## Why does a capacitor take infinite time to charge?

a charging cap proportional to the applied voltage. low voltage equals longer charge. resistance during charging increases, discharged means lower resistance. The voltage level determines the charge rate. never an infinite load.

loading is a time-consuming process (for example, climbing a steep hill). when a voltage is applied to be stored in the capacitor as a charge (s), a force or time is required, once the charge proportional to the applied voltage is stored, it stops.

The magnitude of the current flowing through the capacitor determines the charging speed of the capacitor. in the case of a voltage source, the current is determined by the path resistance and the voltage difference between the input voltage source and the capacitor. this magnitude of the charging current decreases exponentially as the capacitor voltage approaches the input voltage source, which lengthens the time required to fully charge the capacitor.

if a capacitor is charged via a resistor connected, for example, to a battery, the more the capacitor voltage approaches the battery voltage, the lower the charging current. it’s just the ohms law for resistance. but in a time interval which is only a small multiple of the time constant rc, the capacitor voltage is so close to the battery voltage that the difference does not matter.

Because the load circuit current r-c depends on the voltage difference and the resistance. As the voltage on the capacitor approaches the voltage at the source and the Vs-Vc difference approaches zero, the current on the series load resistor approaches zero to never fill completely.

However, if you charge it with a real power source, it will charge in a certain amount of time, limited by the conformity of the current source to the voltage. but of course, if you have an ideal power source with infinite voltage compliance, it will take forever to fully charge!

the short answer is that a capacitor is charging, it creates a repulsive force. the more it is loaded, the more force it takes to charge it at the same rate.