Question

A bulb of resistance 1000 2 is connected across a
potential difference of 100 V. What will be the
percentage change in the electric current if the
potential difference is increased to 200 V?​

Answer 1

Explanation:

Give

Resistance = 1000 ohm

V1 = 100 V

V2 = 200 V

Percentage change in current = i =?

For first case

V= IR

Or

I = V/R = 100/1000

I= 0.1 amperes

For 2nd case

I = V/R = 200/1000

I = 0.2 amperes

Just assume that the in first case current is 100%

initial current = I1 = 100%

Second current = I2 = 100%+ 0.2 = 100.2%

The percentage change in current is

2nd current% - initial current%

100.2% - 100% = 0.2%

:- So the percentage change in current is 0.2% increased.

Hope it helps

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