The current in a device varies with time \( t \) as \( I = 6t \), where \( I \) is in mA and \( t \) is in s. The amount of charge that passes through the device during \( t = 0\,\mathrm{s} \) to \( t = 3\,\mathrm{s} \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 27\,\mathrm{mC} \)
The current is the rate of flow of charge, \( I = dq/dt \), so the charge that passes in a time interval is the integral of the current over time. Since \( I \) is in mA and \( t \) in s, the charge comes out in mC.
\( \displaystyle q = \int_{0}^{3} 6t\,dt = 6\,\left[\frac{t^{2}}{2}\right]_{0}^{3} = 3\,(3^{2} - 0) = 27\,\mathrm{mC} \)