# Transients and Steady State Response

 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   Electric Circuits
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
In the circuit shown below, the switch $S$ is closed at $t=0$. The magnitude of the steady state voltage, in volts, across the $6 \Omega$ resistor is _________. (round off to two decimal places).

 A 5 B 8.25 C 12.55 D 3.35
GATE EE 2022   Electric Circuits
Question 2 Explanation:
Concept: At steady state, capacitor behaves as open circuit.

Using voltage division,
$V=\frac{2}{2+2} \times 10=5V$

 Question 3
A $\text{100 Hz}$ square wave, switching between $\text{0 V}$ and $\text{5 V}$, is applied to a $\text{CR}$ high-pass filter circuit as shown. The output voltage waveform across the resistor is $\text{6.2 V}$ peak-to-peak. If the resistance R is $\text{820 \Omega}$, then the value C is ______________$\mu F$. (Round off to 2 decimal places.)

 A 18.5 B 12.46 C 10.06 D 15.48
GATE EE 2021   Electric Circuits
Question 3 Explanation:

\begin{aligned} v_{0}&=v_{i}-v_{c}\\ \text{For }1^{\text {st }}\text{ half cycle}, \quad v_{0}&=5-v_{c} \\ \text{For }2^{\text {nd }}\text{ half cycle}, \quad v_{0}&=-v_{c}\\ v_{p-p} &=\left(5-V_{c \;\min}\right)-\left(-V_{c}\; \max \right) \\ 6.2 &=5+V_{c \;\max }-V_{c \;\min} \\ \Rightarrow \quad V_{c \max }-V_{c \;\min } &=1.2 \ldots(\alpha) \end{aligned}
For first half cycle i.e. $0 \lt t \lt \frac{T}{2}$
\begin{aligned} v_{c}\left(0^{+}\right) &=v_{c}(0)=v_{c}\left(0^{-}\right)=v_{c} \min \\ v_{c}(\infty) &=5 \mathrm{~V} \\ \therefore \qquad\qquad v_{c}(t) &=v_{c}(\infty)+\left[v_{c}\left(0^{+}\right)-v_{c}(\infty)\right] e^{-t / \tau} \\ v_{c}(t) &=5+\left[V_{c m i n}-5\right] e^{-t / 2 \tau}=V_{c m a x} \\ \Rightarrow \qquad\qquad V_{c} \max &=5\left[1-e^{-T / 2 \tau}\right]+V_{c m i n} e^{-T / 2 \tau} \end{aligned}
$For \frac{T}{2} \lt t \lt T$

\begin{aligned} v_{c}(t) &=v_{c}\left(\frac{T}{2}\right) e^{-t(t-T / 2) \tau} \\ \therefore\qquad v_{c}(t) &=V_{c m a x} e^{-(t-T / 2) \tau} \\ \text{at } t =T,\qquad \qquad v_{c} &=V_{c} \mathrm{~min} \\ \Rightarrow \qquad \qquad V_{C \mathrm{~min}} &=V_{C \max } e^{-T / 2 \tau}\\ \text { As } \quad V_{c \text { max }}-V_{c \text { min }}&=1.2 \qquad \qquad [\text { From }(\alpha)]\\ \therefore \quad V_{\text {cmax }}-V_{c \max } e^{-T / 2 t} &=1.2 \\ V_{c} \max &=\frac{1.2}{1-e^{-T / 2 \tau}} \\ \Rightarrow \qquad\qquad V_{c} \max &=\frac{1.2}{1-e^{-T / 2 \tau}}=5\left[1-e^{-T / 2 \tau}\right]+V_{c \min } e^{-2 \tau} \end{aligned}
From (ii),
\begin{aligned} V_{c \max } &=5\left[1-e^{-T / 2 \tau}\right]+\left(V_{c m a x} e^{-T / 2 \tau}\right) e^{-T / 2 \tau} \\ V_{c \max } &=5\left[1-e^{-T / 2 \tau}\right]+V_{c \max } e^{-T / \tau} \\ \Rightarrow \qquad \qquad V_{c \max }\left[1-e^{-T / \tau}\right] &=5\left[1-e^{-T / 2 \tau}\right] \\ V_{c} \max &=\frac{5\left[1-e^{-T / 2 \tau}\right]}{\left[1+e^{-T / 2 \tau}\right]\left[1-e^{-T / 2 \tau}\right]} \end{aligned}
Using equation (iii)
\begin{aligned} \frac{1.2}{1-e^{-T / 2 \tau}} &=\frac{5}{1+e^{-T / 2 \tau}} \\ \Rightarrow \qquad \qquad 1.2+1.2 e^{-T / 2 \tau} &=5-5 e^{-\pi / 2 t} \\ \Rightarrow \qquad \qquad 6.2 e^{-T / 2 \tau} &=3.8 \\ e^{-T / 2 \tau} &=\frac{3.8}{6.2}=0.6129 \\ \frac{T}{2 \tau} &=0.4895\\ \text { as }\qquad \qquad T&=\frac{1}{f}=\frac{1}{100} \mathrm{sec}\\ \text { and }\qquad \qquad \tau & =R C=820 \mathrm{C} \\ \Rightarrow \qquad \qquad \frac{1}{(100)(2)(820) C} & =0.4895 \\ \therefore \qquad \qquad C & =12.46 \mu \mathrm{F} \end{aligned}
 Question 4
In the circuit, switch 'S' is in the closed position for a very long time. If the switch is opened at time $t = 0$, then in $i_{L} (t)$ amperes, for $t\geq0$ is

 A $8e^{-10t}$ B $10$ C $8+2e^{-10t}$ D $10\left ( 1-e^{-2t} \right )$
GATE EE 2021   Electric Circuits
Question 4 Explanation:
At $t = 0^-$

$i_{L}\left(0^{-}\right)=\frac{10}{1}=10 \mathrm{~A}$
For$t > 0$

At $t = \infty$

$i(\infty)=\frac{40}{5}=8 \mathrm{~A}$
$R_{\text{eq}}:$

\begin{aligned} R_{\mathrm{eq}} &=5 \Omega \\ \tau &=\frac{L}{R_{\mathrm{eq}}}=\frac{0.5}{5}=0.1 \mathrm{sec} \\ i(t) &=8+[10-8] e^{-t / 0.1} \\ &=8+2 e^{-10 t} \mathrm{~A} \end{aligned}
 Question 5
The initial charge in the 1 F capacitor present in the circuit shown is zero. The energy in joules transferred from the DC source until steady state condition is reached equals ______. (Give the answer up to one decimal place.)
 A 100 B 200 C 50 D 400
GATE EE 2017-SET-2   Electric Circuits
Question 5 Explanation:
Consider the following circuit diagram,

After minimizing circuit elements we can have the following circuit,

Here, $\tau =RC=5 sec.$
Now current,
$i(t)=\frac{V}{R}e^{\frac{t}{\tau }}$
$\;\;=\frac{10}{5}e^{-t/5}=2e^{-0.2t}$
Energy supplied by the source,
$E=\int_{0}^{\infty }10 \times 2e^{-0.2t}dt$
$\;\;=100J$

There are 5 questions to complete.