Network Theory Questions and Answers – Series and Parallel Combination of Elements

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 I0 and the voltage at the capacitor is V0. The inductor is represented by a transform impedance _________ in series with a voltage source __________
The inductor is represented by transform impedance Ls in series with a voltage source V0
a) Ls, L V0
b) Ls, LI0
c) 1/Ls, LI0
d) 1/Ls, L V0
View Answer

Answer: a
Explanation: The inductor has an initial current I0. It is represented by a transform impedance Ls in series with a voltage source L V0.

2. In the circuit shown below, the capacitor is replaced by a transform impedance of __________ with an initial voltage ___________
The inductor is represented by transform impedance Ls in series with a voltage source V0
a) 1/Cs, V0/S
b) 1/Cs, I0/S
c) Cs, I0/S
d) Cs, V0/S
View Answer

Answer: a
Explanation: The capacitor has an initial voltage V0 across it. It is represented by a transform impedance of 1/Cs with an initial voltage V0/S.

3. The value of the total voltage after replacing the inductor and capacitor is?
The inductor is represented by transform impedance Ls in series with a voltage source V0
a) V1(S)-LI0-V0/S
b) V1(S)+LI0-V0/S
c) V1(S)+LI0+V0/S
d) V1(S)-LI0+V0/S
View Answer

Answer: b
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) = V1(S)+LI0-V0/S.
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4. The value of the total impedance after replacing the inductor and capacitor is?
The inductor is represented by transform impedance Ls in series with a voltage source V0
a) R-LS-1/CS
b) R-LS+1/CS
c) R+LS+1/CS
d) R+LS-1/CS
View Answer

Answer: c
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 following circuit is?
The inductor is represented by transform impedance Ls in series with a voltage source V0
a) (V1(S)-LI0-V0/S)/(R+LS+1/CS)
b) (V1(S)-LI0+V0/S)/(R+LS+1/CS)
c) (V1(S)+LI0+V0/S)/(R+LS+1/CS)
d) (V1(S)+LI0-V0/S)/(R+LS+1/CS)
View Answer

Answer: d
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) = (V1(S)+LI0-V0/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.
Admittance of last two elements in parallel combination after transformation is 4+s
a) 1+s
b) 2+s
c) 3+s
d) 4+s
View Answer

Answer: d
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 Y1(s) = 4 + s.

7. The impedance of the last two elements in the parallel combination after transformation in the circuit shown below is?
Admittance of last two elements in parallel combination after transformation is 4+s
a) 1/(s+4)
b) 1/(s+3)
c) 1/(s+2)
d) 1/(s+1)
View Answer

Answer: a
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 Z1(s) = 1/(s+4).
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8. The series combination of the last elements after replacing 1/s and 1/4Ω with 1/(S+4) is?
Admittance of last two elements in parallel combination after transformation is 4+s
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

Answer: c
Explanation: We got the impedance of last two elements in parallel combination as Z1(s) = 1/(s+4) and now the impedance of capacitor is 1/2s. So the series combination of the last elements is Z2(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) is?
Admittance of last two elements in parallel combination after transformation is 4+s
a) (4s2-19s+4)/(6s-8)
b) (4s2+19s-4)/(6s+8)
c) (4s2+19s-4)/(6s-8)
d) (4s2+19s+4)/(6s+8)
View Answer

Answer: d
Explanation: The term admittance is defined as the inverse of the term impedance. As the impedance is Z2(s) = 1/2s+1/(s+4)=(3s+4)/2s(s+4), the admittance parallel combination of the last elements is Y2(s) = 1/2+2s(s+4)/(3s+4)=(4s2+19s+4)/(6s+8).
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10. Obtain the transform impedance of the network shown below.
Admittance of last two elements in parallel combination after transformation is 4+s
a) (6s-8)/(4s2+19s-4)
b) (6s+8)/(4s2+19s+4)
c) (6s+8)/(4s2-19s+4)
d) (6s-8)/(4s2+19s+4)
View Answer

Answer: b
Explanation: The term impedance is the inverse of the term admittance. We got admittance as Y2(s) = (4s2+19s+4)/(6s+8). So the transform impedance of the network is
Z (s) = 1/Y2(s) = (6s+8)/(4s2+19s+4).

Sanfoundry Global Education & Learning Series – Network Theory.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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