First slide

Electrical power

Question

Two ends of a copper wire are connected to the terminals of a battery. If the wire is stretched to twice its original length, then power dissipated in the wire. The thermal power dissipated in the wire.

Moderate

A

is doubled
B

is halved
C

becomes one-fourth
D

becomes four times

Solution

When the wire is stretched to n times its original length, its resistance becomes n²R.
Now, $P = \frac{V^{2}}{R} . So P' = \frac{V^{2}}{R^{}} = \frac{V^{2}}{n^{2} R} = \frac{P}{n^{2}} = \frac{P}{{(2)}^{2}} = \frac{P}{4} .$

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