In how much time (in seconds) should a charge of $10$ coulombs flow between two terminals of a circuit so as to maintain a constant current of $100 A ?$

$0.1 s$

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$1 s$

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$11 s$

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$1.1 s$

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Solution

The correct option is A
$0.1 s$

Given: Charge $Q = 10 C$ and current $I = 100 A$

Let time be $t$

As we know, the current is the rate of flow of charge, therefore $I = \frac{Q}{t}$ .

So time, $t = \frac{Q}{I} = \frac{10}{100} = \frac{1}{10} = 0.1 s$

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