This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “The Impulse Function in Circuit Analysis”.

1. For the circuit shown below, find the voltage across the capacitor C_{1} at the time the switch is closed.

a) 0

b) V/4

c) V/2

d) V

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Explanation: We use two different cicuits to illustrate how an impulse function can be created with a switching operation. The capacitor is charged to an initial voltage of V at the time the switch is closed.

2. The charge on capacitor C_{2} in the circuit shown in question 1 is?

a) 0

b) 1

c) 2

d) 3

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Explanation: In the circuit, the initial charge on C

_{2}is zero. So, charge on capacitor C

_{2}= zero. Capacitor circuit and series inductor circuit are two different cicuits to illustrate how an impulse function can be created with a switching operation.

3. The current in the circuit shown in question 1 is?

a) (V/R)/(s+1/C_{e})

b) (V/R)/(s+1/RC_{e})

c) (V/R)/(s-1/RC_{e})

d) (V/R)/(s-1/C_{e})

View Answer

Explanation: The current flowing through the circuit is given by I= (V/s)/(R+1/sC

_{1}+1/sC

_{2}) = (V/R)/(s+1/RC

_{e}) where the equivalent capacitance C

_{1}C

_{2}/(C

_{1}+C

_{2}) and is replaced by C

_{e}.

4. By taking the inverse transform of the current in the question 3, the value of the current is?

a) V/R e^{t/Ce}

b) V/R e^{t/RCe}

c) V/R e^{-t/RCe}

d) V/R e^{-t/Ce}

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Explanation: By taking the inverse transform of the current, we obtain i = V/R e

^{-t/RCe}which indicates that as R decreases, the initial current increases and the time constant decreases.

5. Consider the circuit shown below. On applying the Kirchhoff’s current law, the equation will be?

a) V/(2s-15)+(V-[(100/s)+30])/(3s+10)=0

b) V/(2s-15)+(V-[(100/s)+30])/(3s-10)=0

c) V/(2s+15)+(V-[(100/s)+30])/(3s+10)=0

d) V/(2s+15)+(V-[(100/s)+30])/(3s-10)=0

View Answer

Explanation: The current in the 3H inductor at t = 0 is 10A and the current in 2H inductor at t = 0 is zero. Applying Kirchhoff’s current law, we get V/(2s+15)+(V-[(100/s)+30])/(3s+10)=0.

6. The value of the voltage V in the circuit shown in question 5 is?

a) 40(s+7.5)/s(s+5) -12(s+7.5)/(s-5)

b) 40(s+7.5)/s(s+5) -12(s+7.5)/(s+5)

c) 40(s+7.5)/s(s+5) +12(s+7.5)/(s-5)

d) 40(s+7.5)/s(s+5) +12(s+7.5)/(s+5)

View Answer

Explanation: Solving for V yields V=40(s+7.5)/s(s+5) +12(s+7.5)/(s+5). The current in the 3H inductor at t = 0 is 10A and the current in 2H inductor at t = 0 is zero.

7. The value of the voltage V after taking the partial fractions in the question 5 is?

a) 12+ 60/s+10/(s+5)

b) 12- 60/s+10/(s+5)

c) 12- 60/s-10/(s+5)

d) 12+ 60/s-10/(s+5)

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Explanation: By taking the partial fractions we get V = 60/s-20/(s+5)+12+30/(s+5) and on solving the equation we get V = 12+60/s+10/(s+5).

8. Determine the voltage V after taking the inverse transform in the question 5.

a) 12δ(t)-(60-10e^{-5t})u(t)

b) 12δ(t)+(60+10e^{-5t})u(t)

c) 12δ(t)-(60+10e^{-5t})u(t)

d) 12δ(t)+(60-10e^{-5t})u(t)

View Answer

Explanation: By taking inverse transform of V = 12+60/s+10/(s+5) we have v=12δ(t)+(60+10e

^{-5t})u(t) volts and we have to derive the expression for the current when t > 0.

9. The current equation for the circuit shown in question 5 is?

a) I=4/s-2/(s-5)

b) I=4/s-2/(s+5)

c) I=4/s+2/(s+5)

d) I=4/s+2/(s-5)

View Answer

Explanation: After the switch has been opened, the current in L

_{1}is the same as the current in L

_{1}. The current equation is I=(100/s+30)/(5s+25). On solving we get I = 4/s-2/(s+5).

10. The value of the current after taking the inverse transform of the current in the question 5 is?

a) (4-2e^{5t} )u(t)

b) (4-2e^{-5t})u(t)

c) (4+2e^{5t})u(t)

d) (4-2e^{-5t})u(t)

View Answer

Explanation: By taking the inverse transform of I = 4/s-2/(s+5) gives i = (4-2e

^{-5t})u(t). Before the switch is opened,the current in L

_{1}is 10A and the current in L

_{2}is 0A.

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