Two capacitors with capacities of 4.0 and 1.0 μF are connected in series and connected to a 220V DC voltage source.

Two capacitors with capacities of 4.0 and 1.0 μF are connected in series and connected to a 220V DC voltage source. Determine the total capacitance. How is the voltage distributed between the capacitors?

Given:

C1 = 4.0 μF = 4 * 10-3 F (converted to SI);

C2 = 1.0 μF = 1 * 10-3 F (converted to SI);

U = 220 V

Find: C -? U1, U2 -?

Decision:

The serial connection will:

1 / C = 1 / C1 + 1 / C2 = 1/4 + 1/1 = 5/4;

C = 4/5 = 0.8 μF = 0.8 * 10-3 F;

Let’s find the amount of charge, provided: Q = Q1 = Q2;

Q = C * U = 0.8 * 10-3 * 220 = 0.176 C;

The voltage will be:

U1 = Q / C1 = 0.176 / (4 * 10-3) = 44 B;

U2 = Q / C2 = 0.176 / (1 * 10-3) = 176 B.

Answer: C = 0.8 * 10-3 F; U1 = 44 B; U2 = 60 B.