Question 1 |
A sample and hold circuit is implemented using a resistive switch and a capacitor with a time constant of 1 \mu \mathrm{s}. The time for sampling switch to stay closed to charge a capacitor adequately to a full scale voltage of 1 \mathrm{~V} with 12-bit accuracy is ____ \mu \mathrm{s}. (rounded off to two decimal places)
2.32 | |
6.65 | |
8.32 | |
12.36 |
Question 1 Explanation:
Given: Time constant (\tau) of 1 \mu \mathrm{sec}.
Full scale voltage =1 \mathrm{~V}
The voltage across capacitor is given as
\quad \quad V_{c}(t)=V_{\text {in }}\left(1-e^{-t / \tau}\right)
\therefore \quad V_{c}(t)=\left(1-e^{-t / 1 \mu \mathrm{sec}}\right) \quad ...(i)
To calculate the voltage to stay closed to charge capacitor adequately to a full scale voltage with 12-bit accuracy is given by
\begin{aligned} V_{c}(t) & =V_{\text {ref }}\left[1-\frac{1}{2^{n}}\right] \\ n & =\text { Number of bit }=12 \\ V_{c}(t) & =\left[1-\frac{1}{4096}\right] \quad ...(ii) \end{aligned}
Comparing equations (i) and (ii), we get,
\begin{aligned} e^{-t / 1 \mu \mathrm{sec}} & =\frac{1}{4096} \\ -\frac{t}{1 \mu \mathrm{sec}} & =\ln \left\{\frac{1}{4096}\right\} \\ t & =8.3177 \mu \mathrm{sec} \end{aligned}
Full scale voltage =1 \mathrm{~V}
The voltage across capacitor is given as
\quad \quad V_{c}(t)=V_{\text {in }}\left(1-e^{-t / \tau}\right)
\therefore \quad V_{c}(t)=\left(1-e^{-t / 1 \mu \mathrm{sec}}\right) \quad ...(i)
To calculate the voltage to stay closed to charge capacitor adequately to a full scale voltage with 12-bit accuracy is given by
\begin{aligned} V_{c}(t) & =V_{\text {ref }}\left[1-\frac{1}{2^{n}}\right] \\ n & =\text { Number of bit }=12 \\ V_{c}(t) & =\left[1-\frac{1}{4096}\right] \quad ...(ii) \end{aligned}
Comparing equations (i) and (ii), we get,
\begin{aligned} e^{-t / 1 \mu \mathrm{sec}} & =\frac{1}{4096} \\ -\frac{t}{1 \mu \mathrm{sec}} & =\ln \left\{\frac{1}{4096}\right\} \\ t & =8.3177 \mu \mathrm{sec} \end{aligned}
Question 2 |
The switch S_{1} was closed and S_{2} was open for a long time. At t=0, switch S_{1} is opened and S_{2} is closed, simultaneously. The value of i_{c}\left(0^{+}\right), in amperes, is
1 | |
-1 | |
0.2 | |
0.8 |
Question 2 Explanation:
At t=0^{-} ; S_{1} \rightarrow closed, S_{2} \rightarrow opened
\begin{aligned} i_{L}\left(0^{-}\right)&=\frac{1 \times 25}{100+25}=0.2 \mathrm{~A} \\ v_{C}\left(0^{-}\right)&=\frac{1}{5} \times 100=20 \mathrm{~V} \end{aligned}
At t=0^{+} ; S_{1} \rightarrow opened, S_{2} \rightarrow closed
\begin{aligned} i_{x} & =\frac{20}{25}=\frac{4}{5} \mathrm{~A}=0.8 \mathrm{~A} \\ \text{By KCL: }-i_{c} & =i_{x}+0.2=0.8+0.2 \\ \Rightarrow \quad i_{c} & =-1 \mathrm{~A} \end{aligned}
\begin{aligned} i_{L}\left(0^{-}\right)&=\frac{1 \times 25}{100+25}=0.2 \mathrm{~A} \\ v_{C}\left(0^{-}\right)&=\frac{1}{5} \times 100=20 \mathrm{~V} \end{aligned}
At t=0^{+} ; S_{1} \rightarrow opened, S_{2} \rightarrow closed
\begin{aligned} i_{x} & =\frac{20}{25}=\frac{4}{5} \mathrm{~A}=0.8 \mathrm{~A} \\ \text{By KCL: }-i_{c} & =i_{x}+0.2=0.8+0.2 \\ \Rightarrow \quad i_{c} & =-1 \mathrm{~A} \end{aligned}
Question 3 |
In the circuit shown below, switch S was closed for a long time. If the switch is opened at t=0, the maximum magnitude of the voltage V_{R}, in volts, is ____ (rounded off to the nearest integer).
2 | |
4 | |
6 | |
8 |
Question 3 Explanation:
At t=0^{-}
i_{L}\left(0^{-}\right)=\frac{2}{1}=2 \mathrm{~A}
At t=0^{+}
V_{R}=-2 \times 2=-4,
Magnitude of voltage V_{R},
\left|V_{R}\right|=4
i_{L}\left(0^{-}\right)=\frac{2}{1}=2 \mathrm{~A}
At t=0^{+}
V_{R}=-2 \times 2=-4,
Magnitude of voltage V_{R},
\left|V_{R}\right|=4
Question 4 |
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 4 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 5 |
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 5 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}
There are 5 questions to complete.