Question 1 |
A linear 2-port network is shown in Fig. (a). An ideal DC voltage source of 10 V is
connected across Port 1. A variable resistance R is connected across Port 2. As R is
varied, the measured voltage and current at Port 2 is shown in Fig. (b) as a V_2 versus -I_2 plot. Note that for V_2=5V,I_2=0mA, and for V_2=4V,I_2=-4mA.
When the variable resistance R at Port 2 is replaced by the load shown in Fig. (c), the current I_2 is _______ mA (rounded off to one decimal place).
When the variable resistance R at Port 2 is replaced by the load shown in Fig. (c), the current I_2 is _______ mA (rounded off to one decimal place).
3.2 | |
6.8 | |
4 | |
8 |
Question 1 Explanation:
CASE-1:
I_2=0
V_{oc}=5V
CASE-2: With help of linearity
\frac{4-5}{4-0}=\frac{I_{sc}-5}{I_{sc}-0}\Rightarrow I_{sc}=20mA
R_{th}=\frac{V_{oc}}{I_{sc}}=\frac{5}{20}k=250\Omega =0.25k\Omega
CASE-3
I_2=\frac{10-5}{1.25}=4mA
Question 2 |
Consider the circuit shown in the figure with input V(t) in volts. The sinusoidal
steady state current I(t) flowing through the circuit is shown graphically (where t is
in seconds). The circuit element Z can be ________.
a capacitor of 1 F | |
an inductor of 1 H | |
a capacitor of \sqrt{3} F | |
an inductor of \sqrt{3} H |
Question 2 Explanation:
i(t)=\frac{V(t)}{1+Z} where,
V(t)= \sin t
\because Given i(t) is lagging (from plot)
\begin{aligned} I_{max}&=\frac{V_{max}}{Z}=\frac{1}{Z}\\ \Rightarrow Z&=1/\sqrt{2}\\ Z&=(1+j\omega L)\Rightarrow L=1\\ as\; \omega &=1(\sin \omega t)=\sin t \end{aligned}
\because Given i(t) is lagging (from plot)
\begin{aligned} I_{max}&=\frac{V_{max}}{Z}=\frac{1}{Z}\\ \Rightarrow Z&=1/\sqrt{2}\\ Z&=(1+j\omega L)\Rightarrow L=1\\ as\; \omega &=1(\sin \omega t)=\sin t \end{aligned}
Question 3 |
For the circuit shown, the locus of the impedance Z(j\omega ) is plotted as \omega increases
from zero to infinity. The values of R_1 and R_2 are:
R_1=2k\Omega ,R_2=3k\Omega | |
R_1=5k\Omega ,R_2=2k\Omega | |
R_1=5k\Omega ,R_2=2.5k\Omega | |
R_1=2k\Omega ,R_2=5k\Omega |
Question 3 Explanation:
\begin{aligned}
Z(j\omega )&=R_1+\frac{R_2\cdot 1/j\omega c}{R_2+ 1/j\omega c}\\
&=R_1+\frac{R_2}{\frac{1}{2}+jR_2\omega c}\\
Z(j\omega )_{\omega =0}&=R_1+R_2\\
\Rightarrow R_1+R_2&=5k\Omega \\
Z|j\omega |_{\omega \rightarrow \infty }&=R_1\\
\Rightarrow R_1&=2k\Omega \\
R_2&=3k\Omega
\end{aligned}
Question 4 |
Consider the circuit shown in the figure. The current I flowing through the 10\Omega resistor is _________.
1A | |
0A | |
0.1A | |
-0.1A |
Question 4 Explanation:
Here, there is no any return closed path for Current (I) . Hence I=0
Current always flow in loop.
Current always flow in loop.
Question 5 |
The current I in the circuit shown is ________
1.25 \times 10^{-3}A | |
0.75 \times 10^{-3}A | |
-0.5 \times 10^{-3}A | |
1.16 \times 10^{-3}A |
Question 5 Explanation:
Applying Nodal equation at Node-A
\begin{aligned} \frac{V_A}{2k}+\frac{V_A-5}{2k}&=10^{-3}\\ \Rightarrow 2V_A-5&=2k \times 10^{-3}\\ V_A&=3.5V\\ Again,&\\ I&=\frac{5-V_A}{2k}\\ &=\frac{5-3.5}{2k}\\ &=0.75 \times 10^{-3}A \end{aligned}
Question 6 |
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 6 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 7 |
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 7 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 8 |
Consider the two-port network shown in the figure.
The admittance parameters, in siemens, are
The admittance parameters, in siemens, are
y_{11}=2,\:y_{12}=-4,\:y_{21}=-4,\:y_{22}=2 | |
y_{11}=1,\:y_{12}=-2,\:y_{21}=-1,\:y_{22}=3 | |
y_{11}=2,\:y_{12}=-4,\:y_{21}=-1,\:y_{22}=2 | |
y_{11}=2,\:y_{12}=-4,\:y_{21}=-4,\:y_{22}=3 |
Question 8 Explanation:
Write KCL at V_{1}
\begin{aligned} I_{1}&=\frac{V_{1}}{1}+\frac{V_{1}-V_{2}}{1}-3 V_{2} \\ I_{1}&=2 V_{1}-4 V_{2} \\ \text{Write KCL at }V_{2}\qquad I_{2}&=\frac{V_{2}}{1}+\frac{V_{2}-V_{1}}{1}\\ I_{2}&=-V_{1}+2 V_{2} \\ [y]&=\left[\begin{array}{cc} 2 & -4 \\ -1 & 2 \end{array}\right] \mho \end{aligned}
Question 9 |
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 9 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 10 |
Consider the circuit shown in the figure.
The value of v_{0} (rounded off to one decimal place) is _________ V.
The value of v_{0} (rounded off to one decimal place) is _________ V.
0.5 | |
0.8 | |
2 | |
1 |
Question 10 Explanation:
Write KVL equation first loop,
\begin{aligned} V_{o}-4+1\left(V_{o}+2\right) &=0 \\ V_{o} &=1 \mathrm{~V} \end{aligned}
There are 10 questions to complete.