Q1GATE 2024NAT

In the circuit given below, the switch S was kept open for a sufficiently long time and is closed at time $t=0$ . The time constant (in seconds) of the circuit for $t > 0$ is ______

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Q2GATE 2023NAT

In the circuit shown below, switch $S$ was closed for a long time. If the switch is opened at $t=0$ , the maximum magnitude of the voltage $V_{R}$ , in volts, is ____ (rounded off to the nearest integer).

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📖 Explanation

At $t=0^{-}$ $i_{L}\left(0^{-}\right)=\frac{2}{1}=2 \mathrm{~A}$ At $t=0^{+}$ $V_{R}=-2 \times 2=-4$ , Magnitude of voltage $V_{R}$ , $\left|V_{R}\right|=4$

Q3GATE 2023NAT

A sample and hold circuit is implemented using a resistive switch and a capacitor with a time constant of $1 \mu \mathrm{s}$ . The time for sampling switch to stay closed to charge a capacitor adequately to a full scale voltage of $1 \mathrm{~V}$ with 12-bit accuracy is ____ $\mu \mathrm{s}$ . (rounded off to two decimal places)

🚀 Solve in Practice Mode

📖 Explanation

Given: Time constant $(\tau)$ of $1 \mu \mathrm{sec}$ . Full scale voltage $=1 \mathrm{~V}$ The voltage across capacitor is given as $\quad \quad V_{c}(t)=V_{\text {\in }}\left(1-e^{-t / \tau}\right)$ $\therefore \quad V_{c}(t)=\left(1-e^{-t / 1 \mu \mathrm{sec}}\right) \quad ...(i)$ To calculate the voltage to stay closed to charge capacitor adequately to a full scale voltage with 12-bit accuracy is given by

\begin{aligned} V_{c}(t) & =V_{\text {ref }}\left[1-\frac{1}{2^{n}}\right]\$ \\ n & =\text { Number of bit }=12 \\ V_{c}(t) & =\left[1-\frac{1}{4096}\right] \quad ...(ii) \end{aligned}

Comparing equations (i) and (ii), we get,

\begin{aligned} e^{-t / 1 \mu \mathrm{sec}} & =\frac{1}{4096} \\ -\frac{t}{1 \mu \mathrm{sec}} & =\ln \left\{\frac{1}{4096}\right\} \\ t & =8.3177 \mu \mathrm{sec} \end{aligned}

Q4GATE 2023MCQ

The switch $S_{1}$ was closed and $S_{2}$ was open for a long time. At $t=0$ , switch $S_{1}$ is opened and $S_{2}$ is closed, simultaneously. The value of $i_{c}\left(0^{+}\right)$ , in amperes, is

🚀 Solve in Practice Mode

📖 Explanation

At $t=0^{-} ; S_{1} \rightarrow$ closed, $S_{2} \rightarrow$ opened

\begin{aligned} i_{L}\left(0^{-}\right)&=\frac{1 \times 25}{100+25}=0.2 \mathrm{~A} \\ v_{C}\left(0^{-}\right)&=\frac{1}{5} \times 100=20 \mathrm{~V} \end{aligned}

At $t=0^{+} ; S_{1} \rightarrow$ opened, $S_{2} \rightarrow$ closed

\begin{aligned} i_{x} & =\frac{20}{25}=\frac{4}{5} \mathrm{~A}=0.8 \mathrm{~A} \\ \text{By KCL: }-i_{c} & =i_{x}+0.2=0.8+0.2 \\ \Rightarrow \quad i_{c} & =-1 \mathrm{~A} \end{aligned}

Q5GATE 2021NAT

In the circuit shown in the figure, the switch is closed at time $t=0$ , while the capacitor is initially charged to $-5\:V (i.e., v_{c}(0)=-5V)$ . The time after which the voltage across the capacitor becomes zero (rounded off to three decimal places) is ________________ $\text{ms}$ .

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} V_{c}\left(0^{-}\right)&=-5 \mathrm{~V} \\ V_{c}\left(0^{+}\right)&=-5 \mathrm{~V} \end{aligned}

$t \rightarrow \infty ,$ capacitor acts as a O.C. Write KCL at node

\begin{aligned} \frac{V_{c}(\infty )-5}{250}+\frac{V_{R}}{500}+\frac{V_{c}(\infty )}{250} &=0 \\ V_{c}(\infty ) &=\frac{5}{3} \text { Volts } \\ \text { Time constant }(\tau) &=R_{e q} C \end{aligned}

\begin{aligned} I &=\frac{V}{250}+\frac{V_{R}}{500}+\frac{V}{250} \\ V_{R} &=-V \\ I &=\frac{V}{250}-\frac{V}{500}+\frac{V}{250} \\ \frac{V}{I} &=\frac{500}{3} \Omega ; R_{\mathrm{eq}}=\frac{500}{3} \Omega \\ \tau &=\frac{500}{3} \times 0.6 \mu=0.1 \times 10^{-3} \\ v_{c}(t) &=v_{c}(\infty )+\left(v_{c}(0)-v_{c}(\infty )\right) e^{-\sqrt{2}} \\ v_{c}(t) &=\frac{5}{3}+\left(-5-\frac{5}{3}\right) e^{-t / 0.1 \times 10^{-3}} \\ 0 &=\frac{5}{3}-\frac{20}{3} e^{-10000 t} \\ t &=0.1386 \mathrm{msec} \end{aligned}

Q6GATE 2021NAT

The circuit in the figure contains a current source driving a load having an inductor and a resistor in series, with a shunt capacitor across the load. The ammeter is assumed to have zero resistance. The switch is closed at time t = 0. Initially, when the switch is open, the capacitor is discharged and the ammeter reads zero ampere. After the switch is closed, the ammeter reading keeps fluctuating for some time till it settles to a final steady value. The maximum ammeter reading that one will observe after the switch is closed (rounded off to two decimal places) is _______________ A.

🚀 Solve in Practice Mode

📖 Explanation

Apply Laplace transform,

\begin{aligned} \frac{1}{s} &=\frac{V(s)}{1 / C s}+\frac{V(s)}{R+L s} \\ \frac{1}{s} &=V(s)\left[C s+\frac{1}{R+L s}\right] \\ V(s) &=\frac{(1 / s)}{C s+\frac{1}{R+L s}} \\ V(s) &=\frac{(1 / s)(R+L S)}{L C s^{2}+R C s+1} \\ I(s) &=\frac{(1 / s)(R+L s)}{L C s^{2}+R C s+1} \cdot \frac{1}{(R+L s)} \\ I(s) &=\frac{1 / L C}{s\left(s^{2}+\frac{R}{L} s+\frac{1}{L C}\right)} \\ L &=10 \mathrm{mH} ; C=100 \mathrm{pF} ; R=5 \times 10^{3} \Omega\\ \xi &=\frac{R}{2} \sqrt{\frac{C}{L}}=\frac{5 \times 10^{3}}{2} \sqrt{\frac{100 \times 10^{-12}}{10 \times 10^{-3}}}=0.25 \\ \text{Max}. \text { over shoot } &=e^{-\pi \xi / \sqrt{1-\xi^{2}}}=0.44 \end{aligned}

Maximum value = Steady state + Max. overshoot $=1+0.44$ $=1.44 \mathrm{~A}$

Q7GATE 2021MCQ

The switch in the circuit in the figure is in position P for a long time and then moved to position Q at time t=0. The value of $\dfrac{dv\left ( t \right )}{dt}$ at $t=0^{+}$ is

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📖 Explanation

Inductor and capacitors are connected to the inductance source for a long time, so these elements have reached steady state.

\begin{aligned} i_{L}\left(0^{-}\right) &=\frac{20}{5 \mathrm{k} \Omega+5 \mathrm{k} \Omega+10 \mathrm{k} \Omega}=1 \mathrm{~mA} \\ \mathrm{~V}\left(0^{-}\right) &=10 \mathrm{~V} \\ t &=0^{+} \end{aligned}

\begin{aligned} i\left(0^{+}\right)+\frac{10}{5 k}+1 \mathrm{~m} \mathrm{~A} &=0 \\ i\left(\mathrm{O}^{+}\right) &=-3 \mathrm{~mA} \\ C \frac{d V\left(\mathrm{O}^{+}\right)}{d t} &=-3 \mathrm{~mA} \\ 1 \times 10^{-3} \frac{d V\left(\mathrm{O}^{+}\right)}{d t} &=-3 \mathrm{~mA} \\ \frac{d V\left(\mathrm{O}^{+}\right)}{d t} &=-3 \mathrm{~V} / \mathrm{s} \end{aligned}

Q8GATE 2020NAT

The current in the RL-circuit shown below is $i(t) = 10cos(5t-\pi /4)A$ . The value of the inductor (rounded off to two decimal places) is _________ H.

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📖 Explanation

$Z=\frac{V}{I}=\frac{200\angle 0^{\circ}}{10\angle -45^{\circ}}=20\angle 45^{\circ}$ $Z=10\sqrt{2}+j10\sqrt{2}$ $X_{L}=10\sqrt{2}$ $\omega L=10\sqrt{2}$ $L=\frac{10\sqrt{2}}{5}=2.828\, H$

Q9GATE 2019NAT

The RC circuit shown below has a variable resistance R(t) given by the following expression: $R(t)=R_0\left ( 1-\frac{t}{T} \right ) \; for \;0\leq t \lt T$ where $R_0=1\Omega$ , and C=1F. We are also given that $T=3R_0C$ and the source voltage is $V_s=1V$ . If the current at time t=0 is 1A, then the current I(t), in amperes, at time t=T/2 is ___________(rounded off to 2 decimal places).

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📖 Explanation

\begin{array}{c} T=3 R_{0} C=3 \mathrm{sec} \\ R(t)=\left(1-\frac{t}{3}\right) ; 0 \leq t \leq 3 \mathrm{sec} \end{array}

\begin{aligned} R(t) i(t)+\frac{1}{C} \int i(t) d t&=1 \\ \left(1-\frac{t}{3}\right) i(t)+\int i(t) d t&=1 \end{aligned}

Differentiating both sides, we get

\begin{aligned} \left(1-\frac{t}{3}\right) \frac{d i}{d t}-\frac{i}{3}+i &=0 \\ (3-t) \frac{d i}{d t}+2 i &=0 \\ \frac{d i}{i} &=-\frac{2}{(3-t)} d t \end{aligned}

Integrating on both sides, we get,

\begin{aligned} \ln (i)&=2 \ln (3-t)+\ln (c)\\ i(t)&=c(3-t)^{2} ; t \geq 0\\ \text{Given that,} i(0)&=1 A.\\ \text{So,}\quad c(3-0)^{2} &=1 \mathrm{A} \\ c &=\frac{1}{9} \mathrm{A} \\ i(t) &=\frac{1}{9}(3-t)^{2} \mathrm{A}\\ \text{At }t&=\frac{T}{2}=1.5 \mathrm{sec} \\ i(1.5)&=\frac{1}{9}(1.5)^{2}=0.25 A \end{aligned}

Q10GATE 2018NAT

For the circuit given in the figure, the magnitude of the loop current (in amperes, correct to three decimal places) 0.5 second after closing the switch is _______.

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📖 Explanation

\begin{aligned} \text { Loop current, } i(t)&=\frac{1}{1+1}\left(1-e^{-t / \tau}\right) \mathrm{A} ; t \gt 0 \\ \tau&=\frac{L}{R_{e q}}=\frac{1}{1+1}=\frac{1}{2} \mathrm{sec} \\ i(t)&=\frac{1}{2}\left(1-e^{-2 t}\right) \mathrm{A} ; t \gt 0\\ \text{At }t=0.5 \mathrm{sec} \\ i(t)&=\frac{1}{2}\left(1-e^{-1}\right) \mathrm{A}=0.316 \mathrm{A} \end{aligned}

Q11GATE 2017NAT

The switch in the circuit, shown in the figure, was open for a long time and is closed at t = 0. The current i(t) (in ampere) at t = 0.5 seconds is ________

🚀 Solve in Practice Mode

📖 Explanation

The equivalent circuit at $t = 0^{-}$ is as follows: The Laplace transform model of the circuit for $t \gt O$ is as follows: $I(s)=\frac{10}{s}-\frac{12.5}{5+2.5 s}=\frac{10}{s}-\frac{5}{s+2}$ By taking inverse Laplace transform, $i(t)=\left(10-5 e^{-2 t}\right) u(t) \mathrm{A}$ At t=0.5 seconds, $i(t)=\left(10-\frac{5}{e}\right) A=8.16 \mathrm{A}$

Q12GATE 2017NAT

In the circuit shown, the voltage $V_{IN}(t)$ is described by:

V_{IN}=\left\{\begin{matrix} 0, & for \; t \lt 0\\ 15 \; volts & for \; t\geq 0 \end{matrix}\right.

where t is in seconds. The time (in seconds) at which the current I in the circuit will reach the value 2 Amperes is ___________.

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📖 Explanation

\begin{aligned} i_{s}t)&=\frac{V}{R}\left[1-e^{\frac{-R t}{L}}\right]\$ \\ i_{s}t)&=\frac{15}{1}\left[1-e^{\frac{-3 t}{2}}\right]\$\\ \text{Current through }\\ 2 \mathrm{H} \\ i(t)&=i_{s}t) \frac{1}{1+2}\\ i(t)&=5\left[1-e^{\frac{-3 t}{2}}\right]\$ \mathrm{A}\\ \text{At }\;i(t)=2 \mathrm{A} \\ 2&=5\left[1-e^{\frac{-3 t}{2}}\right]\$\\ \text{By solving,} t&=0.3405 \mathrm{sec} \end{aligned}

Q13GATE 2016NAT

Assume that the circuit in the figure has reached the steady state before time $t= 0$ when the 3 $\Omega$ resistor suddenly burns out, resulting in an open circuit. The current i(t) (in ampere) at $t=0^+$ is__________

🚀 Solve in Practice Mode

📖 Explanation

At $t = 0^{-}$

\begin{aligned} I &=\frac{12}{6}=2 \mathrm{A} \\ V_{3 F} &=10 \times \frac{2}{5}=4 \mathrm{V} \\ V_{2 F} &=10 \times \frac{3}{5}=6 \mathrm{V} \end{aligned}

At $t = 0^{+}$ Note : As the current direction is not mentioned in the question, thus by reversing the current direction 1 A can also be the answer.

Q14GATE 2016MCQ

The switch has been in position 1 for a long time and abruptly changes to position 2 at t=0. If time t is in seconds, the capacitor voltage $V_{C}$ (in volts) for $t \gt 0$ is given by

🚀 Solve in Practice Mode

📖 Explanation

at $t = 0^{-}$ , Switch is at position-1 where, $\quad V_{c}\left(0^{-}\right)=\frac{10 \times 2}{2+3}=4 \mathrm{V} \ldots(1) \\ \therefore \quad V_{c}(0-)=V_{c}\left(0^{+}\right)=4 \mathrm{V}$ $\text{at }t = \infty$ $V_{c}(\infty )=5 \times 2=10 \mathrm{V}$ The time constant of the circuit is

\begin{aligned} \tau &=R_{\mathrm{eq}} C_{\mathrm{eq}} \\ &=(4+2) \times 0.1=0.6 \mathrm{sec} \\ \therefore V_{c}(t)=V_{c}(\infty ) &+\left[V_{c}\left(0^{+}\right)-V_{c}(\infty )\right]\$ e^{-t / \tau} \\ &=10+(4-10) e^{-t / 0.6} \\ V_{c}(t) &=\left(10-6 e^{-t / 0.6}\right) \mathrm{V} \end{aligned}

Q15GATE 2016NAT

The switch S in the circuit shown has been closed for a long time. It is opened at time t=0 and remains open after that. Assume that the diode has zero reverse current and zero forward voltage drop. The steady state magnitude of the capacitor voltage $V_{c}$ (in volts) is ______

🚀 Solve in Practice Mode

📖 Explanation

At $t = 0^{-}$ $i_{L}\left(0^{-}\right)=\frac{10}{1}=10 \mathrm{A}$ For $t>0$ (using Laplace transform)

\begin{array}{l} I(s)=\frac{10 \times 10^{-3}}{10^{-3} s+\frac{10^{6}}{10 s}} \\ V_{c}(s)=I(s) \times \frac{10^{6}}{10 s} \\ V_{c}(s)=\frac{10^{6}}{s^{2}+10^{8}} \end{array}

Taking inverse Laplace, we get $V_{c}(t)=100 \sin 10^{4} t \mathrm{V}$ $\therefore$ Steady state magnitude voltage across capacitor is 100V.

Q16GATE 2015NAT

In the circuit shown, the initial voltages across the capacitors $C_{1}$ and $C_{2}$ are 1V and 3V ,respectively. The switch is closed at time t = 0. The total energy dissipated (in Joules) in the resistor R until steady state is reached, is __________.

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📖 Explanation

\begin{aligned} \text { Initial energy } &=\frac{1}{2}\left(C_{1} V_{1}^{2}+C_{2} V_{2}^{2}\right) \\ &=\frac{1}{2}\left(3 \times 1^{2}+1 \times 3^{2}\right)=6 \mathrm{J} \end{aligned}

Final energy stored in capacitor

\begin{aligned} &= \frac{1}{2}\left(C_{1}+C_{2}\right) V^{2} \\ C_{1} V_{1}+C_{2} V_{2}&=\left(C_{1}+C_{2}\right) V \\ 1 \times 3+3 \times 1&=(1+3) V \\ V&= 1.5 \mathrm{V} \\ \text { Final energy }&=\frac{1}{2}(1+3) \times (1.5)^{2}=4.5 \mathrm{J} \\ \text { Energy dissipated }&=6-4.5=1.5 \mathrm{J} \end{aligned}

Q17GATE 2015MCQ

In the circuit shown, the switch SW is thrown from position A to position B at time t = 0. The energy (in $\mu$ J) taken from the 3 V source to charge the 0.1 $\mu$ F capacitor from 0 V to 3 V is

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Q18GATE 2015NAT

In the circuit shown, switch SW is closed at t = 0. Assuming zero initial conditions, the value of $v_{c}(t)$ (in Volts) at t = 1 sec is ________.

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📖 Explanation

\begin{aligned} & v_{c}\left(0^{-}\right)=0 \mathrm{V} \\ & v_{c}\left(0^{+}\right)=0 \mathrm{V} \end{aligned}

\begin{aligned} v_{c}(\infty )&=\frac{2}{2+3} \times 10 \\ &=4 \mathrm{V}[\text { By voltage divider }] \\ v_{c}(t)&=4\left[1-e^{-t / \tau}\right]\$ \\ \tau &=R_{\mathrm{eq}} C=\frac{3 \times 2}{3+2} \times \frac{5}{6} \\ &=1 \mathrm{sec} \\ v_{c}(1) &=4\left[1-e^{-1 / 1}\right]\$=2.528 \mathrm{Volts} \end{aligned}

Q19GATE 2014NAT

In the circuit shown in the figure, the value of capacitor C (in mF) needed to have critically damped response i(t) is _____.

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📖 Explanation

For critically damped system $\xi=1=\frac{1}{2 Q}\quad \ldots(i)$ where Q= Quality factor For series RLC circuit $Q=\frac{1}{R} \sqrt{\frac{L}{C}}\quad \ldots(ii)$ From equation (i) and (ii),

\begin{aligned} \frac{1}{R} \sqrt{\frac{L}{C}}&=1 \\ \text{or} \quad C&=\left(\frac{2}{R}\right)^{2} L=\left(\frac{2}{40}\right)^{2} \times 4=10 \mathrm{mF} \end{aligned}

Q20GATE 2014MCQ

In the figure shown, the capacitor is initially uncharged. Which one of the following expressions describes the current I(t) (in mA) for t > 0 ?

🚀 Solve in Practice Mode

📖 Explanation

Converting the given circuit into frequency domain and applying KCL at V(s) We get,

\begin{array}{l} \frac{V(s)-\frac{5}{s}}{R_{1}}+\frac{V(s)}{R_{2}}+\frac{V(s)}{\frac{1}{C s}}=0 \quad\ldots(i)\\ \because \quad R_{1}=1 \mathrm{k} \Omega, R_{2}=2 \mathrm{k} \Omega \text { and } C=1 \mu \mathrm{F} \end{array}

Using the components value we get, $V(s)\left[\frac{1}{1}+\frac{1}{2}+s\right]=\frac{5}{s}$ or $\quad V(s)=\frac{5}{s\left(s+\frac{3}{2}\right)} \quad\ldots(ii)$ Using partial fraction on equation (ii) $V(s)=\frac{10}{3 s}-\frac{10}{3\left(s+\frac{3}{2}\right)}\quad\ldots(iii)$ Using inverse Laplace transform $v(t)=\frac{10}{3}\left[1-e^{-\frac{3}{2} t}\right]\$ \quad\ldots(iv)\therefore $Current$ I=\frac{V(t)}{R_{2}}=\frac{5}{3}\left[1-e^{-\frac{3}{2} t}\right]$ m A$

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