# Consider a capacitor of 20 uF charged to 30V. so it has energy of 9mJ. now a 10 uF uncharged capacitor is connected parallel to it. what will be the final total energy left in capacitors? assuming no self leakage.

When two capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. Therefore, the total capacitance of the circuit becomes:

C_total = C1 + C2 = 20 uF + 10 uF = 30 uF

Since the two capacitors are connected in parallel, they will have the same voltage across them. The charge on the capacitors will redistribute such that the voltage across both capacitors is the same.

The total charge on the capacitors before and after connecting them in parallel must be the same. Thus, we can write:

Q_total = C_total * V_final

where V_final is the voltage across both capacitors after they are connected in parallel.

Using the conservation of energy, the total energy of the circuit before and after the capacitors are connected in parallel must be the same. Thus, we can write:

E_total = 1/2 * C1 * V1^2 + 1/2 * C2 * V2^2 = 1/2 * C_total * V_final^2

where V1 and V2 are the voltages across the capacitors before they are connected in parallel.

We know that C1 = 20 uF, V1 = 30 V, C2 = 10 uF, and V2 = 0 V. Solving for V_final, we get:

V_final = (C1 * V1 + C2 * V2) / C_total = (20 uF * 30 V + 10 uF * 0 V) / 30 uF = 20 V

Now we can calculate the total energy of the circuit after the capacitors are connected in parallel:

E_total = 1/2 * C_total * V_final^2 = 1/2 * 30 uF * (20 V)^2 = 6 mJ

Therefore, the total energy of the circuit after the capacitors are connected in parallel is 6 mJ.

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