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Keeping the potential difference constant, if the resistance is doubled, then by what factor does the current change in a circuit?

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0.5

1

0.25

None of the above

answer is A.

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Detailed Solution

By keeping the potential difference constant, if the resistance is doubled then the current changes by a factor of

0.5

times in the circuit.
Initially, let resistance in the circuit be

R_{1}

, the current

I_{1}

in the circuit is:

I_{1} = \frac{V}{R_{1}}

Let the new doubled resistance be

R_{2} .

When the potential difference is kept constant and resistance is doubled then the current

I_{2}

in the circuit is:

I_{2} = \frac{V}{R_{2}}

Therefore, the ratio of current will be,

\Rightarrow \frac{I_{1}}{I_{2}} = \frac{\frac{V}{R_{1}}}{\frac{V}{R_{2}}}

\Rightarrow \frac{I_{1}}{I_{2}} = \frac{R_{2}}{R_{1}}

Since,

R_{2} = 2 R_{1}

\Rightarrow \frac{I_{1}}{I_{2}} = \frac{2 R_{1}}{R_{1}}

\Rightarrow I_{2} = \frac{I_{1}}{2}

Hence, the factor by which the current changes is

\frac{1}{2}

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