Q21GATE 2014MCQ

For maximum power transfer between two cascaded sections of an electrical network, the relationship between the output impedance $Z_{1}$ of the first section to the input impedance $Z_{2}$ of the second section is

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Q22GATE 2014NAT

In the figure shown, the ideal switch has been open for a long time. If it is closed at t = 0, then the magnitude of the current (in mA) through the 4 k $\Omega$ resistance at t = $0^{+}$ is _____.

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📖 Explanation

At steady state $t = 0_{-}$

\begin{aligned} \therefore V_{c}(0) &=V_{c}\left(0^{+}\right)=5 \mathrm{V} \\ I_{L}\left(0^{-}\right) &=I_{L}\left(0^{+}\right)=1 \mathrm{mA} \end{aligned}

At, t=0, switch get closed Thus, the current through $4\Omega$ resistance is $I=\frac{5}{4 \times 10^{3}}=1.25 \mathrm{mA}$

Q23GATE 2014NAT

In the circuit shown in the figure, the value of $v_{0}(t)$ (in volts) for $t\rightarrow \infty$ is____.

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📖 Explanation

At steady state the inductor act as a short circuit. $\therefore \quad V_{x}=5 i_{x}$ By KCL

\begin{aligned} &-10+\frac{V_{x}-2 i_{x}}{5}+i_{x}=0\\ &-10+\frac{5 i_{x}-2 i_{x}}{5}+i_{x}=0\\ \text{or}&\\ \frac{8 i_{x}}{5} &=10 \\ \Rightarrow \quad i_{x} &=\frac{50}{8} \mathrm{A} \\ \therefore v_{0}(t) &=5 i_{x}(t)=\frac{5 \times 50}{8}=31.25 \mathrm{V} \end{aligned}

Q24GATE 2012MCQ

In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current $i(t)$ for all $t$ is

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📖 Explanation

Since there is no resistance so time constant will be zero. That means as soon as the switch will be closed voltages at $C_{1}$ and $C_{2}$ will become equal and capacitor allows sudden change of voltage only if impulse of current will pass through it.

Q25GATE 2011MCQ

In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated. The switch is closed at time $t$ = 0. The current $i(t)$ at a time $t$ after the switch is closed is

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📖 Explanation

\begin{aligned} V(0) &=V\left(0^{+}\right)=\frac{Q}{C} \\ &=\frac{2.5 \times 10^{-3}}{50 \times 10^{-6}}=-50 \mathrm{V} \\ & \quad\because \text{(direction given is opposite)}\\ V(\infty )&=100 \mathrm{V} \end{aligned}

\begin{aligned} V(t) &=V(\infty )+\left[V\left(0^{+}\right)-V(\infty )\right]\$ e^{-\theta \tau} \\ &=100+(-50-100) e^{-j \tau} \\ V(t) &=100-150 e^{-t t} \\ i(t) &=C \frac{d V}{d t}=C \frac{150}{\tau} e^{-t / \tau} \\ \text { and } \tau &=10 \times 50 \times 10^{-6} \\ &=5 \times 10^{-4}\\ i(t) &=50 \times 10^{-6} \times \frac{150}{10 \times 50 \times 10^{-6}} e^{-2 \times 10^{3} t} \\ &=15 e^{-2 \times 10^{3} t} \mathrm{Amp} \end{aligned}

Q26GATE 2010MCQ

In the circuit shown, the switch S is open for a long time and is closed at $t=0$ . The current $i(t)$ for $t\geq 0^{+}$ is

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📖 Explanation

Method 1: At $t = 0^{-}$ . the circuit is shown below: Current will divide equally in both $10\Omega$ resistors as shown in diagram. At $t\geq 0^{+}$ , the circuit ins-domain is shown below: $[L = 15 mH, i(O+) = i(O^{-}) = 0.75 A]$ The above circuit can be rearranged as given below: Applying nodal analysis,

\begin{aligned} &\frac{V(s)-\frac{1.5}{s}}{10}+\frac{V(s)}{10}+\frac{V(s)+15 \times 10^{-3} \times 0.75}{10+15 \times 10^{-3} s}=0 \\ &\frac{\frac{V(s)}{10}+\frac{V(s)}{10}+\frac{V(s)}{10+15 \times 10^{-3} s}}{10 s}-\frac{15 \times 10^{-3} \times 0.75}{10+15 \times 10^{-3} \mathrm{s}} \\ &=\frac{15}{10}\left[\frac{10+15 \times 10^{-3} s+5}{5\left(10+15 \times 10^{-3} \mathrm{s}\right)}\right]\$\\ & \text { V(s) }\left[\frac{10+15 \times 10^{-3} \mathrm{s}+5}{5\left(10+15 \times 10^{-3} \mathrm{s}\right)}\right]\$ \\ =& 15\left[\frac{10+15 \times 10^{-3} \mathrm{s}-10 \mathrm{s} \times 10^{-3} \times 0.75}{10\left(10+15 \times 10^{-3} \mathrm{s}\right)}\right]\$ \\ & \text { (1 }\left.+10^{-3} \mathrm{s}\right) V(s)=\frac{\left(10+7.5 \times 10^{-3} \mathrm{s}\right)}{2 s}\left[\frac{s+1000}{1000}\right] \\ &V(s) =\frac{10000+7.5 \mathrm{s}}{2000 \mathrm{s}} \end{aligned}

\begin{array}{l} V(s)=\frac{10000+7.5 s}{2 s(s+1000)} \\ =\frac{10000}{2 \times 1000 s}+\frac{\{10000+7.5(-1000)\}}{2(-1000)(s+1000)} \\ =\frac{5}{s}-\frac{1.25}{s+1000} \end{array}

Taking inverse Laplace transform,

\begin{aligned} v(t)&=5-1.25 e^{-1000 t} \mathrm{V}\\ i(t) &=\frac{v(t)}{10} \\ &=0.5-0.125 e^{-1000 t} \mathrm{A} \end{aligned}

Method 2:

\begin{aligned} i\left(0^{+}\right)&=\frac{0.75}{2}=0.375 \mathrm{A}\\ i(\infty ) &=0.5 \mathrm{A} \\ i(t) &=i(\infty )-\left\{i(\infty )-i\left(0^{+}\right)\right\} e^{-R t L} \end{aligned}

where, R = equivalent resistance seen across L with current source opened

\begin{aligned} R &=10+(10 \| 10)=15 \Omega \\ i(t) &=0.5-\{0.5-0.375\} e^{-15 t 15 \times 10^{-3}} \\ &=0.5-0.125 e^{-1000 t} \mathrm{A} \end{aligned}

Q27GATE 2009MCQ

The time domain behavior of an RL circuit is represented by $L\frac{di}{dt}+Ri=V_{0}(1+Be^{-Rt/L}sin t)u(t)$ . For an initial current of $f(0)=\frac{V_{0}}{R}$ , the steady state value of the current is given by

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Q28GATE 2009MCQ

The switch in the circuit shown was on position a for a long time, and is move to position b at time $t=0$ . The current $i(t)$ for $t \gt 0$ is given by

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📖 Explanation

$t=0^{-}$ $t\gt 0$

\begin{aligned} C &=0.16 \mu \mathrm{F} \\ i(t) &=\frac{v\left(0^{+}\right)}{R} e^{-t / R C} \cdot u(t) \\ v\left(0^{+}\right) &=100 \mathrm{V} \\ \frac{1}{R C} &=\frac{1}{5 \times 10^{3} \times 0.16 \times 10^{-6}} \\ R &=5 \mathrm{k} \Omega \\ i(t) &=20 \mathrm{e}^{-1250 t} u(t) \mathrm{mA} \end{aligned}

Q29GATE 2008MCQ

In the following circuit, the switch S is closed at t = 0. The rate of change of current $\frac{di}{dt}(0^{+})$ is given by

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📖 Explanation

At t = o, the inductor behaves as an open circuit

\begin{aligned} \text { So, } \quad V_{L}&=I_{S} R_{S} \\ \qquad V_{L}&=L \frac{d i}{d t}\left(0^{+}\right) \\ \Rightarrow \quad \frac{d i}{d t}\left(0^{+}\right)&=\frac{V_{L}}{L}=\frac{I_{s} R_{S}}{L} \end{aligned}

Q30GATE 2008MCQ

The following series RLC circuit with zero initial conditions is excited by a unit impulse function $\delta (t).$ For $t \gt 0$ , the voltage across the resistor is

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📖 Explanation

\begin{array}{l} V_{R}(s)=\frac{1}{\left(s+1+\frac{1}{s}\right)} \cdot 1=\frac{s}{s^{2}+s+1} \\ =\frac{\left(s+\frac{1}{2}\right)}{\left(s+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \frac{1 / 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{2}{\sqrt{3}}}{\left(s+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ V_{R}(t)=e^{-t / 2} \cos \left(\frac{\sqrt{3}}{2} t\right)-\frac{1}{\sqrt{3}} e^{-t / 2} \sin \left(\frac{\sqrt{3}}{2} t\right) \\ V_{A}(t)=e^{-t / 2}\left[\cos \left(\frac{\sqrt{3}}{2} t\right)-\frac{1}{\sqrt{3}} \sin \left(\frac{\sqrt{3}}{2} t\right)\right] \end{array}

Q31GATE 2008MCQ

The circuit shown in the figure is used to charge the capacitor C alternately from two current sources as indicated. The switches $S_{1}$ and $S_{2}$ are mechanically coupled and connected as follows: For 2nT $\leq$ t $\leq$ (2n+1)T, (n=0,1,2,..) $S_{1}$ to $P_{1}$ and $S_{2}$ to $P_{2}$ For (2n+1)T $\leq$ t $\leq$ (2n+2)T, (n=0,1,2,...) $S_{1}$ to $Q_{1}$ and $S_{2}$ to $Q_{2}$ Assume that the capacitor has zero initial charge. Given that u(t) is a unit step function, the voltage $V_{c}(t)$ across the capacitor is given be

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📖 Explanation

The waveform of voltage $V_{c}$ (t) is shown below. In mathematical form,

\begin{aligned} 0 \lt t \lt T, \\ C&=1 \mathrm{F}, I=1 \mathrm{A} \\ V_{c}&=\int_{0}^{t} d t=t\\ at t=T ; \quad V_{c}&=T \\ T \lt t \lt 2 T \\ V_{c}&=T-\int_{T}^{t} d t=2 T-t \\ \text { at } t=2 T ; \quad V_{C}&=0 \\ 2 T \lt t \lt 3 T \\ V_{C}&=\int_{2 T}^{t} d t=t-2 T\\ at t=3 T ; \quad V_{C}&=T \\ 3 T \lt t \lt 4 T \\ V_{C}=& T-\int_{3 T}^{t} d t=4 T-t \\ \therefore \quad V_{c}(t)=& t u(t)-2(t-T) u(t-2 T) \\ &+2(t-2 T) u(t-2 T) \ldots \\ =t u(t)+& 2 \sum_{n=1}^{\infty }(-1)^{n}(t-n T) u(t-n T) \end{aligned}

Q32GATE 2007MCQ

In the circuit shown, $V_{C}$ is 0 volts at t = 0 sec. For $t \gt 0$ , the capacitor current $i_{C}(t)$ , where t is in seconds is given by

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📖 Explanation

At $t=0^{+}$ Capacitor is short circuit and at $t=\infty$ Capacitor is open circuit So, $\quad I_{c}\left(0^{+}\right)=\frac{10 \mathrm{V}}{20 \mathrm{k} \Omega}=0.5 \mathrm{mA}$ $I_{\infty }=0$ Using direct formula $I_{c}(t)=I_{c}(\infty )-\left[I_{c}(\infty )-I_{c}(0)\right] e^{-t / \tau}$ $I_{c}(t)=0-(0-0.5) e^{-t / 40 \mathrm{msec}}$ $I_{c}(t)=0.5 e^{-25 t} \mathrm{mA}$

Q33GATE 2006MCQ

A 2 mH inductor with some initial current can be represented as shown below, where s is the Laplace Transform variable. The value of initial current is

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📖 Explanation

\begin{aligned} V &=\frac{L d I}{d t} \\ V(s) &=s L I(s)-L I\left(0^{+}\right) \\ \Rightarrow \quad-L I\left(0^{+}\right) &=-1 \mathrm{mV} \text { (given \in question)} \\ I\left(0^{+}\right)&=\frac{1 \mathrm{mV}}{2 \mathrm{mH}}=0.5 \mathrm{A} \end{aligned} \\

Q34GATE 2006MCQ

In the figure shown below, assume that all the capacitors are initially uncharged. If $v_{i}(t) = 10u(t)$ Volts, $v_{o}(t)$ is given by

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📖 Explanation

\begin{aligned} z_{1} &=\frac{R_{1} \times \frac{1}{C_{1} S}}{R_{1}+\frac{1}{C_{1} S}}=\frac{R_{1}}{R_{1} C_{1} s+1} \\ &=\frac{1 \mathrm{k}}{4 \times 10^{-3} s+1} \\ Z_{2} &=\frac{R_{2} \times \frac{1}{C_{2} S}}{R_{2}+\frac{1}{C_{2} S}} \\ &=\frac{R_{2}}{R_{2} C_{2} S+1}=\frac{4 \mathrm{k}}{4 \times 10^{-3} s+1} \\ V_{o}(t) &=\left(\frac{Z_{2}}{Z_{1}+Z_{2}}\right) V_{i}(t) \\ Z_{2} &=4 Z_{1} \\ \frac{Z_{2}}{Z_{1}+Z_{2}} &=\frac{4 Z_{1}}{Z_{1}+4 Z_{1}}=\frac{4 Z_{1}}{5 Z_{1}}=0.8 \\ v_{0}(t) &=0.8 v_{1}(t)=0.8 \times 10 u(t) \\ v_{0}(t) &=8 u(t) \end{aligned}

Q35GATE 2005MCQ

A square pulse of 3 volts amplitude is applied to C-R circuit shown in the figure. The capacitor is initially uncharged. The output voltage $V_o$ at time t=2 sec is

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📖 Explanation

\begin{aligned} \mathrm{RC} &=0.1 \times 10^{-6} \times 10^{3}=10^{-4} \\ &=100 \mu \mathrm{s} \end{aligned}

As RC is very small, so steady we reached in 2 sec. state

\begin{array}{l} V_{c}=3 \mathrm{V} \\ V_{2}=-V_{C}=-3 \mathrm{V} \end{array}

Q36GATE 2004MCQ

For the R - L circuit shown in the figure, the input voltage $v_{i}$ (t) = u(t). The current i(t) is

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📖 Explanation

\begin{aligned} I(s)&=\frac{V(s)}{s+2}=\frac{1}{s(s+2)} \\ I(s)&=\frac{1}{s(s+2)}=\frac{1}{2}\left[\frac{1}{s}-\frac{1}{s+2}\right] \\ i(t)&=\frac{1}{2}\left(1-e^{-2 t}\right) \\ \text { At } \quad t&=0, i(t)=0 \\ t&=\infty , i(t)=0.5 \\ t&=\frac{1}{2}, i(t)=0.31 \end{aligned}

Graph (c) satisfies all conditions .

Q37GATE 2004MCQ

The circuit shown in the figure has initial current $i_{L}(0^{-}) = 1 A$ through the inductor and an initial voltage $V_{C}(0^{-})=- 1 V$ across the capacitor. For input v(t)=u(t), the Laplace transform of the current i(t) for t $\geq$ 0 is

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📖 Explanation

$\mathrm{KVL}, \quad v(t)=R i(t)+\frac{L d i(t)}{d t}+\frac{1}{C} \int_{0}^{\infty } i(t) d t$ Taking Laplace transform on both sides, $V(s)=R I(s)+L s I(s)-L I\left(0^{+}\right)+\frac{I(s)}{s C}+\frac{V_{c}\left(0^{+}\right)}{s}$ $\Rightarrow \frac{1}{s}=I(s)+s I(s)-1+\frac{I(s)}{s}-\frac{1}{s}$ $\frac{2}{s}+1=\frac{I(s)}{s}\left[s^{2}+s+1\right]$ $I(s)=\frac{s+2}{s^{2}+s+1}$

Q38GATE 2003MCQ

Assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0. $I_{1}(s)$ and $I_{2}(s)$ are the Laplace transforms of $i_{1}(s)$ and $i_{2}(s)$ respectively. The equations for the loop currents $I_{1}(s)$ and $I_{2}(s)$ for the circuit shown in the figure,after the switch is brought from position 1 to position 2 at t = 0, are

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📖 Explanation

When switch is in position 2, KVL in loop (1), $I_{1}(s) \cdot R+\frac{V}{s}+I_{1}(s) \cdot \frac{1}{s C}+\left[I_{1}(s)-I_{2}(s)\right] s L=0$

\begin{array}{l} \Rightarrow I_{1}(s)\left[R+\frac{1}{s C}+s L\right]-I_{2}(s) \cdot s L=\frac{-V}{s} \\ \text { KVL \in } \operatorname{loop}(2) \\ {\left[I_{2}(s)-I_{1}(s)\right] s L+I_{2}(s) R+I_{2}(s) \cdot \frac{1}{s C}=0} \\ \Rightarrow-I_{1}(s) \cdot s L+I_{2}(s)\left[R+S L+\frac{1}{S C}\right]=0 \\ {\left[\begin{array}{cc} R+s L+\frac{1}{s C} & -s L \\ -s L & R+s L+\frac{1}{s C} \end{array}\right]\left[\begin{array}{c} I_{1}(s) \\ I_{2}(s) \end{array}\right]=\left[\begin{array}{c} -\frac{v}{s} \\ 0 \end{array}\right]} \end{array}

Q39GATE 2002MCQ

In figure, the switch was closed for a long time before opening at t = 0. the voltage $V_x$ at t = $0^{+}$ is

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📖 Explanation

When switch was closed circuit was in steady state, $i_{L}(O-) = 2.5 A$ At, $t = 0^{+}$

\begin{aligned} \Rightarrow & & V=I R \\ & &=2.5 \times 20=50 \mathrm{V} \end{aligned}

$\therefore \quad V_{x}=-50 \mathrm{V}$ (Polarity of $V_{x}$ is given reverse of n)