The charge $q$ (in coulomb) passing through a $10 \mathrm{~ohm}$ resistor as a function of time $t$ (in…

The charge $q$ (in coulomb) passing through a $10 \mathrm{~ohm}$ resistor as a function of time $t$ (in second) is given by $q$ $=3 t^2-2 t+6$. The potential difference across the ends of the resistor at time $t=5 \mathrm{~s}$ is
  1. 120 V
  2. 240 V
  3. 140 V
  4. 280 V

Solution

$\begin{aligned} & \mathrm{q}=3 \mathrm{t}^2-2 \mathrm{t}+6 \\ & \begin{aligned} \therefore & \text { Current, } \mathrm{I}=\frac{\mathrm{dq}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(3 \mathrm{t}^2-2 \mathrm{t}+6\right) \\ & =(6 \mathrm{t}-2) \\ \therefore & A t \mathrm{t}=5 \mathrm{~s}, \mathrm{I}=6 \times 5-2=28 \mathrm{~A} \\ \therefore & \mathrm{~V}=\mathrm{IR}=28 \times 10=280 \mathrm{~V}\end{aligned}\end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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