# Electric Circuits (Test 4)

## Gate Exam : Ee Electrical Engineering

| Home | | Gate Exam | | Ee Electrical Engineering | | Electric Circuits | Electric Circuits
| Electric Circuits |
Q.1
A three-winding transformer is connected to an AC voltage source as shown in the figure. The number of turns are as follows: N= 100, N2 = 50. If the magnetizing current is neglected, and the currents in two windings are and then what is the value of the current in Ampere? A. 1∠90°
B. 1∠270°
C. 4∠90°
D. 4∠270°
Answer : Option A
Explaination / Solution: Workspace
Report
Q.2
A lossy capacitor Cx, rated for operation at 5 kV, 50 Hz is represented by an equivalent circuit with an ideal capacitor Cp in parallel with a resistor Rp. The value Cp is found to be 0.102 µF and the value of Rp = 1.25 MΩ . Then the power loss and tan ∂ of the lossy capacitor operating at the rated voltage, respectively,
A. 10 W and 0.0002
B. 10 W and 0.0025
C. 20 W and 0.025
D. 20 W and 0.04
Answer : Option C
Explaination / Solution:
No Explaination.

Workspace
Report
Q.3
The input voltage given to a converter is The current drawn by the converter is The input power factor of the converter is
A. 0.31
B. 0.44
C. 0.5
D. 0.71
Answer : Option C
Explaination / Solution: Workspace
Report
Q.4
In the following circuit, the switch S is closed at t = 0. The rate of change of current is given by A. 0
B. C. D.
Answer : Option B
Explaination / Solution:

Initially i(0-) = 0 therefore due to inductor i(0+) = 0. Thus all current Is will flow Iin resistor R and voltage across resistor will be IsRs. The voltage across inductor will be equal to voltage across Rs as no current flow through R. Workspace
Report
Q.5
A parallel plate capacitor filled with two dielectrics is shown in the figure below. If the electric field in the region A is 4 kV/cm, the electric field in the region B, in kV/cm, is A. 1
B. 2
C. 4
D. 16
Answer : Option C
Explaination / Solution: Workspace
Report
Q.6
A 230 V rms source supplies power to two loads connected in parallel. The first load draws 10 kW at 0.8 leading power factor and the second one draws 10 kVA at 0.8 lagging power factor. The complex power delivered by the source is
A. (18 + j1.5) kVA
B. (18 - j1.5) kVA
C. (20 + j1.5) kVA
D. (20 - j1.5) kVA
Answer : Option B
Explaination / Solution:

Consider the circuit diagram for given problem as shown below Load delivered to Z1 is

P1 = 10 kW

cos ϕ1 = 0.8, leading

So, we obtain the complex power delivered to Z1 as Again, the delivered power to load Zas

|s1|= 10 kVA

cos ϕ2 = 0.8, lagging

So, we obtain the complex power delivered to load Z2 as Hence, the total complex power delivered by the source is

s1 + s2 = (10 – j7.5) + (8 + j6)

= (18 - j1.5) kVA

Workspace
Report
Q.7
With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed:
(i) 1Ω connected at port B draws a current of 3 A
(ii) 2.5Ω W connected at port B draws a current of 2 A With 10 V dc connected at port A, the current drawn by 7 Ω connected at port B is
A. 3/7 A
B. 5/7 A
C. 1 A
D. 9/7 A
Answer : Option C
Explaination / Solution:
No Explaination.

Workspace
Report
Q.8
A load is supplied by a 230 V, 50 Hz source. The active power P and the reactive power Q consumed by the load are such that 1 kW ≤ P ≤ 2kW and 1kVAR ≤ Q ≤ kVAR . A capacitor connected across the load for power factor correction generates 1 kVAR reactive power. The worst case power factor after power factor correction is
A. 0.447 lag
B. 0.707 lag
C. 0.894 lag
D. 1
Answer : Option B
Explaination / Solution: Workspace
Report
Q.9
A two-port network shown below is excited by external DC source. The voltage and the current are measured with voltmeters V1,V2 and ammeters. A1,A2 (all assumed to be ideal), as indicated Under following conditions, the readings obtained are: The z -parameter matrix for this network is
A. B. C. D. Answer : Option C
Explaination / Solution:

From the problem statement we have  Workspace
Report
Q.10
The circuit shown below is driven by a sinusoidal input Vi = Vpcos(t/RC). The steady state output vo A. (Vp/3)cos(t/RC)
B. (Vp/3)sin(t/RC)
C. (Vp/2)cos(t/RC)
D. (Vp/2)sin(t/RC)
Answer : Option A
Explaination / Solution:

parallel combination of R and C equivalent impedance is Transfer function can be written as Workspace
Report