Question

B) A non-inductive load takes a current of 15 A at 125 V. An inductor is then connected

Answer 1

Answer:

i. 43.5 mH ii.  16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°

Explanation:

i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω

The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.

So, x = 2π(50)L = 100πL Ω = 314.16L Ω

Since the current is the same when the 240 V supply is applied, then

the impedance Z = √(R² + X²) = 240 V/15 A

√(R² + X²) = 16 Ω

8.33² + X² = 16²

69.3889 + X² = 256

X² = 256 - 69.3889

X² = 186.6111

X = √186.6111

X = 13.66 Ω

Since X = 314.16L = 13.66 Ω

L = 13.66/314.16

= 0.0435 H

= 43.5 mH

ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.

So in phasor form Z = (8.33 + j13.66) Ω

iii. The phase difference θ between the current and voltage is  

θ = tan⁻¹X/R

= tan⁻¹(314.16L/R)

= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)

= tan⁻¹(13.66/8.33)

= tan⁻¹(1.6406)

= 58.64°