Calculating the voltage across a capacitor.

The voltage, V, across a capacitor can be expressed as:

A 1.25F capacitor that has an initial voltage of 25.o V is charged with a current that varies with time as i = t (t^{2} + 6.83)^{1/2}. Find the voltage across the capacitor at 1.00seconds.

Write the initial equation.

Now do the substitution for u, du, n.

u = t^{2} + 6.83 du = 2t dt n = 1/2

Now integrate:

V = .8* 2/3u^{2/3} +C = 0.267(t^{2} + 6.83)^{3/2}+ C

To solve for the constant, C, we need a set of initial conditions.

When t = 0, V = 25V.

So, 25 = 0.267((0)^{2} + 6.83)^{3/2}+ C

C = 20.2 V

Then V = 0.267(t^{2} + 6.83)^{2/3}+ 20.2.

And when t = 1 sec, V = 26.0V