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Inductors in Circuits

The network s...

Question

The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is 5 A and it is decreasing at a rate of $10^{3} A s^{- 1}$ then $V_{B} - V_{A}$ equals

5 V

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10 V

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15 V

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20 V

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Solution

The correct option is C
15 V

$\frac{d I}{d t} = - 10^{3}$ A/s

emf $= - L \frac{d I}{d t} = - 5 \times 10^{- 3} \times - 10^{3}$

$= + 5 V$

from KVL,

$V_{A} - 1 \times 5 + 15 + 5 = V_{B}$

$V_{A} - V_{B} = - 15 V$

(or) $V_{B} - V_{A} = + 15 V$

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The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is 5 A and it is decreasing at a rate of $10^{3} A s^{- 1}$ then $V_{B} - V_{A}$ equals

The correct option is C
15 V

$\frac{d I}{d t} = - 10^{3}$ A/s

emf $= - L \frac{d I}{d t} = - 5 \times 10^{- 3} \times - 10^{3}$

$= + 5 V$

from KVL,

$V_{A} - 1 \times 5 + 15 + 5 = V_{B}$

$V_{A} - V_{B} = - 15 V$

(or) $V_{B} - V_{A} = + 15 V$