(a) The original current through the lamp was Io = P/V = 0.833 A. The lamp filament resistance is
R = V/Io = 128.6 ohms.
Then an additional series resistance R' is added to the lamp to reduce the current in half.
Assuming that the lamp filament resistance R remains the same (which is not quite true),
R' = R = 128.6 ohms
The voltages across the lamp and the series resistor are both the same, and equal to half the externally applied voltage. That would be 60 V.
(b) 128.6 ohms
(c)With half as much current gling through the lamp, and assuming the lamp resistance is unchanged, I^2*R is reduced by a factor of four, making it 25 Watts.
A 100 W lamp is connected to 120 V, and its current is measured. A resistor is added to the lamp to reduce the current to half its original value.
(a) What is the potential difference across the lamp?
(b) How much resistance was added to the circuit?
(c) How much power is now dissipated in the lamp?
1 answer
4 answers