This set of Network Theory online quiz focuses on “Series and Parallel Combination of Elements”.

1. In the circuit shown below, switch K is moved from position to position 2 at time t = 0. At time t = 0-, the current in the inductor is I_{0} and the voltage at the capacitor is V_{0}. The inductor is represented by a transform impedance _________ in series with a voltage source __________

a) Ls, L V_{0}

b) Ls, LI_{0}

c) 1/Ls, LI_{0}

d) 1/Ls, L V_{0}

View Answer

Explanation: The inductor has an initial current I

_{0}. It is represented by a transform impedance Ls in series with a voltage source L V

_{0}.

2. In the circuit shown in question 1, the capacitor is replaced by a transform impedance of __________ with an initial voltage ___________

a) 1/Cs, V_{0}/S

b) 1/Cs, I_{0}/S

c) Cs, I_{0}/S

d) Cs, V_{0}/S

View Answer

Explanation: The capacitor has an initial voltage V

_{0}across it. It is represented by a transform impedance of 1/Cs with an initial voltage V

_{0}/S.

3. The value of the total voltage after replacing the inductor and capacitor in question 1 is?

a) V_{1}(S)-LI_{0}-V_{0}/S

b) V_{1}(S)+LI_{0}-V_{0}/S

c) V_{1}(S)+LI_{0}+V_{0}/S

d) V_{1}(S)-LI_{0}+V_{0}/S

View Answer

Explanation: The current I(s) is given as the total transform voltage in the circuit divided by the total transform impedance. The value of the total voltage after replacing the inductor and capacitor is V (s) = V

_{1}(S)+LI

_{0}-V

_{0}/S.

4. The value of the total impedance after replacing the inductor and capacitor in question 1 is?

a) R-LS-1/CS

b) R-LS+1/CS

c) R+LS+1/CS

d) R+LS-1/CS

View Answer

Explanation: The value of the total impedance after replacing the inductor and capacitor is

Z (s) = R+LS+1/CS. By knowing the V(s) and Z(s) we can calculate I(s) as I(s) is given as the total transform voltage in the circuit divided by the total transform impedance.

5. The current flowing in the circuit in question 1 is?

a) (V_{1}(S)-LI_{0}-V_{0}/S)/( R+LS+1/CS)

b) (V_{1}(S)-LI_{0}+V_{0}/S)/( R+LS+1/CS)

c) (V_{1}(S)+LI_{0}+V_{0}/S)/( R+LS+1/CS)

d) (V_{1}(S)+LI_{0}-V_{0}/S)/( R+LS+1/CS)

View Answer

Explanation: The current I(s) is given as the total transform voltage in the circuit divided by the total transform impedance. The current flowing in the circuit is I (s) = V(s)/I(s) =(V

_{1}(S)+LI

_{0}-V

_{0}/S)/( R+LS+1/CS).

6. Obtain the admittance of the last two elements in the parallel combination after transformation in the circuit shown below.

a) 1+s

b) 2+s

c) 3+s

d) 4+s

View Answer

Explanation: The term admittance is defined as the inverse of impedance. The admittance of capacitor is 1/s and the admittance of resistor is 1/4 mho. So the admittance of the last two elements in the parallel combination is Y

_{1}(s) = 4 + s.

7. The impedance of the last two elements in the parallel combination after transformation in the circuit shown in question 6 is?

a) 1/(s+4)

b) 1/(s+3)

c) 1/(s+2)

d) 1/(s+1)

View Answer

Explanation: The impedance of resistor is 4Ω and the impedance of capacitor is s. So the impedance of the last two elements in the parallel combination is Z

_{1}(s) = 1/(s+4).

8. The series combination of the last elements after replacing 1/s and 1/4Ω with 1/(S+4) in the question 6 is?

a) (3s+4)/2s(s-4)

b) (3s-4)/2s(s-4)

c) (3s+4)/2s(s+4)

d) (3s-4)/2s(s+4)

View Answer

Explanation: We got the impedance of last two elements in parallel combination as Z

_{1}(s) = 1/(s+4) and now the impedance of capacitor is 1/2s. So the series combination of the last elements is Z

_{2}(s) =1/2s+1/(s+4)=(3s+4)/2s(s+4).

9. Determine the admittance parallel combination of the last elements after replacing with (3s+4)/2s(s+4) in the question 6 is?

a) (4s^{2}-19s+4)/(6s-8)

b) (4s^{2}+19s-4)/(6s+8)

c) (4s^{2}+19s-4)/(6s-8)

d) (4s^{2}+19s+4)/(6s+8)

View Answer

Explanation: The term admittance is defined as the inverse of the term impedance. As the impedance is Z

_{2}(s) =1/2s+1/(s+4)=(3s+4)/2s(s+4) , the admittance parallel combination of the last elements is Y

_{2}(s) = 1/2+2s(s+4)/( 3s+4)=(4s

^{2}+19s+4)/(6s+8).

10. Obtain the transform impedance of the network shown in question 6.

a) (6s-8)/( 4s^{2}+19s-4)

b) (6s+8)/( 4s^{2}+19s+4)

c) (6s+8)/( 4s^{2}-19s+4)

d) (6s-8)/( 4s^{2}+19s+4)

View Answer

Explanation: The term impedance is the inverse of the term admittance. We got admittance as Y

_{2}(s) = (4s

^{2}+19s+4)/(6s+8). So the transform impedance of the network is

Z (s) = 1 / Y

_{2}(s) = (6s+8)/( 4s

^{2}+19s+4).

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

To practice all areas of Network Theory for online Quizzes, __here is complete set of 1000+ Multiple Choice Questions and Answers__.