Q2GATE 2022MCQ

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:

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📖 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}

Q3GATE 2022MCQ

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 ________.

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📖 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}

Q4GATE 2019MCQ

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

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📖 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}

Q5GATE 2018MCQ

For the circuit given in the figure, the voltage $V_{C}$ (in volts) across the capacitor is

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📖 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}

Q6GATE 2017NAT

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 _________.

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📖 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}

Q7GATE 2017NAT

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

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📖 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}

Q8GATE 2017NAT

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

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📖 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}

Q9GATE 2016MCQ

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

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📖 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}

Q10GATE 2016NAT

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 ________

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📖 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}

Q11GATE 2015NAT

In the circuit shown, the average value of the voltage $V_{ab}$ (in Volts) in steady state condition is________.

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📖 Explanation

Applying superposition: $V_{ab}$ = 5 V [ open circuited in steady state] $V_{ab}$ will be sinusoid with average value zero $\Rightarrow$ Average $V_{ab} = 5V.$

Q12GATE 2015MCQ

The damping ratio of a series RLC circuit can be expressed as

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📖 Explanation

\begin{aligned} &\xi=\frac{1}{2 Q} ; Q=\frac{1}{R} \sqrt{\frac{L}{C}}\\ &\therefore \text { Damping ratio }=\xi=\frac{R}{2} \sqrt{\frac{C}{L}} \end{aligned}

Q13GATE 2015NAT

In the circuit shown, at resonance, the amplitude of the sinusoidal voltage (in Volts) across the capacitor is ________.

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📖 Explanation

\begin{array}{l} \text { At resonance, } I=\frac{10}{4} \\ \qquad \begin{aligned} \omega&=\frac{1}{\sqrt{0.1 \times 10^{-3} \times 10^{-6}}}=10^{5} \text { rad/sec } \\ X_{C}&=\frac{1}{\omega C}=\frac{1}{10^{5} \times 1 \times 10^{-6}}=10 \Omega \\ V_{C}&=I X_{C}=10 \times \frac{10}{4}=25 \mathrm{V} \end{aligned} \end{array}

Q14GATE 2015NAT

The voltage ( $V_{c}$ ) across the capacitor (in Volts) in the network shown is ______ .

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📖 Explanation

\begin{aligned} (80)^{2}+\left(40-V_{c}\right)^{2} &=100^{2} \\ \left(40-V_{C}\right)^{2} &=100^{2}-80^{2}=3600 \\ \left|40-V_{C}\right| &=60 \\ V_{C} &=100 \mathrm{V} \end{aligned}

Q15GATE 2015NAT

At very high frequencies, the peak output voltage $V_{0}$ (in Volts) is ________.

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📖 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}

Q16GATE 2015NAT

In the circuit shown, the current I flowing through the 50 $\Omega$ resistor will be zero if the value of capacitor C (in $\mu$ F) is ______.

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📖 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}

Q17GATE 2015MCQ

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

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📖 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}

Q18GATE 2014NAT

A periodic variable x is shown in the figure as a function of time. The root-mean-square (rms) value of x is _____.

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📖 Explanation

$\text { RMS value }=\sqrt{\frac{1}{T} \int_{0}^{T} x^{2}(t) d t}$ where, T= time period for the given signal,

\begin{array}{l} \text { RMS value }=\sqrt{\frac{1}{T} \int_{0}^{T / 2}\left(\frac{2}{T} t\right)^{2} d t+\int_{T / 2}^{T}(0)^{2} d t}\\ =\sqrt{\frac{1}{T} \int_{0}^{1 / 2} \frac{4}{T^{2}} t^{2} d t}=\sqrt{\frac{1}{T} \times \frac{4}{T^{2}} \times \left(\frac{t^{3}}{3}\right)_{0}^{T / 2}} \\ =\sqrt{\frac{4}{T^{3}}\left(\frac{T^{3}}{24}\right)=\sqrt{\frac{1}{6}}}=0.408 \end{array}

Q19GATE 2014MCQ

A 230 V rms source supplies power to two loads connected in parallel. The first load draws 10 kW at 0.8 leading power factor and the second one draws 10 kVA at 0.8 lagging power factor. The complex power delivered by the source is

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📖 Explanation

Real power $(\mathrm{kW}) \mathrm{P}=\mathrm{VI} \cos \phi \qquad \ldots(i)$ $\therefore \quad P_{1}=V \times I_{1} \cos \phi$ Thus, $l_{1}=\frac{10 \times 10^{3}}{230 \times 0.8}=54.34 \angle 36.86 \qquad \ldots(ii)$ Total power (kVA) $P=V \cdot I_{2}$ $\therefore \quad I_{2}=43.47 \angle-36.86 \qquad \ldots(iii)$ Therefore, $\vec{I}=\vec{I}_{1}+\vec{I}_{2}=78.25+j 6.25 \qquad \ldots(iv)$ So, the complex power delivered by source

\begin{aligned} S&=V I^{*}=230(78.25-j 6.25) \\ \text{or}\quad S&=(18-j 1.5) \mathrm{kVA} \end{aligned}

Q20GATE 2014MCQ

A series RC circuit is connected to a DC voltage source at time t = 0. The relation between the source voltage $V_{S}$ , the resistance R, the capacitance C, and the current $i(t)$ is given below : $V_{S}=Ri(t)+\frac{1}{C}\int_{0}^{t}i(u)du$ Which one of the following represents the current $i(t)$ ?

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📖 Explanation

Given that: $V_{s}=R i(t)+\frac{1}{C} \int_{0}^{t} i(t) d t\qquad\ldots(i)$ Using Laplace transform,

\begin{aligned} \text { WS) }&=R I(s)+\frac{1}{C s} I(s) &\ldots(ii)\\ \text { or } I(s)&=\frac{V(s)}{\left(R+\frac{1}{C S}\right)} &\ldots(iii)\\ \text { For, } V(s)&=\frac{1}{s}&\ldots(iv) \end{aligned}

From equation (iii) and (iv), $I(s)=\frac{C}{(R C s+1)}=\frac{1}{R\left(s+\frac{1}{R C}\right)}\qquad\ldots(v)$ Using inverse Laplace transform in equation (v), we get, $i(t)=\frac{1}{R} e^{-t / R C}$ Thus, option (A) is correct.

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