Physics RC Circuit

Question

A 14.5-uF capacitor is charged to a potential of 40.0 V and then discharged through a 65.0 ohm­ resistor.

How long after discharge begins does it take for the capacitor to lose 90.0% of its initial charge? 

How long after discharge begins does it take for the capacitor to lose 90.0% of its initial energy?

What is the current through the resistor at the time when the capacitor has lost 90.0% of its initial charge? 

What is the current through the resistor at the time when the capacitor has lost 90.0% of its initial energy? 

Answer 1

For part a)

In an RC circuit the charge at time t during capacitor discharge is given by the equation

Q = Q₀e^-t/RC where Q₀ is the initial charge, R is resistance, and C is capacitance

We can rearrange this equation to

e^-t/RC = Q/Q₀

Since the capacitor has lost 90% of it's initial charge there's only 10% left, so Q = 0.1 Q₀, so we can write the equation as

e^-t/RC = 0.1 Q₀/Q₀ OR

e^-t/RC = 0.1

Now we can solve for t

-t/RC = ln 0.1

t = -RC ln 0.1

t = -65(14.5 * 10^-6 F) ln (0.1)

t = 2.17 * 10^-3 s or 2.17 ms