Physics
An electric bulb of resistance 500 Ω draws current 0.4 A from the source. Calculate: (a) the power of bulb and (b) the potential difference at it's end.
Answer
(a) Given,
Resistance (R) = 500 Ω
Current (I) = 0.4 A
Power = ?
From Ohm's law:
V = IR
Substituting the values in the formula above, we get,
V = 0.4 x 500 = 200 V
Power (P) = VI
Substituting the values in the power formula we get,
P = 200 x 0.4 = 80 W
Hence, the power of bulb = 80 W
(b) The potential difference at it's end = 200 V
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