In shown diagram- voltage of battery is doubled and resistance kept constant then what is the new Power dissipated resistor-Consider P was the initial power dissipation- - dfrac-P-4-dfrac-P-2-P2P4P

Question

Toppr Admin · Accepted Answer

Power- P-V2RAs R is kept constant so P-x221D-V2P-x2032-is new power dissipation-So- P-x2032-P-2V-2V2-4V2V2 or P-x2032-4PThus- option E is correct-xA0-

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) ?

$\frac{P}{4}$
$\frac{P}{2}$
P
2P
4P

Power, $P = \frac{V^{2}}{R}$
As R is kept constant so $P \propto V^{2}$
$P^{'}$ is new power dissipation.
So, $\frac{P^{'}}{P} = \frac{(2 V)^{2}}{V^{2}} = \frac{4 V^{2}}{V^{2}}$ or $P^{'} = 4 P$
Thus, option E is correct.

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) ? P4P2P2P4P

Power, P=V2RAs R is kept constant so P∝V2P′is new power dissipation.So, P′P=(2V)2V2=4V2V2 or P′=4PThus, option E is correct.

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) ?

$\frac{P}{4}$
$\frac{P}{2}$
P
2P
4P

Power, $P = \frac{V^{2}}{R}$
As R is kept constant so $P \propto V^{2}$
$P^{'}$ is new power dissipation.
So, $\frac{P^{'}}{P} = \frac{(2 V)^{2}}{V^{2}} = \frac{4 V^{2}}{V^{2}}$ or $P^{'} = 4 P$
Thus, option E is correct.