If a capacitor is introduced into this circuit, it will gradually charge until the the voltage across it is also approximately 5V, and the current in this circuit will become zero. What is now preventing us from suddenly changing the voltage from 5V to let's say 10V (again like a step increase - instantaneously)?
We now apply a voltage of 5V to the circuit (like a step increase - instantaneously). The voltage across the resistor changes instantaneously to 5V. If a capacitor is introduced into this circuit, it will gradually charge until the the voltage across it is also approximately 5V, and the current in this circuit will become zero.
The circuit is at steady state when the voltage and the current reach their final values and stop changing. In steady state, the capacitor has a voltage across it, but no current flows through the circuit: the capacitor acts like an open circuit. How do you calculate steady state current in a capacitor? Is a capacitor fully charged in steady state?
Capacitors do not so much resist current; it is more productive to think in terms of them reacting to it. The current through a capacitor is equal to the capacitance times the rate of change of the capacitor voltage with respect to time (i.e., its slope).
Answer: Since the voltage across the capacitor approaches the voltage across the terminals, the electric field in the wires approaches zero, and so the current approaches zero. Therefore no more charge will flow to or from the plates of the capacitor. Do capacitors act as open circuits?
Given a fixed voltage, the capacitor current is zero and thus the capacitor behaves like an open. If the voltage is changing rapidly, the current will be high and the capacitor behaves more like a short. Expressed as a formula: i = Cdv dt (6.1.2.5) (6.1.2.5) i = C d v d t Where i i is the current flowing through the capacitor,