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

It seems like there's some confusion with your question as it's not fully formed. However, I can tell you that the voltage across a capacitor, given a current impulse, can be found using the formula:

V(t) = 1/C ∫ i(t) dt

Where:

• V(t) is the voltage across the capacitor
• C is the capacitance of the capacitor
• i(t) is the current at time t
• ∫ i(t) dt is the integral of the current with respect to time

Given a current impulse i(t) = 5δ(t), where δ(t) is the Dirac delta function, the voltage across the capacitor would be:

V(t) = 1/C ∫ 5δ(t) dt

The integral of the Dirac delta function is a step function, so the result would be:

V(t) = 5/C for t > 0

This means that the voltage across the capacitor jumps to 5/C at the instant the current impulse is applied, and remains at that level for t > 0.