Question

Consider the arrangement of the four initially uncharged capacitors shown in Fig. 25-36. Capacitors A, B, C, and D have capacitances 5.4 muF, 4.3 muF, 3.2 muF, and 2.1 muF, respectively. Suppose that a battery applies a potential difference of 3000 V across the circuit, which is then disconnected from the battery. What is the potential difference across each capacitor?

Answer 1

First Get total C: Here Given that C_A = 5.4 mu f C_B = 4.3 mu f C_C = 3.2 mu f C_D = 2.1 mu f V = 3000 V Here C_B and C_D are in parallel so add uo C_S = C_B + C_D = 43 mu f + 2.1 mu f C_s = 6.40 mu f C_A and C_s and C_C are in series then 1/C_total = 1/C_A + 1/C_s + 1/C_C 1/C_total = 1/5.4 + 1/6.40 + 1/3.2 1/C_total = 0.653 mu f C_total = 1.52 mu f so Q = C_total v = 1.52 times 3000 V = 4560 mu c Parallel capacitance will share equal voltage os V_B = V_D = Q/C_s = 4560 mu_c/6.40 = 712.5 V \r\n Voltage value on capacitor A = V_A = 4560 mu c/5.4 mu f V_A = 844.44 v Voltage value on capacitor c V_c = Q/C_C = 4560 mu c/32 mu f V_C = 1425 V