Question 1 |
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).
0.87 | |
1.57 | |
1.88 | |
2.26 |
Question 1 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}
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}
Question 2 |
The admittance parameters of the passive resistive two-port network shown in the figure are
\mathrm{y}_{11}=5 \mathrm{~S}, \mathrm{y}_{22}=1 \mathrm{~S}, \mathrm{y}_{12}=\mathrm{y}_{21}=-2.5 \mathrm{~S}
The power delivered to the load resistor R_{L} in Watt is ____ (Round off to 2 decimal places).
\mathrm{y}_{11}=5 \mathrm{~S}, \mathrm{y}_{22}=1 \mathrm{~S}, \mathrm{y}_{12}=\mathrm{y}_{21}=-2.5 \mathrm{~S}
The power delivered to the load resistor R_{L} in Watt is ____ (Round off to 2 decimal places).
238 | |
452.25 | |
632.12 | |
145.25 |
Question 2 Explanation:
Y =[Y]_{A}+[Y]_{B}
\begin{aligned} Y_{A} & =\left[\begin{array}{cc} \frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{array}\right] \\ Y_{B} & =\left[\begin{array}{cc} 5 & -2.5 \\ -2.5 & 1 \end{array}\right] s \\ Y & =\left[\begin{array}{cc} \frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{array}\right]+\left[\begin{array}{cc} 5 \\ -2.5 & 1 \end{array}\right] \\ & =\left[\begin{array}{cc} \frac{16}{3} & -\frac{8.5}{3} \\ -\frac{8.5}{3} & \frac{4}{3} \end{array}\right] \\ I_{1} & =\frac{16}{3} V_{1}-\frac{8.5}{3} V_{2} \\ I_{2} & =-\frac{8.5}{3} V_{1}+\frac{4}{3} V_{2} \end{aligned}
Put I_{2}=0 and V_{1}=20 \mathrm{~V}
0=-\frac{8.5}{3} \times 20+\frac{4}{3} V_{2} \begin{aligned} V_{\text {th }} & =V_{2}=\frac{8.5 \times 20}{4}=42.5 \mathrm{~V} \\ R_{\mathrm{th}} & =\frac{V_{2}}{I_{2}} \\ I_{2} & =\frac{4}{3} V_{2} \\ \frac{V_{2}}{I_{2}} & =\frac{3}{4} \Omega \end{aligned}
Equivalent circuit
I=\frac{42.5}{\frac{3}{4}+6}=6.296 \mathrm{~A}
\begin{aligned} P & =I^{2} R_{L}=(6.296)^{2} \times 6 \\ & =238 \mathrm{~W} \end{aligned}
\begin{aligned} Y_{A} & =\left[\begin{array}{cc} \frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{array}\right] \\ Y_{B} & =\left[\begin{array}{cc} 5 & -2.5 \\ -2.5 & 1 \end{array}\right] s \\ Y & =\left[\begin{array}{cc} \frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{array}\right]+\left[\begin{array}{cc} 5 \\ -2.5 & 1 \end{array}\right] \\ & =\left[\begin{array}{cc} \frac{16}{3} & -\frac{8.5}{3} \\ -\frac{8.5}{3} & \frac{4}{3} \end{array}\right] \\ I_{1} & =\frac{16}{3} V_{1}-\frac{8.5}{3} V_{2} \\ I_{2} & =-\frac{8.5}{3} V_{1}+\frac{4}{3} V_{2} \end{aligned}
Put I_{2}=0 and V_{1}=20 \mathrm{~V}
0=-\frac{8.5}{3} \times 20+\frac{4}{3} V_{2} \begin{aligned} V_{\text {th }} & =V_{2}=\frac{8.5 \times 20}{4}=42.5 \mathrm{~V} \\ R_{\mathrm{th}} & =\frac{V_{2}}{I_{2}} \\ I_{2} & =\frac{4}{3} V_{2} \\ \frac{V_{2}}{I_{2}} & =\frac{3}{4} \Omega \end{aligned}
Equivalent circuit
I=\frac{42.5}{\frac{3}{4}+6}=6.296 \mathrm{~A}
\begin{aligned} P & =I^{2} R_{L}=(6.296)^{2} \times 6 \\ & =238 \mathrm{~W} \end{aligned}
Question 3 |
For the circuit shown, if \mathrm{i}=\sin 1000 t, the instantaneous value of the Thevenin's equivalent voltage (in Volts) across the terminals a-b at time \mathrm{t}=5 \mathrm{~ms} is ___ (Round off to 2 decimal places).
-12.25 | |
12.25 | |
11.97 | |
-11.97 |
Question 3 Explanation:
By source transformation, the circuit become
Apply KVL in loop,
\begin{aligned} & (10+j 1)+4 i_{x}-(10+j 10+10-j 10) i_{x}=0 \\ & 10+\mathrm{j} 10+4 \mathrm{i}_{x}-20 \mathrm{i}_{x}=0 \\ & \Rightarrow \quad \mathrm{i}_{\mathrm{x}}=0.884 \angle 45^{\circ} \mathrm{A} \\ & \text { Now, } \quad \mathrm{V}_{\text {Th }}=\mathrm{i}_{x}(10-\mathrm{j10}) \\ & =0.884 \angle 45^{\circ} \times 14.142 \angle-45^{\circ} \\ & =12.5 \angle 0^{\circ} \mathrm{V} \\ & \therefore \quad \mathrm{V}_{\text {Th }}=12.5 \sin 1000 \mathrm{t} \\ & \text { At } \quad \mathrm{t}=5 \mathrm{msec} \\ & \mathrm{V}_{\text {Th }}=12.5 \times \sin \left(\frac{5 \times 180^{\circ}}{\pi}\right) \\ & =-11.986 \mathrm{~V} \end{aligned}
Apply KVL in loop,
\begin{aligned} & (10+j 1)+4 i_{x}-(10+j 10+10-j 10) i_{x}=0 \\ & 10+\mathrm{j} 10+4 \mathrm{i}_{x}-20 \mathrm{i}_{x}=0 \\ & \Rightarrow \quad \mathrm{i}_{\mathrm{x}}=0.884 \angle 45^{\circ} \mathrm{A} \\ & \text { Now, } \quad \mathrm{V}_{\text {Th }}=\mathrm{i}_{x}(10-\mathrm{j10}) \\ & =0.884 \angle 45^{\circ} \times 14.142 \angle-45^{\circ} \\ & =12.5 \angle 0^{\circ} \mathrm{V} \\ & \therefore \quad \mathrm{V}_{\text {Th }}=12.5 \sin 1000 \mathrm{t} \\ & \text { At } \quad \mathrm{t}=5 \mathrm{msec} \\ & \mathrm{V}_{\text {Th }}=12.5 \times \sin \left(\frac{5 \times 180^{\circ}}{\pi}\right) \\ & =-11.986 \mathrm{~V} \end{aligned}
Question 4 |
For the circuit shown in the figure, \mathrm{V}_{1}=8 \mathrm{~V}, DC and I_{1}=8 A, DC. The voltage V_{a b} in Volts is ___ (Round off to 1 decimal place).
4.2 | |
6 | |
8 | |
10.6 |
Question 4 Explanation:
Reraw the circuit:
Now, using voltage division,
\mathrm{V}_{\mathrm{ab}}=\frac{1.5 \times 8}{1.5+0.5}=6 \mathrm{~V}
Now, using voltage division,
\mathrm{V}_{\mathrm{ab}}=\frac{1.5 \times 8}{1.5+0.5}=6 \mathrm{~V}
Question 5 |
The value of parameters of the circuit shown in the figure are
\mathrm{R}_{1}=2 \Omega, \mathrm{R}_{2}=2 \Omega, \mathrm{R}_{3}=3 \Omega, \mathrm{L}=10 \mathrm{mH}, \mathrm{C}=100 \mu \mathrm{F}
For time t \lt 0, the circuit is at steady state with the switch 'K' in closed condition. If the switch is opened at t=0, the value of the voltage across the inductor \left(\mathrm{V}_{\mathrm{L}}\right) at t=0^{+}in Volts is _____ (Round off to 1 decimal place).
\mathrm{R}_{1}=2 \Omega, \mathrm{R}_{2}=2 \Omega, \mathrm{R}_{3}=3 \Omega, \mathrm{L}=10 \mathrm{mH}, \mathrm{C}=100 \mu \mathrm{F}
For time t \lt 0, the circuit is at steady state with the switch 'K' in closed condition. If the switch is opened at t=0, the value of the voltage across the inductor \left(\mathrm{V}_{\mathrm{L}}\right) at t=0^{+}in Volts is _____ (Round off to 1 decimal place).
6.5 | |
8 | |
12.5 | |
16 |
Question 5 Explanation:
Case (i) \mathbf{t} \lt 1
At steady state, capacitor behaves as open circuit and inductor behaves as short circuit.
Redraw the circuit:
Using current division,
\begin{aligned} \mathrm{i}_{\mathrm{L}}\left(0^{-}\right) & =\frac{3}{3+2} \times 10=6 \mathrm{~A} \\ \therefore \quad \mathrm{V}_{\mathrm{C}}\left(0^{-}\right) & =5 \times 2=12 \mathrm{~V} \end{aligned}
Case (ii) \mathbf{t} \gt 0 :
Switch is opened.
\because \quad \mathrm{V}_{\mathrm{C}}\left(0^{-}\right)=\mathrm{V}_{\mathrm{C}}\left(0^{+}\right)=12 \mathrm{~V}
and \quad \mathrm{i}_{\mathrm{L}}\left(0^{-}\right)=\mathrm{i}_{\mathrm{L}}\left(0^{+}\right)=6 \mathrm{~A}
Redraw the circuit \mathrm{t}=0^{+}:
Apply KVL in loop,
\begin{array}{rlrl} 12+8-12-\mathrm{V}_{\mathrm{L}} & =0 \\ \Rightarrow & \mathrm{V}_{\mathrm{L}} & =8 \mathrm{~V} \end{array}
At steady state, capacitor behaves as open circuit and inductor behaves as short circuit.
Redraw the circuit:
Using current division,
\begin{aligned} \mathrm{i}_{\mathrm{L}}\left(0^{-}\right) & =\frac{3}{3+2} \times 10=6 \mathrm{~A} \\ \therefore \quad \mathrm{V}_{\mathrm{C}}\left(0^{-}\right) & =5 \times 2=12 \mathrm{~V} \end{aligned}
Case (ii) \mathbf{t} \gt 0 :
Switch is opened.
\because \quad \mathrm{V}_{\mathrm{C}}\left(0^{-}\right)=\mathrm{V}_{\mathrm{C}}\left(0^{+}\right)=12 \mathrm{~V}
and \quad \mathrm{i}_{\mathrm{L}}\left(0^{-}\right)=\mathrm{i}_{\mathrm{L}}\left(0^{+}\right)=6 \mathrm{~A}
Redraw the circuit \mathrm{t}=0^{+}:
Apply KVL in loop,
\begin{array}{rlrl} 12+8-12-\mathrm{V}_{\mathrm{L}} & =0 \\ \Rightarrow & \mathrm{V}_{\mathrm{L}} & =8 \mathrm{~V} \end{array}
There are 5 questions to complete.