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The network shown in figure is a part of a complete circuit. If at a certain instant, the current ‘i’ is 4 A and is increasing at a rate of $10^{3} A S^{- 1}$ . Then $(V_{B} - V_{A})$ will be

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– 3 V

3 V

- 6 V

12 V

answer is B.

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Detailed Solution

$V_{A} - I R + E - L \frac{d i}{d t} - V_{B} = 0$

$V_{B} - V_{A} = - 4 + 12 - 5 \times 10^{- 3} \times 10^{3}$

= +3V

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