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?
can anybody answer this?
To find the final voltages across the capacitor C2, we need to understand how capacitors behave when connected in series.
When capacitors are connected in series, the total capacitance (C_series) can be calculated using the formula:
1 / C_series = 1 / C1 + 1 / C2
Let's plug in the values:
1 / C_series = 1 / 1uF + 1 / 1uF = 2uF
1 / C_series = 1 / 2uF
C_series = 2uF
Next, we need to apply the principles of capacitor charging and discharging. In this case, the capacitor C2 starts with an initial voltage of 1V, and the other capacitor C1 has no charge initially.
When capacitors are in series with a voltage source, the charges across the capacitors must be the same. Therefore, the charges Q1 and Q2 on capacitors C1 and C2, respectively, are equal:
Q1 = Q2
The charge Q on a capacitor is given by the formula:
Q = C * V
where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.
For capacitor C1:
Q1 = C1 * V1 = C1 * (5V)
For capacitor C2:
Q2 = C2 * V2 = C2 * (V2) [Note: We're trying to find V2, the final voltage across C2.]
Since the charges on the two capacitors are equal, we can equate Q1 and Q2:
C1 * (5V) = C2 * (V2)
Rearranging the equation to solve for V2:
V2 = (C1 / C2) * 5V
Plugging in the values:
V2 = (1uF / 1uF) * 5V = 5V
Therefore, the final voltage across the capacitor C2 is 5V.