Question

Two resistors, the first 12 ? and the second 6 2, are connected in parallel to a 48 V battery What is the power dissipated by each resistor? A) P1 144 W, P2-288 W B) P1-162 W, P2-324 W C) P1-180 W, P2-360 W D) P1-192 W, P2 384 W

Answer 1

Answer:

P₁ = 192 W and P₂ = 384 W

Explanation:

It is given that,

Resistor 1, $R_1=12\ \Omega$

Resistor 2, $R_1=6\ \Omega$

Voltage, V = 48 V

Power dissipated by resistor 1 is given by :

$P_1=\dfrac{V^2}{R_1}$

$P_1=\dfrac{(48)^2}{12}$

P₁ = 192 watts

Power dissipated by resistor 2 is given by :

$P_2=\dfrac{V^2}{R_1}$

$P_2=\dfrac{(48)^2}{6}$

P₂ = 384 watts

So, the power dissipated by both the resistors is 192 watts and 384 watts respectively. Hence, this is the required solution.