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.