SSC JE Electrical Engineering 23 Jan 2018 Evening Paper

Question : 117 of 200

Marks: +1, -0

Determine the power (in W) dissipated by a 16 ohms resistor, when the potential difference between the ends of the resistor is 32 V.

Solution:

Power dissipated by resistor

P=I^2R=V^2/R

=32^2/16=64\watt

Go to Question:

Prev Question

Next Question