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SSC JE Electrical 2018 with solution SET-2

SSC JE Electrical 2018 with solution SET-2
 Determine the value of maximum power (in W) transferred from the source to the load in the circuit given below

37.5
30
25
20

ANSWER DOWNLOAD EXAMIANS APP

SSC JE Electrical 2018 with solution SET-2
 How much power (in W) will be dissipated by a 5 Ohm resistor in which the value of current is 2 A?

20
30
40
10

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Given
Resistance R = 5Ω
Current I = 2 A
Power dissipated by the resistor is
P = I2R
P = 22 × 5
P = 20 watts
 

SSC JE Electrical 2018 with solution SET-2
 What is the equivalent capacitance (in μF) for the circuit given below?

4.56
7.5
54.56
54.28

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

In the given circuit the capacitance C1 and C2 are parallel with the capacitance C3 i.e
(C1 || C2) + C3
∴(20 × 30) ⁄ (20 + 30) + 20
CA= 12 + 20 = 32 μF
Now capacitance CA, C4, & C5 are in the series therefore
Ceqv = (1/30 + 1/20 + 1/20)
Ceqv = (60/8) = 7.5 μF
 

SSC JE Electrical 2018 with solution SET-2
Determine the value of current (in A) through both the resistor of the given circuit

2, −1.5
−2, −1.5
−2, 1.5
2, 1.5

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

 
Current through the 10Ω resistance
I1 = V/R = 20/10 = 2A
I1 = 2A
Now current through the 20Ω resistance
I2 = V − (-10)/R = 20 + 10/30 = 1.5 A
I2 = 1.5 A
 


SSC JE Electrical 2018 with solution SET-2
Determine the value of current I1 (in A) and V1 (in V) respectively, for the given circuit below.
 

4, 32
−6, 30
6, 30
−4, 32

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

According to Kirchhoff’s Current Law: At any point in an electrical circuit, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.
∴ I1 = 1 + 3 = 4A
V = IR
∴ V1 = I1R = 8 × 4
V1 = 32Ω
 

SSC JE Electrical 2018 with solution SET-2
What is the value of Norton resistance (in Ω) between the terminal A and B for the given Norton’s equivalent circuit?

5.6
2
4.66
4

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Norton equivalent resistance for the given network is
R = (R1 || R2) + R3
R = (4 || 8) + 2 = (4 x 8) ⁄ (4 + 8) + 2 = 5.6Ω
Norton equivalent resistance = 5.6Ω

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