Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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 6 |
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 6 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 7 |
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 7 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.
Question 8 |
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
If time t is in seconds, the capacitor voltage V_{C} (in volts) for t \gt 0 is given by
4(1 - exp(-t/0.5)) | |
10 - 6 exp(-t/0.5) | |
4(1 - exp(-t/0.6)) | |
10 - 6 exp(-t/0.6) |
Question 8 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}
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}
Question 9 |
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 __________.
0 | |
0.5 | |
1 | |
1.5 |
Question 9 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}
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}
Question 10 |
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 ________.
1 | |
2 | |
2.5 | |
3 |
Question 10 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}
\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}
There are 10 questions to complete.