Question

927 The potential difference across the 1.00 μF capacitor in Fig. 18-58 is 10.0 V. Find (a) the equivalent capacitance between A and B: (b) the potential difference between C and D 1.00 uF 2.00 pF 2.00 pF 3.00 uF 32 Fig. 18-58

Answer 1

a) The equivalent capacitance of capacitor 1.00 mu F, 2.00 mu F and 2.00 mu F is C_s = (1/(1.00 mu F) + 1/(2.00 mu F) + 1/(1.00 mu F))^-1 = 0.50 mu F The equivalent capacitance between C and D is C_C D = 0.50 mu F + 3.00 mu F = 3.50 mu F The equivalent capacitance between A and B is C_A B = (1/3.50 mu F + 1/7.00 mu F)^-1 = 2.33 mu F b) The total charge is Q = C_A B delta V = (2.33 mu F) (10.0 V) = 23.3 mu C The potential difference across the CD is V_CD = Q/C_CD = 23.3 mu C/3.50 mu F = 6.657 V