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

Answer:

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

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:

$Q(t)=Q_0\,e^{-\frac{t}{RC} }$

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:

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

In seconds this is approximately 0.009 seconds