# A fully charged 6.2 μF capacitor is connected in series with a 2.0×105 Ω resistor. What percentage of the original charge is left on the capacitor after 1.6 s of discharging?

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### 1 Answer

If a charged capacitor is connected in series to a resistor, the current will begin to flow in the circuit and the capacitor is going to discharge.

When a capacitor is discharging, the charge on it depends on time as

`Q=Q_0*e^(-t/(RC))` , where Q_0 is the original charge on the capacitor, R is the resistance in the circuit, and C is the capacitance. The charge will approach zero exponentially.

For the given capacitor, RC can be calculated:

`RC = 6.2*10^(-6)*2*10^5 = 1.24 s`

So the ratio of the charge on the capacitor after t = 1.6 s of discharging to the original charge can be found:

`Q=Q_0*e^(-1.6/1.24) = Q_0*0.275`

`Q/Q_0 = 0.275`

**So the charge on the capacitor at the time t = 1.6 s is 0.275, or 27.5% of the original charge on the capacitor.**

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