This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Circuit Elements in the S-Domain”.

1. The resistance element __________ while going from the time domain to frequency domain.

a) does not change

b) increases

c) decreases

d) increases exponentially

View Answer

Explanation: The s-domain equivalent circuit of a resistor is simply resistance of R ohms that carries a current I ampere seconds and has a terminal voltage V volts-seconds. The resistance element does not change while going from the time domain to the frequency domain.

2. The relation between current and voltage in case of inductor is?

a) v=Ldt/di

b) v=Ldi/dt

c) v=dt/di

d) v=di/dt

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Explanation: Consider an inductor with an initial current I

_{o}. The time domain relation between current and voltage is v=Ldi/dt.

3. The s-domain equivalent of the inductor reduces to an inductor with impedance?

a) L

b) sL

c) s^{2}L

d) s^{3}L

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Explanation: If the initial energy stored in the inductor is zero, the equivalent circuit of the inductor reduces to an inductor with impedance sL ohms.

4. The voltage and current in a capacitor are related as?

a) i=Cdt/dv

b) v=Cdv/dt

c) i=Cdv/dt

d) v=Cdt/dv

View Answer

Explanation: Consider an initially charged capacitor and the initial voltage on the capacitor is V

_{o}. The voltage current relation in the time domain is i=Cdv/dt.

5. The s-domain equivalent of the capacitor reduces to an capacitor with impedance?

a) sC

b) C

c) 1/C

d) 1/sC

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Explanation: The s-domain equivalent of the capacitor can be derived for the charged capacitor and it reduces to an capacitor with impedance 1/sC.

6. From the circuit shown below, find the value of current in the loop.

a) (V/R)/(s+1/RC)

b) (V/C)/(s+1/R)

c) (V/C)/(s+1/RC)

d) (V/R)/(s+1/R)

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Explanation: Applying Kirchhoff’s law around the loop, we have V/s=1/sC I+RI. Solving above equation yields I=CV/(RCS+1)=(V/R)/(s+1/RC).

7. After taking the inverse transform of current in the circuit shown in question 6, the value of current is?

a) i=(V/C)e^{-t/R}

b) i=(V/C)e^{-t/RC}

c) i=(V/R)e^{-t/RC}

d) i=(V/R)e^{-t/R}

View Answer

Explanation: We had assumed the capacitor is initially charged to V

_{o}volts. By taking the inverse transform of the current, we get i=(V/R) e

^{-t/RC}.

8. The voltage across the resistor in the circuit shown in question 6 is?

a) Ve^{t/R}

b) Ve^{-t/RC}

c) Ve^{-t/R}

d) Ve^{t/RC}

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Explanation: We can determine the voltage v by simply applying the ohm’s law from the circuit. And applying the Ohm’s law from the circuit v = Ri = Ve

^{-t/RC}.

9. The voltage across the resistor in the parallel circuit shown is?

a) V/(s-1/R)

b) V/(s-1/RC)

c) V/(s+1/RC)

d) V/(s+1/C)

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Explanation: The given circuit is converted to parallel equivalent circuit. By taking the node equation, we get v/R+sCv=CV. Solving the above equation, v=V/(s+1/RC).

10. Taking the inverse transform of the voltage across the resistor in the circuit shown in question 9.

a) Ve^{-t/τ}

b) Ve^{t/τ}

c) Ve^{tτ}

d) Ve^{-tτ}

View Answer

Explanation: By taking the inverse transform, we get v=Ve

^{-t/RC}=Ve

^{-t/τ}, where τis the time constant and τ = RC. And v is the voltage across the resistor.

**Sanfoundry Global Education & Learning Series – Network Theory.**

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