geq 1 \quad ; \quad R \geq 2 \sqrt{\frac{L}{C}}$

Q29GATE 2005MCQ

For the circuit shown in the figure, the instantaneous current $i_{1}$ (t) is

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📖 Explanation

When $5 \angle 0^{\circ}$ is acting along. $i_{1}(t)=-5 \angle 0^{\circ}$ $\qquad \left(10 \angle 60^{\circ} \text { is kept open }\right)$ When $10 \angle 60^{\circ}$ is acting alone.

\begin{aligned} i_{1}(t)&=10 \angle 60^{\circ}\\ &\qquad(5 \angle 0^{\circ} \text{ is kept open})\\ i_{1}(t)&=10 \angle 60^{\circ}-5 \angle 0^{\circ}=5+8.66 j-5 \\ i_{1}(t)&=8.66 j \\ i(t)&=5 \sqrt{3} \angle 90^{\circ}=\frac{10}{2} \sqrt{3} \angle 90^{\circ} \end{aligned}

Q30GATE 2005MCQ

In a series RLC circuit, R=2k $\Omega$ , L=1H, and $C=\frac{1}{400}\mu F$ The resonant frequency is

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📖 Explanation

\begin{aligned} f_{0} &=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{1 \times \frac{1}{400} \times 10^{-6}}} \\ &=\frac{10^{3} \times 20}{2 \pi}=\frac{10^{4}}{\pi} \mathrm{Hz} \end{aligned}

Q31GATE 2004MCQ

Consider the following statements S1 and S2 S1 : At the resonant frequency the impedance of a series RLC circuit is zero. S2 : In a parallel GLC circuit, increasing the conductance G results in increase in its Q factor. Which one of the following is correct?

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📖 Explanation

$S_{1}$ : Impedance of series RLC circuit at resonant frequency is minimum $Z=R+j\left(\omega L-\frac{1}{\omega C}\right)$ $\omega L-\frac{1}{\omega C}=0$ $Z=R \text{(Purely resistive)}$ $S_{2}: Q=R \sqrt{\frac{C}{L}}$ $G=\frac{1}{R} \Rightarrow Q=\frac{1}{G} \sqrt{\frac{C}{L}}$ $G \uparrow$ then $Q \downarrow$ if $C \& L$ are same

Q32GATE 2004MCQ

The circuit shown in the figure, with $R=\frac{1}{3}\Omega ,L=\frac{1}{4}H$ and C = 3 F has input voltage v(t) = sin2t. The resulting current i(t) is

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📖 Explanation

\begin{aligned} i(t)=& V(t), Y \\ Y=& V(t)\left[\frac{1}{R_{1}}+\frac{1}{j \omega L}+j \omega C\right] \\ =& \sin 2 t\left[3+\frac{4}{2 j}+j \times 2 \times 3\right] \\ =& \sin 2 t[3-2 j+6 j]=\sin 2 t[3+4 j] \\ =& 5 \sin 2 t \angle \tan ^{-1} \frac{4}{3}=5 \sin \left(2 t+53.1^{\circ}\right) \end{aligned}

Q33GATE 2004MCQ

The transfer function $H(s)=\frac{V_{o}(s)}{V_{i}(s)}$ of an RLC circuit is given by $H(s)=\frac{10^{6}}{s^{2}+20s+10^{6}}$ . The Quality factor (Q-factor) of this circuit is

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📖 Explanation

\begin{array}{l} \text { Characteristic equation }=s^{2}+20 s+10^{6} \\ Q=\frac{\omega_{0}}{B W}, \omega_{0}=\sqrt{10^{6}} \\ Q=\frac{10^{3}}{20}=\frac{1000}{20}=50 \end{array}

Q34GATE 2004MCQ

For the circuit shown in the figure, the time constant RC = 1 ms. The input voltage is $v_{i}(t)=\sqrt{2}sin10^{3}t$ . The output voltage $v_{o}(t)$ is equal to

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📖 Explanation

\begin{aligned} v_{0}(t)&=\frac{\frac{1}{j \omega C}}{R+\frac{1}{j \omega C}}\; v_{i}(t)=\frac{1}{1+j \omega C R} \sqrt{2} \sin 10^{3} t \\ &=\frac{1}{1+j \times 10^{3} \times 10^{-3}} \sqrt{2} \sin 10^{3} t \\ v_{0}(t) &=\sin \left(10^{3} t-45^{\circ}\right) \end{aligned}

Q35GATE 2003MCQ

An input voltage $v(t)=10\sqrt{5}\cos (t+10^{\circ})+10\sqrt{5}\cos (2t+10^{\circ})V$ is applied to a series combination of resistance R = 1 $\Omega$ and an inductance L = 1 H. The resulting steady-state current i(t) in ampere is

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📖 Explanation

\begin{array}{l} \text { if }(t)=\frac{v(t)}{R+f(u L}=\frac{10 \sqrt{2} \cos \left(t+10^{\circ}\right)}{1+1 j}+\frac{10 \sqrt{5} \cos \left(2 t+10^{\circ}\right)}{1+2 j} \\ i(t)=\frac{10 \sqrt{2} \cos \left(t+10^{\circ}\right)}{\sqrt{2} \angle 45^{\circ}}+\frac{10 \sqrt{5} \cos \left(2 t+10^{\circ}\right)}{\sqrt{5} \angle \tan ^{-1} 2} \\ \therefore i(t)=10 \cos \left(t-35^{\circ}\right)+10 \cos \left(2 t+10-\tan ^{-1} 2\right) \end{array}

Q36GATE 2003MCQ

A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q=100. If each of R, L and C is doubled from its original value, the new Q of the circuit is

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📖 Explanation

\begin{aligned} Q=& \frac{f_{o}}{B W} \\ f_{0} &=\frac{1}{2 \pi \sqrt{L C}} \\ B W &=\frac{R}{L} \\ \text { (Characteristic equation } &\left.=s^{2}+\frac{R s}{L}+\frac{1}{L C}\right)\\ \text{or}\qquad&=\frac{1}{R} \sqrt{\frac{L}{C}} \\ \text{When R, L, C are doubled,}\\ Q&=\frac{1}{2} Q=50 \end{aligned}

Q37GATE 2001MCQ

When the angular frequency $\omega$ in figure is varied from 0 to $\infty$ , the locus of the current phasor $I_{2}$ is given by

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📖 Explanation

$i_{2}(t)=\frac{E_{m} \cos \omega t}{R_{2}+\frac{1}{j \omega C}}=E_{m} \angle 0 \frac{j \omega C}{1+j \omega C R_{2}}$ $\angle i_{2}(t)=\frac{E_{m} \angle 0 \omega C \angle 90^{\circ}}{\sqrt{1+\omega^{2} C^{2} R^{2}} \angle \tan ^{-1} \omega C R_{2}}$ $i_{2}(t)=\frac{E_{m} \omega C}{\sqrt{1+\omega^{2} C^{2} R^{2}}} \angle 90^{\circ}-\tan ^{-1} \omega C R$ at $\omega=0, i_{2}(t)=0, \omega=\infty , i_{2}(t)=\frac{E_{m}}{R_{2}}$ Option(a) satisfies both conditions.