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CR Discharging

<^< Worksheet: C-R Charging | Course Index | Worksheet: C-R Discharging >^>

Having considered the situation when a capacitor is being charged, let's also consider what happens when an already charged capacitor is discharged. When the battery is removed and the fully charged capacitor is connected to the resistor, the capacitor will discharge through the resistor as soon as the switch is closed and the capacitor voltage (V_C) will fall exponentially. The current in the circuit (I_C) will also decay. The rate of discharge (i.e., the rate of decay of voltage with time) will once again be governed by the time constant of the circuit (C x R).

The voltage developed across the discharging capacitor (V_C) varies with time (t) according to the relationship:

V_C = V_S e^-t/CR

The capacitor voltage will fall to approximately 37% of the initial voltage in a time equal to the time constant. At the end of the next interval of time equal to the time constant (i.e., after an elapsed time equal to 2CR) the voltage will have fallen by 37% of the remainder, and so on.

In theory, the capacitor will never become fully discharged. However, after a period of time equal to 5CR, the capacitor voltage will to all intents and purposes be zero. At this point the capacitor voltage will have fallen below 1% of its initial value. At this point we can consider it to be fully discharged.

As with charging, the current in the capacitor (I_C) varies with time (t) according to the relationship:

I_C = (V_S/R) x (e^t/CR)

where V_S, is the supply voltage ( the voltage to which the capacitor was initially charged) , t is the time, C is the capacitance, and R is the resistance.

The current will fall to approximately 37% of the initial current in a time equal to the time constant. At the end of the next interval of time equal to the time constant (i.e., after a total time of 2CR has elapsed) the voltage will have fallen by a further 37% of the remainder, and so on.

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