A current impulse 5δ(t) is forced through a capacitor C. The voltage vc(t) across the capacitor is given by

To analyze the voltage $v_c(t)$ across a capacitor $C$ when subjected to a current impulse $5\delta(t)$, we can follow these steps:

1. ### Break Down the Problem

We start with the relationship between the current through a capacitor and the voltage across it. The current $i(t)$ through a capacitor is related to the voltage $v_c(t)$ by the equation:

$i(t) = C \frac{dv_c(t)}{dt}$

Given that the current impulse is $i(t) = 5\delta(t)$, we will substitute this into the equation.

2. ### Relevant Concepts

From the relationship stated above, we can modify it to find $v_c(t)$:

$5\delta(t) = C \frac{dv_c(t)}{dt}$

3. ### Analysis and Detail

To find $v_c(t)$, we can rearrange the equation:

$\frac{dv_c(t)}{dt} = \frac{5\delta(t)}{C}$

Now, we integrate both sides with respect to time $t$:

$v_c(t) = \frac{5}{C} \int \delta(t) dt$

The integral of the delta function $\delta(t)$ is:

$\int \delta(t) dt = u(t)$

where $u(t)$ is the unit step function. Therefore, we have:

$v_c(t) = \frac{5}{C} u(t)$

4. ### Verify and Summarize

To summarize, the voltage across the capacitor given the impulse current is:

$v_c(t) = \frac{5}{C} u(t)$

Final Answer

Thus, the voltage across the capacitor is:

$v_c(t) = \frac{5}{C} u(t)$