Answer:
Explanation:
let resistance = R and power = P
power dissipated in the circuit, P = V² / R
let the new resistance = 2 R
then new power P₁ = V² / 2 R = 1/2 ( V² / R) = [tex]\frac{1}{2}[/tex] P
therefore the power dissipated by this circuit will decrease to one-half its original value