Consider a time-dependent current across a 1 F capacitor given by iC(t) = { 1, 0

Question

Consider a time-dependent current across a [latex]1 \mathrm{~F}[/latex] capacitor given by

[latex]i_{C}(t)=\left\{\begin{aligned}1,\quad & 0

Assuming [latex]v_{C}(0)=0[/latex], find and plot the voltage across the capacitor, as well as the power of the capacitor and the stored energy.

Accepted Answer

We again have [latex]v_{C}(t)=0[/latex] for [latex]t<0[/latex]. In addition,

[latex]v_{C}(t)=(t) \mathrm{V}[/latex]

for [latex]0the previous example. Then, for [latex]1

[latex]v_{C}(t)=v_{C}(1)+\int_{1}^{t}(-1) d t^{\prime}=(-t+2) \mathrm{V},[/latex]

leading to [latex]v_{C}(3)=-1 \mathrm{~V}[/latex] and [latex]v_{C}(t)=-1 \mathrm{~V}[/latex] for [latex]t>3 \mathrm{~s}[/latex]. Overall, we have

[latex]v_{C}(t)=\left\{\begin{array}{rl}0, & t<0 \t, & 03 \mathrm{~s}\end{array} \quad(\mathrm{V}) . ight.[/latex]

Consequently, the current and voltage across the capacitor can be plotted as follows.

Using the expressions for the current and voltage, the power of the capacitor can be found as

[latex]p_{C}(t)=v_{C}(t) i_{C}(t)=\left\{\begin{array}{rl}0, & t<0 \t, & 03 \mathrm{~s}\end{array} \quad(\mathrm{~W}) . ight.[/latex]

Considering the expression above, it can be deduced that the capacitor stores energy (positive power) in the time periods [latex]0

[latex]w_{C}(t)=\frac{1}{2} C\left[v_{C}(t) ight]^{2}=\left\{\begin{array}{rll}0, & t<0 \t^{2} / 2, & 03 \mathrm{~s}\end{array} \quad(\mathrm{~J}) . ight.[/latex]

The plots of the power and energy of the capacitor are as follows.

Question 6.100: Consider a time-dependent current across a 1 F capacitor giv...

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