An ideal coil of 10H is connected in series with a resistance of 5- and a battery of 5V- 2 seconds after the connection is made- the current flowing in amperes in the circuit is -

The correct option is C (1 e^ 1) i(2) = 55 ( 1 e^( 2 × 5/10)) = 1 e^ 1 ...

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Byju's Answer

Standard XII

Physics

Mutual Induction

An ideal coil...

Question

An ideal coil of $10 H$ is connected in series with a resistance of $5 Ω$ and a battery of $5 V$ . $2$ seconds after the connection is made, the current flowing in amperes in the circuit is -

(1 - e)

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e

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e^{- 1}

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(1 - e^{- 1})

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Solution

The correct option is C $(1 - e^{- 1})$
$i (2) = \frac{5}{5} (1 - e^{(- 2 \times 5 / 10)}) = 1 - e^{- 1}$

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An ideal coil of 10H is connected in series with a resistance of 5Ω and a battery of 5V. 2 seconds after the connection is made, the current flowing in amperes in the circuit is -

The correct option is C (1−e−1)i(2)=55(1−e(−2×5/10))=1−e−1

An ideal coil of $10 H$ is connected in series with a resistance of $5 Ω$ and a battery of $5 V$ . $2$ seconds after the connection is made, the current flowing in amperes in the circuit is -

The correct option is C $(1 - e^{- 1})$
$i (2) = \frac{5}{5} (1 - e^{(- 2 \times 5 / 10)}) = 1 - e^{- 1}$