Q2GATE 2023NAT

The circuit shown in the figure is initially in the steady state with the switch $\mathrm{K}$ in open condition and $\bar{K}$ in closed condition. The switch $\mathrm{K}$ is closed and $\overline{\mathrm{K}}$ is opened simultaneously at the instant $t=t_{1}$ , where $t_{1} \gt 0$ . The minimum value of $t_{1}$ in milliseconds, such that there is no transient in the voltage across the $100 \mu \mathrm{F}$ capacitor, is ___ (Round off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

Case (i): Switch $K$ is open and $\bar{K}$ is closed. Redraw the circuit : From circuit, using current division,

\begin{aligned} \mathrm{i}_{\mathrm{C}} & =\frac{10}{10-\mathrm{j} 10} \times 1 \angle 0^{\circ} \\ \therefore \quad \mathrm{V}_{\mathrm{C}} & =\frac{10}{10-\mathrm{j} 10} \times (-\mathrm{j} 10)=7.07 \angle-45^{\circ} \mathrm{V} \\ \therefore \quad \mathrm{V}_{\mathrm{C}}\left(\mathrm{t}_{1}\right) & =7.07 \sin \left(1000 \mathrm{t}-45^{\circ}\right) \mathrm{V} \end{aligned}

Case (ii) : Switch $K$ is closed and $\bar{K}$ is open. Current source and $10 \Omega$ resistor becomes short circuited. Redraw the circuit : From circuit,

\begin{aligned} \mathrm{V}_{\mathrm{C}}(\infty ) & =5 \mathrm{~V} \\ \tau & =\mathrm{RC}=10 \times 100 \times 10^{-6}=1 \mathrm{msec} \end{aligned}

We have, $\mathrm{V}_{\mathrm{C}}(\mathrm{t})=\mathrm{V}_{\mathrm{C}}(\infty )+\left[\mathrm{V}_{\mathrm{C}}(0)-\mathrm{V}_{\mathrm{C}}(\infty )\right] \mathrm{e}^{-\mathrm{t} / \tau}$ $=5+\left(7.07 \sin \left(1000 t_{1}-45^{\circ}\right)-5\right) e^{-t / \tau}$ For transient free voltage,

\begin{aligned} 7.07 \sin \left(1000 \mathrm{t}_{1}-45^{\circ}\right)&=5 \\ 1000 \mathrm{t}_{1}-\frac{\pi}{4}&=\frac{\pi}{4} \\ \Rightarrow \quad t_{1}&=1.57 \mathrm{msec} . \end{aligned}

Q3GATE 2022NAT

In the circuit shown below, the switch $S$ is closed at $t=0$ . The magnitude of the steady state voltage, in volts, across the $6 \Omega$ resistor is _________. (round off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

Concept: At steady state, capacitor behaves as open circuit. Using voltage division, $V=\frac{2}{2+2} \times 10=5V$

Q4GATE 2021MCQ

In the circuit, switch 'S' is in the closed position for a very long time. If the switch is opened at time $t = 0$ , then in $i_{L} (t)$ amperes, for $t\geq0$ is

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📖 Explanation

At $t = 0^-$ $i_{L}\left(0^{-}\right)=\frac{10}{1}=10 \mathrm{~A}$ For $t > 0$ At $t = \infty$ $i(\infty )=\frac{40}{5}=8 \mathrm{~A}$ $R_{\text{eq}}:$

\begin{aligned} R_{\mathrm{eq}} &=5 \Omega \\ \tau &=\frac{L}{R_{\mathrm{eq}}}=\frac{0.5}{5}=0.1 \mathrm{sec} \\ i(t) &=8+[10-8] e^{-t / 0.1} \\ &=8+2 e^{-10 t} \mathrm{~A} \end{aligned}

Q5GATE 2021NAT

A $\text{100 Hz}$ square wave, switching between $\text{0 V}$ and $\text{5 V}$ , is applied to a $\text{CR}$ high-pass filter circuit as shown. The output voltage waveform across the resistor is $\text{6.2 V}$ peak-to-peak. If the resistance R is $\text{820$ \Omega $}$ , then the value C is ______________ $\mu F$ . (Round off to 2 decimal places.)

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} v_{0}&=v_{i}-v_{c}\\ \text{For }1^{\text {st }}\text{ half cycle}, \quad v_{0}&=5-v_{c} \\ \text{For }2^{\text {nd }}\text{ half cycle}, \quad v_{0}&=-v_{c}\\ v_{p-p} &=\left(5-V_{c \;\min}\right)-\left(-V_{c}\; \max \right) \\ 6.2 &=5+V_{c \;\max }-V_{c \;\min} \\ \Rightarrow \quad V_{c \max }-V_{c \;\min } &=1.2 \ldots(\alpha) \end{aligned}

For first half cycle i.e. $0 \lt t \lt \frac{T}{2}$

\begin{aligned} v_{c}\left(0^{+}\right) &=v_{c}(0)=v_{c}\left(0^{-}\right)=v_{c} \min \\ v_{c}(\infty ) &=5 \mathrm{~V} \\ \therefore \qquad\qquad v_{c}(t) &=v_{c}(\infty )+\left[v_{c}\left(0^{+}\right)-v_{c}(\infty )\right]\$ e^{-t / \tau} \\ v_{c}(t) &=5+\left[V_{c m i n}-5\right] e^{-t / 2 \tau}=V_{c m a x} \\ \Rightarrow \qquad\qquad V_{c} \max &=5\left[1-e^{-T / 2 \tau}\right]\$+V_{c m i n} e^{-T / 2 \tau} \end{aligned}

$For \frac{T}{2} \lt t \lt T$

\begin{aligned} v_{c}(t) &=v_{c}\left(\frac{T}{2}\right) e^{-t(t-T / 2) \tau} \\ \therefore\qquad v_{c}(t) &=V_{c m a x} e^{-(t-T / 2) \tau} \\ \text{at } t =T,\qquad \qquad v_{c} &=V_{c} \mathrm{~min} \\ \Rightarrow \qquad \qquad V_{C \mathrm{~min}} &=V_{C \max } e^{-T / 2 \tau}\\ \text { As } \quad V_{c \text { max }}-V_{c \text { min }}&=1.2 \qquad \qquad [\text { From }(\alpha)]\\ \therefore \quad V_{\text {cmax }}-V_{c \max } e^{-T / 2 t} &=1.2 \\ V_{c} \max &=\frac{1.2}{1-e^{-T / 2 \tau}} \\ \Rightarrow \qquad\qquad V_{c} \max &=\frac{1.2}{1-e^{-T / 2 \tau}}=5\left[1-e^{-T / 2 \tau}\right]\$+V_{c \min } e^{-2 \tau} \end{aligned}

From (ii),

\begin{aligned} V_{c \max } &=5\left[1-e^{-T / 2 \tau}\right]\$+\left(V_{c m a x} e^{-T / 2 \tau}\right) e^{-T / 2 \tau} \\ V_{c \max } &=5\left[1-e^{-T / 2 \tau}\right]\$+V_{c \max } e^{-T / \tau} \\ \Rightarrow \qquad \qquad V_{c \max }\left[1-e^{-T / \tau}\right]\$ &=5\left[1-e^{-T / 2 \tau}\right]\$ \\ V_{c} \max &=\frac{5\left[1-e^{-T / 2 \tau}\right]\$}{\left[1+e^{-T / 2 \tau}\right]\$\left[1-e^{-T / 2 \tau}\right]\$} \end{aligned}

Using equation (iii)

\begin{aligned} \frac{1.2}{1-e^{-T / 2 \tau}} &=\frac{5}{1+e^{-T / 2 \tau}} \\ \Rightarrow \qquad \qquad 1.2+1.2 e^{-T / 2 \tau} &=5-5 e^{-\pi / 2 t} \\ \Rightarrow \qquad \qquad 6.2 e^{-T / 2 \tau} &=3.8 \\ e^{-T / 2 \tau} &=\frac{3.8}{6.2}=0.6129 \\ \frac{T}{2 \tau} &=0.4895\\ \text { as }\qquad \qquad T&=\frac{1}{f}=\frac{1}{100} \mathrm{sec}\\ \text { and }\qquad \qquad \tau & =R C=820 \mathrm{C} \\ \Rightarrow \qquad \qquad \frac{1}{(100)(2)(820) C} & =0.4895 \\ \therefore \qquad \qquad C & =12.46 \mu \mathrm{F} \end{aligned}

Q6GATE 2017MCQ

The switch in the figure below was closed for a long time. It is opened at t=0. The current in the inductor of 2 H for $t \geq 0$ , is

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📖 Explanation

From the given circuit, consider the following circuit diagram, After rearrangement For $t \geq 0$ $I_0=i(0^-)=2.5A$ We can write, $i(t)=I_0e^{-\frac{Rt}{L}}$ $i(t)=2.5e^{-4t}A$

Q7GATE 2017NAT

The initial charge in the 1 F capacitor present in the circuit shown is zero. The energy in joules transferred from the DC source until steady state condition is reached equals ______. (Give the answer up to one decimal place.)

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📖 Explanation

Consider the following circuit diagram, After minimizing circuit elements we can have the following circuit, Here, $\tau =RC=5 sec.$ Now current, $i(t)=\frac{V}{R}e^{\frac{t}{\tau }}$ $\;\;=\frac{10}{5}e^{-t/5}=2e^{-0.2t}$ Energy supplied by the source, $E=\int_{0}^{\infty }10 \times 2e^{-0.2t}dt$ $\;\;=100J$

Q8GATE 2016NAT

In the circuit shown, switch $S_2$ has been closed for a long time. At time t=0 switch $S_1$ is closed. At $t = 0^{+}$ , the rate of change of current through the inductor, in amperes per second, is _____.

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📖 Explanation

KCL at node A, $\frac{V_A-3}{1}+\frac{3}{2}+\frac{V_A-3}{2}=0$ $2(V_A-3)+3+(V_A-3)=0$ $3V_A=6$ $V_A=2$ $V_A=L\frac{di(0^+)}{dt}=2$ $\frac{di(0^+)}{dt}=\frac{2}{L}=\frac{2}{1}=2A/sec$

Q9GATE 2016NAT

In the circuit shown below, the initial capacitor voltage is 4 V. Switch $S_1$ is closed at t=0. The charge (in $\mu$ C) lost by the capacitor from t=25 $\mu s$ to t=100 $\mu s$ is ____________.

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📖 Explanation

$i(t)=\frac{4}{5}e^{-t/\tau }$ $\tau =RC$ $\;\;=20 \times 10^{-6}sec$ Change lost by capacitor from $t=25\mu s$ to $100\mu s$ is $\int_{25 \mu s}^{100\mu s}i(t)dt=6.99 \times 10^{-6}C$

Q10GATE 2015MCQ

A series RL circuit is excited at t = 0 by closing a switch as shown in the figure. Assuming zero initial conditions, the value of $\frac{d^{2}i}{dt^{2}}\; at \; t=0^{+}$ is

🚀 Solve in Practice Mode

📖 Explanation

Initially $(t=0^-)$ the inductor would be uncharged. So, $I(0^+)=0$ The KVL in th loop will be $V=RI+L\frac{dI}{dt}$ At $t=0^+$ $V=RI(0^+)+L\frac{dI}{dt}(0^+)$ Since, $I(0^+)=0$ So, $\frac{dI}{dt}(0^+)=\frac{V}{L}$ Now, lets differentiate the above equation So, $\frac{dV}{dt}=R\frac{dI}{dt}+L\frac{d^2I}{dt^2}$ $\;\;0=R\frac{dI}{dt}+L\frac{d^2I}{dt^2}$ At $t=0^+$ $0=R\frac{dI}{dt}(0^+)+L\frac{d^2I}{dt^2}(0^+)$ So, $\frac{d^2I}{dt^2}(0^+)=-\frac{R}{L^2}\cdot V$

Q11GATE 2014MCQ

The switch SW shown in the circuit is kept at position '1' for a long duration. At t=0+, the switch is moved to position '2'. Assuming $|V_{o2} | \gt |V_{o1}|$ , the voltage $v_{c}(t)$ across the capacitor is

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Q12GATE 2014MCQ

A combination of 1 $\mu$ F capacitor with an initial voltage $v_c(0)=-2V$ in series with a 100 $\Omega$ resistor is connected to a 20 mA ideal dc current source by operating both switches at t=0s as shown. Which of the following graphs shown in the options approximates the voltage $v_s$ across the current source over the next few seconds ?

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📖 Explanation

Given $C=1\mu F, V_c(0)=-2V$ , $R=100\Omega , I=20mA$ . Circuit fot the given condition at time $t \gt 0$ is shown below: Applying KVL, we have, $V_s(s)=\left ( \frac{-2}{s} \right )+\frac{I}{s} \left ( R+\frac{I}{Cs} \right )$ $\;\;=\frac{1}{s}\left[ -2+IR+\frac{I}{Cs} \right]$ $\;\;=\frac{1}{s}\left[ (IR-2) +\frac{I}{Cs}\right]$ Putting values of R, C and I, we get, $V_s(s)=\frac{1}{s}\left[ (20 \times 10^{-3} \times 200-2)+\left ( \frac{20 \times 10^{-3}}{10^{-6}} \right ) \times \frac{1}{s} \right]$ $\;\;=\frac{1}{s}\left[ (2-2)+20 \times 10^3 \times \frac{1}{s} \right]$ $\;\;=\frac{20 \times 10^3}{s^2}$ $\therefore \;\; V_s(s)=\frac{20 \times 10^3}{s^2}$ $V_s(t)=20000 t u(t)$ $\therefore \;\; V_s(s)=(20000)tu(t)$ Which is equation of a straight line passing through origin. Hence option (C) is correct.

Q13GATE 2012MCQ

In the following figure, $C_1 \; and \; C_2$ are ideal capacitors. $C_1$ 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

Circuit is s-domain By applying KVL, $\frac{12}{s}+\frac{I(s)}{s}\left ( \frac{1}{C_1} +\frac{1}{C_2}\right )=0$ $I(s)=-\frac{12C_1C_2}{C_1+C_2}=k(constant)$ $\Rightarrow \;\; i(t)=k\delta (t)$ $\therefore$ Current $i(t)$ is an implulse function.

Q14GATE 2010MCQ

The L-C circuit shown in the figure has an inductance L=1mH and a capacitance C= $\mu$ F. The initial current through the inductor is zero, while the initial capacitor voltage is 100 V. The switch is closed at t=0. The current i through the circuit

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📖 Explanation

Initial current through the inductor is zeroand capacitor voltage is charged upto voltage $V_c(0^-)=100V$ As current through inductor and voltage across capacitor can not change abruptly So, after closing the switch, $i_L(0^+)=i_L(0^-)=0$ and $V_c(0^+)=V_c(0^-)=100V$ The circuit id s-domain $I(s)=\frac{100/s}{\left ( sL+\frac{1}{sC} \right )}$ $\;\;=\frac{100}{L}\left ( \frac{1}{s^{2+}\frac{1}{LC}} \right )$ $\;\;=100\sqrt{\frac{C}{L}}\left ( \frac{1/\sqrt{LC}}{s^{2+}(1/\sqrt{LC})^2} \right )$ taking inverse laplace transform $i(t)=L^{-1}[I(s)]$ $\;\;=100\sqrt{\frac{C}{L}} \sin \frac{1}{\sqrt{LC}}t$ $\;\;=100 \times \sqrt{\frac{10 \times 10^3}{1 \times 10^{-3}}}\times$ $\sin\left ( \frac{1}{\sqrt{1 \times 10^{-3} \times 10 \times 10^{-6}}} \right )$ $i(t)=10\sin (10^4 t)A$

Q15GATE 2010MCQ

The switch in the circuit has been closed for a long time. It is opened at t = 0. At t = $0^+$ , the current through the 1 $\mu$ F capacitor is

🚀 Solve in Practice Mode

📖 Explanation

As the switch has been closed for a long time, the circuit is in steady state. At steadystate, capacitor is open circuit, Using KVL, $5-I-4I=0 I=1A$ $V_c(0^-)=4 \times 1=4V$ As the voltage across capacitorcan not change abruptly, So, $V_c(0^+)=V_c(0^-)=4V$ Circuit at $t=0^+$ Current through capacitor at $t=0^+$ $I_c(0^+)=\frac{4}{4}=1A$

Q16GATE 2009MCQ

In the figure shown, all elements used are ideal. For time $t \lt 0$ , $S_1$ remained closed and $S_2$ open. At t=0, $S_1$ is opened and $S_2$ is closed. If the voltage $V_{C2}$ across the capacitor $C_2$ at t=0 is zero, the voltage across the capacitor combination at t= $0^+$ will be

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📖 Explanation

At $t=0^-, S_1$ is closed, $S_2$ is open $C_1$ gets charged upto 3 V Charge stored in $C_1$ $Q_0=C_1V=1 \times 3=3C$ Voltage across $C_2$ is zero at $t=0^-$ , so no charge is stored in $C_2$ . At $t \gt 0,\;S_1$ is open and $S_2$ is closed. Charge stored ( $Q_0$ ) initially in $C_2$ gets redistributed between $C_1$ and $C_2$ . Let charge stored in $C_1=Q_1$ Charge stored in $C_2=Q_1$ According to conservation of charge $Q_1+Q_2=Q_0=3\;\;...(i)$ Voltage across $C_1$ =Voltage across $C_2$ $\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$ $\rightarrow \;\; \frac{Q_1}{1}=\frac{Q_2}{2}$ $Q_2=2Q_1\;\;...(ii)$ Solving equation (i) and (ii) we get $Q_1=1 C$ and $Q_2 =2 C$ Voltage across combination $= \frac{Q_1}{C_1}=\frac{1}{1}=1$

Q17GATE 2008MCQ

The current i(t) sketched in the figure flows through a initially uncharged 0.3 nF capacitor. The capacitor charged upto 5 $\mu$ s, as per the current profile given in the figure, is connected across an inductor of 0.6 mH. Then the value of voltage across the capacitor after 1 $\mu$ s will approximately be

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📖 Explanation

Capacitor charge upto $5\mu s$ , so total charge stored in capacitor = Q = 13 nC. Voltage across the capacitor before connecting to inductor $V_0=\frac{Q}{C}$ $\;\;=\frac{13 \times 10^{-9}}{0.3 \times 10^{-9}}=43.33V$ Voltage across the capacitor at time $t$ $V_c(t)$ at $t= 1\mu s$ $V_c(t)|_{t=1\mu s}=[V_0 \cos \omega _0 t]_{t=1\mu s}$ $\omega _0t=\frac{1}{\sqrt{0.6 \times 10^3 \times 0.3 \times 10^{-9}}}\times 1 \times 10^{-6}$ $\;\; =2.357 rad=135^{\circ}$ $V_c(t)|_{t=1\mu s}=43.33 \times \cos 135^{\circ}$ $\;\;\approx -30.6 V$

Q18GATE 2008MCQ

The time constant for the given circuit will be

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📖 Explanation

For finding time constant, we neglect current source as a open circuit. Therefore, circuit becomes Therefore, Time constant $=R_{eq}C_{eq}$ $=6 \times \frac{2}{3}=4sec$

Q19GATE 2007MCQ

In the circuit shown in figure. Switch $SW_1$ is initially closed and $SW_2$ is open. The inductor L carries a current of 10 A and the capacitor charged to 10 V with polarities as indicated. $SW_2$ is closed at t= $0^{-}$ and $SW_1$ is opened at t=0. The current through C and the voltage across L at t= $0^{+}$ is

🚀 Solve in Practice Mode

📖 Explanation

By KCL, $\frac{V_L}{10}-10+\frac{V_L-10}{10}=0$ $2V_L=110$ $V_L=55V$ $I_C=\frac{55-10}{10}=4.5A$

Q20GATE 2007MCQ

In the figure, transformer $T_1$ has two secondaries, all three windings having the same number of turns and with polarities as indicated. One secondary is shorted by a 10 $\Omega$ resistor R, and the other by a 15 $\mu$ F capacitor. The switch SW is opened (t=0) when the capacitor is charged to 5 V with the left plate as positive. At t=0+ the voltage $V_P$ and current $I_R$ are

🚀 Solve in Practice Mode

📖 Explanation

All the three windings has same number of turns, so magnitude of induced emf's in all the three windings will be same i.e. $|V_P|=|V_s|=|V_T|$ Polarity of the winding is decided on the basis of dot-convention. As capacitor is carged to 5V with left plate as positive So, $T_1$ is positive w.r.t. $T_2$ $V_T=V_{T_1}-V{T_2}=5V$ As $T_2$ has negative polarity. So $P_1$ has negative polarity. Therefore , $V_P=V_{P_1}-V{P_2}=-5V$ Similarly, $S_1$ has negative polarity So, $V_s=V_{s_1}-V{s_2}=-5V$ $I_R=\frac{V_s}{R}=\frac{-5}{10}=-0.5A$

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