Because the charge (Q) is equal and constant, the voltage drop or potential difference across the capacitor is dependent on the capacitor value, V = Q/C. A lower capacitance value results in a bigger voltage drop, whereas a large capacitance value results in a smaller voltage drop. A capacitor is a critical component in an electrical circuit.
As the charge, ( Q ) is equal and constant, the voltage drop across the capacitor is determined by the value of the capacitor only as V = Q ÷ C. A small capacitance value will result in a larger voltage while a large value of capacitance will result in a smaller voltage drop.
The voltage drop is the same over both capacitors. The voltage level is not. For instance, if there is a total voltage of 2 V across the whole circuit, and there is nothing in the circuit other than the capacitors and the voltage source, then both capacitors will have a voltage drop of 1 V.
The voltage drop across an uncharged capacitor is zero. Because, for an uncharged capacitor, Q=0 and hence, the voltage V=0. During charging an AC capacitor of capacitance C with a series resistor R, the equation for the voltage across a charging capacitor at any time t is, V (t) = V s (1 – e -t/τ) …….. (1)
: If you connect two uncharged capacitors in series to a battery, there will be a current in the circuit until equilibrium is reached. As current flows, the capacitors will start charging, and there will be a voltage drop along each capacitor. In equilibrium, the net voltage drop in the two capacitors will be equal to the voltage in the battery.
The voltage across a capacitor changes due to a change in charge on it. So, during the charging of a capacitor, the voltage across it increases. When the capacitor is completely charged, the voltage across the capacitor becomes constant. Now, if we remove the external battery, the discharging of the capacitor begins.