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(D) From the graph,the potential difference $V$ as a function of time $t$ is a linear equation: $V(t) = mt + c$.At $t=0$,$V=2\, V$,so the intercept $c

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capacitance of $0.5\, F$

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(D) From the graph,the potential difference $V$ as a function of time $t$ is a linear equation: $V(t) = mt + c$.At $t=0$,$V=2\, V$,so the intercept $c=2$.At $t=3\, s$,$V=8\, V$,so the slope $m = \frac{8-2}{3-0} = \frac{6}{3} = 2\, V/s$.Thus,$V(t) = 2t + 2$.For a capacitor,$V = \frac{q}{C} = \frac{1}{C} \int i \, dt$.Given $i = 1\, A$,we have $V = \frac{1}{C} \int 1 \, dt = \frac{t}{C} + V_0$,where $V_0$ is the initial voltage at $t=0$.Comparing $V(t) = \frac{1}{C}t + 2$ with $V(t) = 2t + 2$,we get $\frac{1}{C} = 2$,which implies $C = 0.5\, F$.

$A$ circuit element is placed in a closed box. At time $t=0$,a constant current generator supplying a current of $1\, A$ is connected across the box. The potential difference across the box varies according to the graph shown in the figure. The element in the box is:

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