# Why does a capacitor discharge a percentage of the original energy in the same time?

If I charge a capacitor ($220\mu{F}$) using a 6V battery, and then measure the time it takes to discharge 90% of the initial energy over a resistor (${100k}\Omega$), and then charge the same capacitor using a 12V battery and measure the time it takes to discharge 90% of its initial energy again (over the same resistor).

Why are both times the same? Especially given that the second time there is 4 times more starting energy that the first time. ($E=\frac{1}{2}CV^2$.)

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## 1 Answer

Why are both times the same?

Because the time constant of the circuit hasn't changed. For an RC circuit, the time constant $\tau$ is just the product of the resistance and the capacitance: $\tau = RC$.

When you write and solve the differential equation for the RC circuit with an initial voltage across the capacitor $V_0$, the solution is:

$v_C(t) = V_0 e^{-t/\tau}$

By inspection, the time to decay to a percentage of the initial value is independent of the initial value.

More intuitively, the power delivered to the resistor goes as the square of the voltage. So, when the initial voltage is doubled, the initial energy stored is squared, but the initial power delivered by the capacitor is also squared, i.e., the initial rate of change of energy is also squared.

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