The charge q (in coulomb) passing through a 1 | vedclass

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(D) The charge passing through the resistor is given by $q = 3t^2 - 2t + 6$.To find the current $I$, we differentiate the charge with respect to time

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$280$

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(D) The charge passing through the resistor is given by $q = 3t^2 - 2t + 6$.To find the current $I$, we differentiate the charge with respect to time $t$:$I = \frac{dq}{dt} = \frac{d}{dt}(3t^2 - 2t + 6) = 6t - 2$.At time $t = 5 \, s$, the current is:$I = 6(5) - 2 = 30 - 2 = 28 \, A$.The resistance is given as $R = 10 \, \Omega$.Using Ohm's law, the potential difference $V$ is:$V = I 	imes R = 28 \, A 	imes 10 \, \Omega = 280 \, V$.

The charge $q$ (in coulomb) passing through a $10 \, \Omega$ resistor as a function of time $t$ (in second) is given by $q = 3t^2 - 2t + 6$. The potential difference across the ends of the resistor at time $t = 5 \, s$ is: (in $V$)

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