Question

An electric bulb of resistance 500 Ω draws current 0.4 A from the source. Calculate:

1. the power of bulb and
2. the potential difference at its end.

Answer 1

1. Resistance of electric bulb (R) = 500Ω
   current drawn from the source (I) = 0.4 A
   Power of the bulb (P) = VI
   V = I × R
   V = 0.4 × 500
   V = 200 V
2. The potential difference at its end is 200 V.
   Hence,
   Power (P) = VI
   P = 200 × 0.4
   P = 80 W
   The power of the bulb is 80 Watt.