Electrical Circuits Objective Questions: Part-3
OBJECTIVE QUESTIONS FROM CIRCUIT THEORY:
[1] A balanced RYB-sequence,Y-connected(Star Connected) source with VRN=100 volts is connected to a Δ-connected (Delta connected) balanced load of (8+j6) ohms per phase.Then the phase current and line current values respectively,are[IES2010]
A.10A;30A
B.10√3A;30A
C.10A;10A
D.10√3A;10√3A
Ans: B
[2] Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is [GATE2012]
a. 0.8Ω
b. 1.4Ω
c. 2Ω
d. 2.8Ω
Ans:A
[3] The voltage gain Av of the circuit shown below is [GATE2012]
a. |Av| ≈200
b. |Av|≈100
c. |Av|≈20
d. |Av|≈10
Ans:D
[4] If VA-VB=6V, then VC-VD is [GATE2012]
a. -5V
b. 2V
c. 3V
d. 6V
Ans:A
Solution Hint:
As Va-Vb =6 => I = V/R = 6/2 = 3A
entering current = leaving current [ KCL]
so current from Vd to Vc is same as 3A
convert the current source to voltage source
V = IR =2 x 1 = 2V
Vd = Vc + 2V + (3x1)V = Vc + 5V
Vc-Vd = -5V
[5]The circuit shown is a [GATE2012]
a. Low pass filter with f3db=[1/(R1+R2)C] rad/s
b.High pass filter with f3db=[1/(R1C)] rad/s
c. Low pass filter with f3db=[1/(R1C)] rad/s
d. High pass filter with f3db=[1/(R1+R2)C] rad/s
Ans:B
[6] In the circuit shown, an ideal switch S operated at 100kHz with a duty ratio or 50%. Given thatΔic 1.6A peak-to-peak and I0 is 5A dc, the peak current in S is [GATE2012]
a. 6.6A
b. 5.0A
c. 5.8A
d. 4.2A
Ans:C
Solution Hint:
Δic = 1.6A peak-to-peak
The positive peak = Δic /2
The current flows through the switch has a peak of = Io + Δic /2
For your information...The current waveform is like this
The lowest point will be 5A....highest point will be 5+ 0.8
[7] The i-v characteristics of the diode in the circuit given below are
i={ [ V- 0.7 /500]A, V > or = 0.7 V
0A, V< 0.7 V
The current in the circuit is [GATE 2012]
a. 10 mA
b. 9.3 mA
c. 6.67 mA
d. 6.2 mA
Ans:D
Sol. Hint:
we have to find v and then apply in ct. equ...
KVL...10 = 1K * i + v
10 = 1000i +v = 1000(v-0.7/500) + v
solve it ....v= 0.8...
[8] In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12V before the ideal switch S is closed at t=0. The current i(t) for all t is [GATE 2012]
a. Zero
b. A step function
c. An exponentially decaying function
d. An impulse function
Ans:D
Sol. Hint:
If it is RC ckt...ie any Resistance in series with capacitor, it will be an exponentially decaying function....here no resistance...so charge instantly....ie impulse function...
[9] The impedance looking into nodes 1 and 2 in the given circuit is [GATE 2012]
a. 50 Ω
b. 100 Ω
c. 5 KΩ
d. 10.1 KΩ
Ans:A
[10] In the circuit given below,the current through the inductor is [GATE 2012]
a. (2/1+j)A
b. (-1/1+j)A
c. (1/1+j)A
d. 0A
Ans:C
Sol Hint: consider top half of the circuit
R & L are in parallel with ct. source....other end connected to low potential...ie GND... convert ct. to V source...:
[1] A balanced RYB-sequence,Y-connected(Star Connected) source with VRN=100 volts is connected to a Δ-connected (Delta connected) balanced load of (8+j6) ohms per phase.Then the phase current and line current values respectively,are[IES2010]
A.10A;30A
B.10√3A;30A
C.10A;10A
D.10√3A;10√3A
Ans:A
Ans:B
a. Zero
Sol Hint: consider top half of the circuit
R & L are in parallel with ct. source....other end connected to low potential...ie GND... convert ct. to V source...:
A.10A;30A
B.10√3A;30A
C.10A;10A
D.10√3A;10√3A
Ans: B
[2] Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is [GATE2012]
a. 0.8Ω
b. 1.4Ω
c. 2Ω
d. 2.8Ω
Ans:A
[3] The voltage gain Av of the circuit shown below is [GATE2012]
a. |Av| ≈200
b. |Av|≈100
c. |Av|≈20
d. |Av|≈10
Ans:D
[4] If VA-VB=6V, then VC-VD is [GATE2012]
a. -5V
b. 2V
c. 3V
d. 6V
Ans:A
Solution Hint:
As Va-Vb =6 => I = V/R = 6/2 = 3A
entering current = leaving current [ KCL]
so current from Vd to Vc is same as 3A
convert the current source to voltage source
V = IR =2 x 1 = 2V
Vd = Vc + 2V + (3x1)V = Vc + 5V
Vc-Vd = -5V
As Va-Vb =6 => I = V/R = 6/2 = 3A
entering current = leaving current [ KCL]
so current from Vd to Vc is same as 3A
convert the current source to voltage source
V = IR =2 x 1 = 2V
Vd = Vc + 2V + (3x1)V = Vc + 5V
Vc-Vd = -5V
[5]The circuit shown is a [GATE2012]
a. Low pass filter with f3db=[1/(R1+R2)C] rad/s
b.High pass filter with f3db=[1/(R1C)] rad/s
c. Low pass filter with f3db=[1/(R1C)] rad/s
d. High pass filter with f3db=[1/(R1+R2)C] rad/s
Ans:B
[6] In the circuit shown, an ideal switch S operated at 100kHz with a duty ratio or 50%. Given thatΔic 1.6A peak-to-peak and I0 is 5A dc, the peak current in S is [GATE2012]
a. 6.6A
b. 5.0A
c. 5.8A
d. 4.2A
Ans:C
Solution Hint:
Δic = 1.6A peak-to-peak
The positive peak = Δic /2
The current flows through the switch has a peak of = Io + Δic /2
For your information...The current waveform is like this
Δic = 1.6A peak-to-peak
The positive peak = Δic /2
The current flows through the switch has a peak of = Io + Δic /2
For your information...The current waveform is like this
The lowest point will be 5A....highest point will be 5+ 0.8
[7] The i-v characteristics of the diode in the circuit given below are
i={ [ V- 0.7 /500]A, V > or = 0.7 V
0A, V< 0.7 V
The current in the circuit is [GATE 2012]
a. 10 mA
b. 9.3 mA
c. 6.67 mA
d. 6.2 mA
Ans:D
Sol. Hint:
we have to find v and then apply in ct. equ...
KVL...10 = 1K * i + v
10 = 1000i +v = 1000(v-0.7/500) + v
solve it ....v= 0.8...
[8] In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12V before the ideal switch S is closed at t=0. The current i(t) for all t is [GATE 2012]
we have to find v and then apply in ct. equ...
KVL...10 = 1K * i + v
10 = 1000i +v = 1000(v-0.7/500) + v
solve it ....v= 0.8...
[8] In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12V before the ideal switch S is closed at t=0. The current i(t) for all t is [GATE 2012]
b. A step function
c. An exponentially decaying function
d. An impulse function
Ans:D
Sol. Hint:
If it is RC ckt...ie any Resistance in series with capacitor, it will be an exponentially decaying function....here no resistance...so charge instantly....ie impulse function...
[9] The impedance looking into nodes 1 and 2 in the given circuit is [GATE 2012]
If it is RC ckt...ie any Resistance in series with capacitor, it will be an exponentially decaying function....here no resistance...so charge instantly....ie impulse function...
[9] The impedance looking into nodes 1 and 2 in the given circuit is [GATE 2012]
a. 50 Ω
b. 100 Ω
c. 5 KΩ
d. 10.1 KΩ
Ans:A
[10] In the circuit given below,the current through the inductor is [GATE 2012]
a. (2/1+j)A
b. (-1/1+j)A
c. (1/1+j)A
d. 0A
Ans:C
Sol Hint: consider top half of the circuit
R & L are in parallel with ct. source....other end connected to low potential...ie GND... convert ct. to V source...:
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