Networks Analysis And Synthesis

Question: A dc machine has 6 poles all connected in series. Each pole has 9 H inductance and field current is 2 A. Total inductance and total energy stored are
[A].

9 H and 18 J

[B].

18 H and 36 J

[C].

54 H and 108 J

[D].

108 H and 216 J

Answer: Option C

Explanation:

Total L = 6 x 9 = 54 H.

Energy = x 54 x 22 = 108 J.

Question: A series resonant circuit has R = 2 Ω, L = 0.1 H and C =10 μF. At resonance applied voltage = 10 ∠0 V. Voltage across inductance is likely to be
[A].

10 ∠0 V

[B].

10 ∠45° V

[C].

100 ∠90° V

[D].

100 ∠-90° V

Answer: Option C

Explanation:

Voltage across inductance = QV and leads the applied voltage.

Question: A two branch parallel tuned circuits has a coil of resistance R and inductance L in one branch and capacitance C in the second branch. The bandwidth is 200 radians/sec. A load of resistance RL is connected in parallel. The new bandwidth will be
[A].

200 rad/sec

[B].

more than 200 rad/sec

[C].

less than 200 rad/sec

[D].

200 rad/sec or less

Answer: Option B

Explanation:

Loading increases bandwidth of parallel tuned circuii.

Question: A wave trap is a
[A].

series tuned circuit

[B].

a parallel tuned circuit

[C].

either a series tuned circuit or parallel tuned circuit

[D].

none of the above

Answer: Option B

Explanation:

It has R and L in one branch and a variable capacitance in second branch.

Question: A resistance R, inductance L and capacitance C are in series. The source frequency is adjusted to be equal to resonant frequency. The lower half power frequency is ω1. Another resistance R is added in series with the circuit. The new lower half power frequency will be
[A].

ω1

[B].

less than ω1

[C].

more than ω1

[D].

either more than ω1 or less then ω1

Answer: Option C

Explanation:

As R increases, Q decreases, bandwidth = . Therefore, bandwidth increases. Hence lower half power frequency will be less than ωr.

Question: A coil of resistance R and inductance L has a capacitor C in parallel with it. The source frequency is adjusted to a value equal to resonant frequency. Another capacitor C is added in parallel with the above circuit, the source current will
[A].

increase

[B].

decrease

[C].

remain the same

[D].

may increase or decrease

Answer: Option B

Explanation:

The circuit will not in resonance. In parallel resonant circuit the current is minimum at resonance. Therefore, current will increase.

Question: A two branch parallel resonant circuit has a coil of resistance R and inductance L, in one branch and a capacitors C in the other branch. The source voltage is V, At resonance the circulating current is
[A].

[B].

[C].

[D].

Answer: Option D

Explanation:

No answer description available for this question.

Question: A coil of resistance R and inductance L is connected in series with 5 μF capacitor. The source frequency is equal to resonance frequency and circuit current is 0.5 A. Another 5 F capacitor is connected in parallel with the above capacitor. The circuit current will be
[A].

0.5 A

[B].

more than 0.5 A

[C].

less than 0.5 A

[D].

0.5 A or less

Answer: Option C

Explanation:

Since C increases, the circuit will not be in resonance. Therefore, current decreases (In series resonant circuit current is maximum at resonance).

Question: A coil of resistance R and inductance L is connected in series with 10 mF capacitor. The resonant frequency is 1000 Hz. Another 10 F capacitor is connected in parallel with the above capacitor. The new resonance frequency will be
[A].

1000 Hz

[B].

more than 1000 Hz

[C].

less than 1000 Hz

[D].

may be more or less than 10000 Hz

Answer: Option C

Explanation:

Total capacitance increases and therefore ωr decreases.

Question: A series resonant circuit has inductance L. If L is increased, the resonant frequency
[A].

increases

[B].

decreases

[C].

remains constant

[D].

may increases or decreases

Answer: Option B

Explanation:

. As L increases ωr decreases.