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SSC JE Electrical 2018 with solution SET-2
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Determine the Norton’s current (in A) and Norton’s resistance (in ohms) respectively, for the given electrical circuit across the load resistance RL.

Determine the resistance RN of the network as seen from the network terminals. (Its value is the same as that of Rth).

RN = (4Ω || 8Ω) + 5Ω = (4 × 8)/(4 + 8) + 5

RN = 7.66

The value I for the current used in Norton’s Theorem is found by determining the open circuit voltage at the terminals AB and dividing it by the Norton resistance r.

According to voltage Division Rule

VAB = V1R3 ⁄ (R1 + R3)

= 24 × 8 ⁄ (4 + 8)

VAB = 16 V

Now Norton Current IN is

IN = VAB ⁄ RN

IN = 16 ⁄ 7.66

IN = 2.08

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

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
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What is the value of Norton resistance (in Ω) between the terminal A and B for the given Norton’s equivalent circuit?

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

SSC JE Electrical 2018 with solution SET-2
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What will be the equivalent resistance (in Ω) for the circuit given below?

In the given circuit the resistance R9 and R8 are Parallel with Resistance R10 therefore

(R9 + R8) || R10

= {(4 + 2) × 12} ⁄ {(4 + 2) + 12}

= (6 × 12) ⁄ (6 + 12)

RA = 4Ω

Now resistance RA and R7 is parallel with Resistance R6

∴ {(4 + 8) × 6} ⁄ {(4 + 8) + 6}

= (12 × 6) ⁄ (12 + 6)

RB = 4Ω

Now the circuit becomes as shown in the figure

Resistance RB and R7 is parallel with Resistance R4

{(4 + 8) × 6} ⁄ {(4 + 8) + 6}

= (12 × 6) ⁄ (12 + 6)

RC = 4Ω

Now resistance RC and R3 is parallel with Resistance R2

{(4 + 2) × 6} ⁄ {(4 + 2) + 6}

RD = 3

Now our final circuit becomes as shown in the figure

Therefore the equivalent resistance is

R1 + RD

Req = 4 + 3 = 7Ω

SSC JE Electrical 2018 with solution SET-2
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What is the resistivity (in ohms-m) of a 2 ohm cylindrical wire when the length and the diameter of the wire are 10 m and 0.4 m respectively?

The resistance of the conductor is determined by the

R = ρL/A

Where

ρ = Resistivity of the conductor

L = Length of the conductor = 10 m

A = Area of the conductor = πR2 = 3.14 × 0.22 = 0.125 m2

R = Resistance = 2 Ω

Therefore the resistivity is

ρ = A × R ⁄ L

= 0.125 × 2 ⁄ 10

ρ = 0.025 Ω-m