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Q.

In the figure shown the value of $V_{a b}$ at t = 1s if, the charge changes according to the time is given by $q (t) = 4 t^{2}$

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a

30V

b

– 30V

c

20V

d

– 20V

answer is B.

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Detailed Solution

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$I = \frac{d q}{d t} = \frac{d}{d t} (4 t^{2}) = 8 t$

$= \frac{d I}{d t} = 8$

$V_{a} - V_{b} = - \frac{q}{C} - L \frac{d I}{d T} + ε - I R$

$= - \frac{4}{2} - (2) (8) + 4 - 8 (2)$

$= - 2 - 16 + 4 - 16 = - 30 volt$

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In the figure shown the value of Vab at t = 1s if, the charge changes according to the time is given by q(t)=4t2