For the given equation of charge \( Q = 3t^{2} + 6t \), where Q is in coulomb and t is time in sec, the initial value of current is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6 A
The current is the rate of flow of charge, \( I = dQ/dt \).
\( \displaystyle I = \frac{d}{dt}\left(3t^{2} + 6t\right) = 6t + 6 \)
The initial value is at \( t = 0 \):
\( I(0) = 6 \times 0 + 6 = 6\,\mathrm{A} \)