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.