Question

In shown diagram, if voltage of battery is doubled and resistance kept constant then what is the new Power dissipated at resistor(Consider $P$ was the initial power dissipation) ?

• A. $\frac{P}{4}$
• B. $\frac{P}{2}$
• C. P
• D. 2P
• E. 4P

Answer 1

The correct option is E 4P

Power, $P=\frac{V^2}{R}$

As R is kept constant so $P\propto V^2$

$P^{\prime }$is new power dissipation.

So, $\frac{P^{\prime }}{P}=\frac{(2V{)}^2}{V^2}=\frac{4V^2}{V^2}$ or $P^{\prime }=4P$

Thus, option E is correct.