Question

A 10.0 μFμF capacitor initially charged to 30.0 μCμC is discharged through a 1.30 kΩkΩ resistor. How long does it take to reduce the capacitor's charge to 15.0 μCμC ? Express your answer with the appropriate units. 0.052s

Answer 1

It takes approximately 0.009 seconds for the capacitor to discharge to half its original charge.

Step-by-step explanation:

Recall that the capacitor discharges with an exponential decay associated with the time constant for the circuit (
`\tau_0=RC`) which in our case is;

`\tau_0=1.3\,k\Omega\,*\,10.0 \mu F=13\,\,10^(-3)\,s` (13 milliseconds)

Recall as well the expression for the charge in the capacitor (from it initial value
`Q_0`, as it discharges via a resistor R:

`Q(t)=Q_0\,e^{-(t)/(RC) }`

So knowing that the capacitor started with a charge of 30
`\mu C` and reduced after a time "t" to 30
`\mu C` , and knowing from our first formula what the RC of the circuit is, we can solve for the time elapsed using the charge as function of time equation shown above:

`Q(t)=Q_0\,e^{-(t)/(RC) }\\15 \mu C=30 \mu C\,\,e^{-(t)/(13\,ms) }\\e^{-(t)/(13\,ms) }=(1)/(2) \\-(t)/(13\,ms)=ln((1)/(2))\\t=-13\,ms\,*\,ln((1)/(2))\\t=9.109\,\, ms`

In seconds this is approximately 0.009 seconds