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Byju's Answer

Standard XIII

Physics

Inductors in Circuits

When the curr...

Question

When the current in the portion of the circuit shown in the figure is $2 A$ and increasing at the rate of $1 A/s$ , the measured potential difference $V_{a b} = 8 V$ . However when the current is $2 A$ and decreasing at the rate of $1 A/s$ , the measured potential difference $V_{a b} = 4 V$ . The value of $R$ and $L$ are

3 Ω

and

2 H

respectively

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2 Ω

and

3 H

respectively

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10 Ω

and

6 H

respectively

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6 Ω

and

1 H

respectively.

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Solution

The correct option is A $3 Ω$ and $2 H$ respectively
$V_{a b} = i R + L \frac{d i}{d t}$
Now, $8 = 2 R + L . . . (i)$
and $4 = 2 R - L . . . (i i)$
Solving Eq. $(i)$ and Eq. $(i i)$ , we get
$R = 3 Ω$
and $L = 2 H$

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