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Having $V$ as constant, the resistance $R$ of an electric circuit is doubled, then the reading on the ammeter becomes [[1]] of its original value.

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Detailed Solution

Having

V

as constant, the resistance

R

of an electric circuit is doubled, then the reading on the ammeter becomes half of its original value.
As given, having V (potential difference) as constant, the resistance of an electric circuit is doubled,

R' = 2 R

, then using Ohm's law, we have

I = \frac{V}{R}

.
Now the current flowing in the circuit will be

I^{'} = \frac{V}{R^{'}}

= \frac{V}{2 R}

\Rightarrow I^{'} = \frac{I}{2}

I^{'}

is reduced by half from its previous values. Hence the reading in the ammeter becomes half of its original value.

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