# Electric Circuits (Test 8)

## Gate Exam : Ee Electrical Engineering

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Electric Circuits
| Electric Circuits |
Q.1
For parallel RLC circuit, which one of the following statements is NOT correct ?
A. The bandwidth of the circuit decreases if R is increased
B. The bandwidth of the circuit remains same if L is increased
C. At resonance, input impedance is a real quantity
D. At resonance, the magnitude of input impedance attains its minimum values.
Explaination / Solution:

A parallel RLC circuit is shown below :
Input impedance

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Q.2
In the circuit shown below, the value of RL such that the power transferred to RL is maximum is

A.
B. 10Ω
C. 15Ω
D. 20Ω
Explaination / Solution:

Power transferred to RL will be maximum when RL is equal to the thevenin resistance. We determine thevenin resistance by killing all source as follows :

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Q.3
In the circuit shown below, if the source voltage Vs = 100∠53.13° V then the Thevenin’s equivalent voltage in Volts as seen by the load resistance RL is

A. 100∠90°
B. 800∠0°
C. 800∠90°
D. 100∠60°
Explaination / Solution:
No Explaination.

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Q.4
In the circuit shown, the power supplied by the voltage source is

A. 0 W
B. 5 W
C. 10 W
D. 100 W
Explaination / Solution:

Applying nodal analysis

Current from voltage source is

Since current through voltage source is zero, therefore power delivered is zero.

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Q.5
If the transfer function of the following network is

A. R/4

B. R/2
C. R
D. 2R
Explaination / Solution:

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Q.6
The voltage across the capacitor, as sown in the figure, is expressed as

The value of A1 and A2 respectively, are
A. 2.0 and 1.98
B. 2.0 and 4.20
C. 2.5 and 3.50
D. 5.0 and 6.40
Explaination / Solution:

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Q.7
In the circuit shown below, if the source voltage  then the Thevenin’s equivalent voltage in Volts as seen by the load resistance RL is

A. 100∟900
B. 800∟00
C. 800∟900
D. 100∟600
Explaination / Solution:
No Explaination.

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Q.8
For the circuit shown in the figure, the Thevenin voltage and resistance looking into X − Y are

A.
B.
C.
D. 4 V, 2 Ω
Explaination / Solution:

Here Vth is voltage across node also. Applying nodal analysis we get

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Q.9
In the given network V1 = 100∠0° V, V2 = 100∠ − 120° V, V3 = 100∠ + 120° V. The phasor current 𝑖 (in Ampere) is

A. 0
B.
C.
D.