Question 1 |
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.)
100 | |
200 | |
50 | |
400 |
Question 1 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
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
Question 2 |
The switch in the figure below was closed for a long time. It is opened at t=0. The current in
the inductor of 2 H for t \geq 0, is
2.5e^{-4t} | |
5e^{-4t} | |
2.5e^{-0.25t} | |
5e^{-0.25t} |
Question 2 Explanation:
From the given circuit, consider the following circuit diagram,
After rearrangement
For t \geq 0
I_0=i(0^-)=2.5A
We can write,
i(t)=I_0e^{-\frac{Rt}{L}}
i(t)=2.5e^{-4t}A
After rearrangement
For t \geq 0
I_0=i(0^-)=2.5A
We can write,
i(t)=I_0e^{-\frac{Rt}{L}}
i(t)=2.5e^{-4t}A
Question 3 |
In the circuit shown below, the initial capacitor voltage is 4 V. Switch S_1 is closed at t=0. The charge (in \muC) lost by the capacitor from t=25\mu s to t=100\mu s is ____________.
5 | |
6 | |
7 | |
8 |
Question 3 Explanation:
i(t)=\frac{4}{5}e^{-t/\tau }
\tau =RC
\;\;=20 \times 10^{-6}sec
Change lost by capacitor from t=25\mu s to 100\mu s is
\int_{25 \mu s}^{100\mu s}i(t)dt=6.99 \times 10^{-6}C
Question 4 |
In the circuit shown, switch S_2 has been closed for a long time. At time t=0 switch S_1 is closed. At t = 0^{+}, the rate of change of current through the inductor, in amperes per second, is _____.
1 | |
2 | |
3 | |
4 |
Question 4 Explanation:
KCL at node A,
\frac{V_A-3}{1}+\frac{3}{2}+\frac{V_A-3}{2}=0
2(V_A-3)+3+(V_A-3)=0
3V_A=6
V_A=2
V_A=L\frac{di(0^+)}{dt}=2
\frac{di(0^+)}{dt}=\frac{2}{L}=\frac{2}{1}=2A/sec
Question 5 |
A series RL circuit is excited at t = 0 by closing a switch as shown in the figure. Assuming zero initial conditions, the value of \frac{d^{2}i}{dt^{2}}\; at \; t=0^{+} is
\frac{V}{L} | |
\frac{-V}{R} | |
0 | |
\frac{-RV}{L^{2}} |
Question 5 Explanation:
Initially (t=0^-) the inductor would be uncharged.
So, I(0^+)=0
The KVL in th loop will be
V=RI+L\frac{dI}{dt}
At t=0^+
V=RI(0^+)+L\frac{dI}{dt}(0^+)
Since, I(0^+)=0
So, \frac{dI}{dt}(0^+)=\frac{V}{L}
Now, lets differentiate the above equation
So, \frac{dV}{dt}=R\frac{dI}{dt}+L\frac{d^2I}{dt^2}
\;\;0=R\frac{dI}{dt}+L\frac{d^2I}{dt^2}
At t=0^+
0=R\frac{dI}{dt}(0^+)+L\frac{d^2I}{dt^2}(0^+)
So, \frac{d^2I}{dt^2}(0^+)=-\frac{R}{L^2}\cdot V
Question 6 |
The switch SW shown in the circuit is kept at position '1' for a long duration. At t=0+, the switch is moved to position '2'. Assuming |V_{o2} | \gt |V_{o1}|, the voltage v_{c}(t) across the capacitor is
v_{c}(t)=-V_{o2}(1-e^{-t/2RC})-V_{o1} | |
v_{c}(t)=V_{o2}(1-e^{-t/2RC})+V_{o1} | |
v_{c}(t)=-(V_{o2}+V_{o1})(1-e^{-t/2RC})-V_{o1} | |
v_{c}(t)=(V_{o2}-V_{o1})(1-e^{-t/2RC})+V_{o1} |
Question 7 |
A combination of 1 \muF capacitor with an initial voltage v_c(0)=-2V in series with a 100 \Omega resistor is connected to a 20 mA ideal dc current source by operating both switches at t=0s as shown. Which of the following graphs shown in the options approximates the voltage v_s across the current source over the next few seconds ?
A | |
B | |
C | |
D |
Question 7 Explanation:
Given C=1\mu F, V_c(0)=-2V, R=100\Omega , I=20mA. Circuit fot the given condition at time t \gt 0 is shown below:
Applying KVL, we have,V_s(s)=\left ( \frac{-2}{s} \right )+\frac{I}{s} \left ( R+\frac{I}{Cs} \right )
\;\;=\frac{1}{s}\left [ -2+IR+\frac{I}{Cs} \right ]
\;\;=\frac{1}{s}\left [ (IR-2) +\frac{I}{Cs}\right ]
Putting values of R, C and I, we get,
V_s(s)=\frac{1}{s}\left [ (20 \times 10^{-3} \times 200-2)+\left ( \frac{20 \times 10^{-3}}{10^{-6}} \right ) \times \frac{1}{s} \right ]
\;\;=\frac{1}{s}\left [ (2-2)+20 \times 10^3 \times \frac{1}{s} \right ]
\;\;=\frac{20 \times 10^3}{s^2}
\therefore \;\; V_s(s)=\frac{20 \times 10^3}{s^2}
V_s(t)=20000 t u(t)
\therefore \;\; V_s(s)=(20000)tu(t)
Which is equation of a straight line passing through origin. Hence option (C) is correct.
Applying KVL, we have,V_s(s)=\left ( \frac{-2}{s} \right )+\frac{I}{s} \left ( R+\frac{I}{Cs} \right )
\;\;=\frac{1}{s}\left [ -2+IR+\frac{I}{Cs} \right ]
\;\;=\frac{1}{s}\left [ (IR-2) +\frac{I}{Cs}\right ]
Putting values of R, C and I, we get,
V_s(s)=\frac{1}{s}\left [ (20 \times 10^{-3} \times 200-2)+\left ( \frac{20 \times 10^{-3}}{10^{-6}} \right ) \times \frac{1}{s} \right ]
\;\;=\frac{1}{s}\left [ (2-2)+20 \times 10^3 \times \frac{1}{s} \right ]
\;\;=\frac{20 \times 10^3}{s^2}
\therefore \;\; V_s(s)=\frac{20 \times 10^3}{s^2}
V_s(t)=20000 t u(t)
\therefore \;\; V_s(s)=(20000)tu(t)
Which is equation of a straight line passing through origin. Hence option (C) is correct.
Question 8 |
In the following figure, C_1 \; and \; C_2 are ideal capacitors. C_1 has been charged to
12 V before the ideal switch S is closed at t=0. The current i(t) for all t is
zero | |
a step function | |
an exponentially decaying function | |
an impulse function |
Question 8 Explanation:
Circuit is s-domain
By applying KVL,
\frac{12}{s}+\frac{I(s)}{s}\left ( \frac{1}{C_1} +\frac{1}{C_2}\right )=0
I(s)=-\frac{12C_1C_2}{C_1+C_2}=k(constant)
\Rightarrow \;\; i(t)=k\delta (t)
\therefore Current i(t) is an implulse function.
Question 9 |
The L-C circuit shown in the figure has an inductance L=1mH and a
capacitance C=\muF.
The initial current through the inductor is zero, while the initial capacitor voltage is 100 V. The switch is closed at t=0. The current i through the circuit
The initial current through the inductor is zero, while the initial capacitor voltage is 100 V. The switch is closed at t=0. The current i through the circuit
5 cos(5\times 10^{3}t)A | |
10 sin(10^{4}t)A | |
10 cos(5\times 10^{3}t)A | |
10 sin(10^{4}t)A |
Question 9 Explanation:
Initial current through the inductor is zeroand capacitor voltage is charged upto voltage V_c(0^-)=100V
As current through inductor and voltage across capacitor can not change abruptly
So, after closing the switch,
i_L(0^+)=i_L(0^-)=0
and V_c(0^+)=V_c(0^-)=100V
The circuit id s-domain
I(s)=\frac{100/s}{\left ( sL+\frac{1}{sC} \right )}
\;\;=\frac{100}{L}\left ( \frac{1}{s^2+\frac{1}{LC}} \right )
\;\;=100\sqrt{\frac{C}{L}}\left ( \frac{1/\sqrt{LC}}{s^2+(1/\sqrt{LC})^2} \right )
taking inverse laplace transform
i(t)=L^{-1}[I(s)]
\;\;=100\sqrt{\frac{C}{L}} \sin \frac{1}{\sqrt{LC}}t
\;\;=100 \times \sqrt{\frac{10 \times 10^3}{1 \times 10^{-3}}}\times \sin\left ( \frac{1}{\sqrt{1 \times 10^{-3} \times 10 \times 10^{-6}}} \right )
i(t)=10\sin (10^4 t)A
As current through inductor and voltage across capacitor can not change abruptly
So, after closing the switch,
i_L(0^+)=i_L(0^-)=0
and V_c(0^+)=V_c(0^-)=100V
The circuit id s-domain
I(s)=\frac{100/s}{\left ( sL+\frac{1}{sC} \right )}
\;\;=\frac{100}{L}\left ( \frac{1}{s^2+\frac{1}{LC}} \right )
\;\;=100\sqrt{\frac{C}{L}}\left ( \frac{1/\sqrt{LC}}{s^2+(1/\sqrt{LC})^2} \right )
taking inverse laplace transform
i(t)=L^{-1}[I(s)]
\;\;=100\sqrt{\frac{C}{L}} \sin \frac{1}{\sqrt{LC}}t
\;\;=100 \times \sqrt{\frac{10 \times 10^3}{1 \times 10^{-3}}}\times \sin\left ( \frac{1}{\sqrt{1 \times 10^{-3} \times 10 \times 10^{-6}}} \right )
i(t)=10\sin (10^4 t)A
Question 10 |
The switch in the circuit has been closed for a long time. It is opened at t = 0. At t =0^+, the current through the 1 \muF capacitor is
0A | |
1A | |
1.25A | |
5A |
Question 10 Explanation:
As the switch has been closed for a long time, the circuit is in steady state. At steadystate, capacitor is open circuit,
Using KVL,
5-I-4I=0 I=1A
V_c(0^-)=4 \times 1=4V
As the voltage across capacitorcan not change abruptly,
So, V_c(0^+)=V_c(0^-)=4V
Circuit at t=0^+
Current through capacitor at t=0^+
I_c(0^+)=\frac{4}{4}=1A
Using KVL,
5-I-4I=0 I=1A
V_c(0^-)=4 \times 1=4V
As the voltage across capacitorcan not change abruptly,
So, V_c(0^+)=V_c(0^-)=4V
Circuit at t=0^+
Current through capacitor at t=0^+
I_c(0^+)=\frac{4}{4}=1A
There are 10 questions to complete.
