# Three capacitors with capacities C1 = 1.0 μF, C2 = 1.0 μF, C3 = 2.0 μF are connected according

**Three capacitors with capacities C1 = 1.0 μF, C2 = 1.0 μF, C3 = 2.0 μF are connected according to the diagram shown in Figure 11.38, connected to a 120 V DC voltage source. What is their total capacitance? Determine the charge and voltage across each of the capacitors.**

When capacitors are connected in parallel, their total capacitance is equal to the sum of the capacitors. Therefore, the capacitance of the section C2 and C3 is equal to C2 + C3, and the capacitor C1 is connected in series with them, which means: 1 / C1 + 1 / (C2 + C3) = 1 / C; 1 / C = 1/1 + 1 / (1 + 2) ó1 / C = 1 + 1/3 ó1 / C = 4/3 ó C = 0.75 * 10-6 F. The total charge is determined by Q = C * U = 0.75 * 10-6 F * 120 V = 9 * 10-5 Cl. The charge on the capacitor C1 will be Q = 9 * 10-5 C. Then the voltage on it will be U1 = Q / C1; U1 = 9 * 10-5 / 1 * 10-6 = 90 V. Then on C2 and C3 there will be a voltage U2 = U3 = U – U1 = 120 – 90 = 30 V. The charges on them will be: Q2 = C2 * U2 = 1 * 10-6 * 30 = 3 * 10-5 Cl. Similarly, Q3 = C3 * U3 = 2 * 10-6 * 30 = 6 * 10-5 Cl.