Answer:
b.The power dissipated is reduced by a factor of 2.
Explanation:
The power dissipated in the circuit is given by
[tex]P=\frac{V^2}{R}[/tex]
where
V is the voltage
R is the resistance
In this problem:
- The voltage V is kept constant
- The resistance is doubled, so R' = 2R
Therefore, the new power dissipated is
[tex]P'=\frac{V^2}{R'}=\frac{V^2}{2R}=\frac{1}{2}\frac{V^2}{R}=\frac{1}{2}P[/tex]
so, the power dissipated is reduced by a factor of 2.
