# 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. ($E=\frac{1}{2}CV^2$.)

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