Question

A series circuit has a voltage supply of 12 V and a resistor of 2.4 kΩ. How much power is dissipated by the resistor?

Answer 1

Answer:

(B) 0.06W

Explanation:

power absorbed by the resistor is given by $\frac{V^2}{R}=I^2R$

Where V = voltage

           R= resistance

            I =  current through the circuit

we have given V =12 Volt and resistance =2.4 K$\Omega$

current $I=\frac{V}{R}=\frac{12}{2.4\times 1000}=5\times 10^{-3}A=5mA$

power using voltage and resistance equation

$p=\frac{12^2}{2.4\times 1000}=60mW$ =0.06W

using current equation $P=I^2R=5^2\times 12=60mW$= 0.06W