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Find I in 4 Ω resistor.

1 A

0.5 A

0.75 A

0.95 A

Answer : Option C

Explanation :

Apply KVL to first to first loop

6i₁ - 2i₂ = 10

(4 + R)i₂ - 2i₁ - 2i₃ = 0

4i₃ - 2i₂ = 0

but i₃ = 0.5 A

2 - 2i₂ = 0

∴ i₂ = 1 A, 6i₁ - 2 = 10

∴ i₁ = 2 A, (4 + R) - 2 X 2 -1 = 0

∴ R = 1 Ω

Using Nodal analysis for loop 2

At node A,

V_A - 20 + V_A + V_A - V_B = 0

3V_A - V_B - 20 = 0

3V_A - V_B = 20 ...(i)

At node B,

2V_B - 2V_A + 2V_B + V_B = 0

5V_B - 2V_A = 0 ...(ii)

Multiplying (i) by 2 and (ii) by 3

6V_A - 2V_B = 40

- 6V_A - 15V_B = 0

13V_B = 40

V_B ≅ 3 V

∴

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