SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

Question : 177 of 200

Marks: +1, -0

A 500 W discharge lamp takes a current of 4 A at unity p.f. Find the inductance of a choke required to enable the lamp to work on 250 V, 50 Hz main.

Solution:

P = 500 W
I = 4 A
V = 250 V
f = 50 Hz
pf = unity

\R = \P / \I^2 = 500 / 16 = 31.25 \Ω

\Z = \V / \I = 250 / 4 = 62.5 \Ω

\X_\L = 54.125 \Ω

\L = \X_\L / (2\π \f) = 0.1732 \H

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