Question 1 |
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.
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.
1.44 | |
2.56 | |
8.65 | |
7.26 |
Question 1 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}
\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}
Question 2 |
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}.
The time after which the voltage across the capacitor becomes zero (rounded off to three decimal places) is ________________ \text{ms}.
0.258 | |
0.842 | |
0.139 | |
0.241 |
Question 2 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}
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}
Question 3 |
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
The value of \dfrac{dv\left ( t \right )}{dt} at t=0^{+} is
0\:V/s | |
3\:V/s | |
-3\:V/s | |
-5\:V/s |
Question 3 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}
\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}
Question 4 |
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.
4.82 | |
5.24 | |
2.83 | |
1.32 |
Question 4 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
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
Question 5 |
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).
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).
0.25 | |
0.12 | |
0.68 | |
0.72 |
Question 5 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}
\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}
Question 6 |
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 _______.
0.3 | |
0.4 | |
0.5 | |
0.2 |
Question 6 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}
Question 7 |
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 ________
1.8 | |
6.3 | |
8.16 | |
3.2 |
Question 7 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}
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}
Question 8 |
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 ___________.
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 ___________.
0.1 | |
0.34 | |
0.24 | |
0.4 |
Question 8 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}
Question 9 |
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__________
0 | |
1 | |
3 | |
4 |
Question 9 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.
\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.
Question 10 |
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 ______
The steady state magnitude of the capacitor voltage V_{c} (in volts) is ______
90 | |
100 | |
110 | |
120 |
Question 10 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.
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.
There are 10 questions to complete.