Answer:
[tex]\frac{P_1}{P_2} =\frac{1}{4}[/tex]
That means the heat loss will be one fourth
Explanation:
The resistance R is the same in both cases
First case
[tex]P_1=\frac{V_1^2}{R}[/tex]
Second case
[tex]P_2=\frac{(V_2)^2}{R}[/tex]
[tex]P_2=\frac{(2V_1)^2}{R} \\\\P_2=\frac{4V_1^2}{R}[/tex]
Since power is double here
Therefore ratio of the power loss is
[tex]\frac{P_1}{P_2} =\frac{1}{4}[/tex]
That means the heat loss will be one fourth