# Electric Circuits

 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).

 A 0.87 B 1.57 C 1.88 D 2.26
GATE EE 2023      Transients and Steady State Response
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}
 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).

 A 238 B 452.25 C 632.12 D 145.25
GATE EE 2023      Two Port Network and Network Functions
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}

 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).

 A -12.25 B 12.25 C 11.97 D -11.97
GATE EE 2023      Network Theorems
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}
 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).

 A 4.2 B 6 C 8 D 10.6
GATE EE 2023      Basics of Electric Circuit
Question 4 Explanation:
Reraw the circuit:

Now, using voltage division,
$\mathrm{V}_{\mathrm{ab}}=\frac{1.5 \times 8}{1.5+0.5}=6 \mathrm{~V}$
 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).

 A 6.5 B 8 C 12.5 D 16
GATE EE 2023      Steady state AC Analysis
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,
$\begin{array}{rlrl} 12+8-12-\mathrm{V}_{\mathrm{L}} & =0 \\ \Rightarrow & \mathrm{V}_{\mathrm{L}} & =8 \mathrm{~V} \end{array}$

There are 5 questions to complete.