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.
Then we get Q = CV0. This is a popular formula for the voltage across a capacitor. If the external battery is removed, the capacitor switches to discharging mode and the voltage drop across the capacitor starts to decrease. The voltage across the discharging capacitor becomes, V (t) = V 0 e -t/τ ……… (3) τ = RC is the time constant.
This capacitive reactance produces a voltage drop across each capacitor, therefore the series connected capacitors act as a capacitive voltage divider network. The result is that the voltage divider formula applied to resistors can also be used to find the individual voltages for two capacitors in series. Then:
The voltage of C1 and C2 must sum to 6V. Use q=CV and solve for the voltages. Reworked by RM: Take 3: The same current flows in C1 & C2. the charge on C1 and C2 must be equal. But, also by definition Charge = capacitance x Voltage (Q = C x V). So, for equal charges in each, capacitor voltage will be inversely proportional to capacitance.
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)
Like other components (resistors, inductors), a capacitor also offers opposition to the current flow (Direct current only) through it. That means it generates impedance. Ohm’s law tells us that an impedance causes a voltage drop. Now, the question is, “ Is there any voltage drop across a capacitor? ” The answer is, “Yes”.