lets say two capacitors C1 and C2 1uF each are connected in series with a voltage source 5v. One of the capacitors C2 has an initial voltage of 1v and the other has no charge on it initially.What is the final voltages across the capacitor C2?
Ct = (1*1)/(1+1) = 1/2 = 0.5uF. = Tot.
capacitance.
Qt=Q1 = Q2 = Ct*E = 0.5*5 = 2.5uC.
V2 = Q2/C2 = 2.5/1 = 2.5 Volts.
what about the initial 1v on capacitor c2. Since the source is at 5v ,this capacitor charge cannot get redistributed.Hence should it not add to the final value across c2? can somebody throw some light on this.?
To find the final voltage across capacitor C2, we need to calculate the charge on each capacitor and then calculate the voltage across each capacitor.
First, let's calculate the charge on each capacitor using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.
For capacitor C1:
Q1 = C1 * V = (1uF) * (5V) = 5μC
For capacitor C2:
Q2 = C2 * V = (1uF) * (5V) = 5μC
Since both capacitors are connected in series, they share the same charge. Therefore, the charge on C1 and C2 will be the same.
Now, let's calculate the final voltage across capacitor C2. We can use the formula V = Q / C, where V is the voltage, Q is the charge, and C is the capacitance.
For capacitor C2:
V2 = Q2 / C2 = (5μC) / (1uF) = 5V
Therefore, the final voltage across capacitor C2 is 5V.