Animated Solution for Physics - Current Electricity: Capacitor $C_1$ of capacitance $1\mu \text{F}$ and capacitor $C_2$ of capacitance $2\mu \text{F}$ are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time $t=0$.

Imagine two separate electrical circuits, each containing a fully charged capacitor ready to release its stored energy. We have capacitor $C_1$ with a capacitance of $1\mu \text{F}$ and capacitor $C_2$ with a capacitance of $2\mu \text{F}$. Both were charged by the same battery, meaning they start with the exact same initial voltage, $V$. At $t=0$, both are connected to identical resistors, $R$, and begin to discharge.

When a capacitor discharges through a resistor, it acts like a temporary voltage source. The current at any given time $t$ is described by the equation:

Here, $i_0$ is the initial current. According to Ohm's law, at the exact moment the switch is closed ($t=0$), the current is simply the initial voltage divided by the resistance:

Notice that this initial current depends only on the initial voltage $V$ and the resistance $R$. It does not depend on the capacitance! Since both circuits have the same $V$ and the same $R$, their initial currents are perfectly equal and non-zero.

While the initial currents are the same, the rate at which the current and charge decay is different. The charge on a discharging capacitor follows an exponential decay:

The crucial factor here is the time constant, $\tau$, which is defined as the product of capacitance and resistance ($\tau =RC$). The time constant tells us how sluggish the circuit is; a larger $\tau$ means the capacitor takes longer to discharge.

Let's compare the time constants for our two circuits:

Clearly, ${\tau }_1<{\tau }_2$. Because $C_1$ has a smaller time constant, it dumps its charge much more rapidly than $C_2$. Consequently, $C_1$ will reach the 50% mark of its initial charge significantly sooner than $C_2$.