A Capacitor is connected to a voltage source of 6V in series. A load is connected in parallel with the capacitor. The load will remain as an open circuit until 5V is put across it. The capacitor is then charged. Once the voltage across the load reaches 5V the load will no longer act as an open circuit and the capacitor will discharge. Once the load voltage has droped to 0.2 the load will act as an open circuit and the capactior will begin to charge once more.
Write an equation for the Capacitor to only discharg to 0.2V due to the load acting as an open circuit at 0.2V