Question

3. 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 1

Explanation:

Solution

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Resistance R=V

2

/P=48400/200=242Ω

I

rms

=V

rms

/R=220/242=0.909A

Peak Value of current =I

rms

×

2

2

=0.909×1.414=1.285A

(2)

Power P=V

1

2

/R=200×200/242=165.28W