Consider the system of capacitors shown in the figure below (C_1 = 2.00 mu F, C_

Question

Consider the system of capacitors shown in the figure below (C_1 = 2.00 mu F, C_2 = 7.00 mu F). (a) Find the equivalent capacitance of the system. 3.055 mu F (b) Find the charge on each capacitor. 135 mu C (on C_1) 140 mu C (on C_2) 135 mu C (on the 6.00 mu F capacitor) 140 mu C (on the 2.00 mu F capacitor) (c) Find the potential difference across each capacitor. 67.5 V (across C_1) 22.5 V (across C_2) 67.5 V (across the 6.00 mu F capacitor) 67.5 V (across the 2.00 mu F capacitor) (d) Find the total energy stored by the group. mJ

Explanation / Answer

C.

Potential difference on C2 = Q2/C2 = 140 uC/7 uF = 20 V

Potential difference across 6.00 uF = 135/6 = 22.5 V

Potential difference across 2.00 uF = 140/2 = 70 V

D.

Total energy = U = 0.5*C*V^2

U = 0.5*3.055*10^-6*90^2

U = 12.37*10^-3 J

U = 12.37 mJ

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