Two capacitors with a capacity of 10 μF each were connected in series and connected to a power source

Two capacitors with a capacity of 10 μF each were connected in series and connected to a power source, the voltage of which at the output was 2V. Determine the energy of this battery capacitors.

These tasks: C1 (capacity of the first capacitor) = C2 (capacity of the second capacitor) = 10 μF (10 * 10-6 F); taken capacitors were connected in series; U (voltage at the output of the power supply) = 2 V.

The energy of the resulting capacitor bank can be calculated using the formula: W = 0.5 * C * U ^ 2 = 0.5 * (C1 / n) * U ^ 2.

Calculation: W = 0.5 * (10 * 10 ^ -6 / 2) * 2 ^ 2 = 10 * 10 ^ -6 J (10 μJ).

Answer: The energy of the resulting capacitor bank is 10 μJ.