Question

Find the voltage of power supply to which a bulb of rating 200W-

220V is connected to consume energy at the rate of 100 J/s
Answer with Explanation.
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Answer 1

Answer:

Resistance R = V2/P = 48400/200 = 2422

Irms Vrms/R = 220/242 = 0.909A =

Peak Value of current=Irms x 2 = 0.909 ×

1.414 1.285A =

(2)

Power P = V₁/R = 200 × 200/242 = 165.28W