First slide

Electrical power

Question

In the arrangement shown initially switch s is open. When switch s is closed,

Moderate

A

reading of voltmeter is halned
B

reading of voltmeter becomes 4 times the initial reading
C

power dissipated in left resistor becomes 4 times the initial power
D

power dissipated in the left resister is doubled

Solution

When s is open, i = E/2R
$∴$ reading of voltmeter = i R= E/2 .
Initial power dissipated in the left resistor = $i^{2} R = {(\frac{E}{2 R})}^{2} R = E^{2} / 4 R$ .
When s is closed, reading of voltmeter = E.
Power dissipated in left resistor = $\frac{E^{2}}{R}$ .

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