Consider the system of capacitors shown in the figure below (C_1 = 9.00 mu F, C_

Question

Consider the system of capacitors shown in the figure below (C_1 = 9.00 mu F, C_2 = 2.00 mu F F). Find the equivalent capacitance of the system. _____________ mu F Find the charge on each capacitor. ______________ mu C (on C_1) ___________ mu C (on C_2) ___________ mu C (on the 6.00 mu F capacitor) ___________ mu C (on the 2.00 mu F capacitor) Find the potential difference across each capacitor. _____________ V (across C_1) _____________ V (across C_2) _____________ V (across the 6.00 mu F capacitor) _____________ V (across the 2.00 mu F capacitor) Find the total energy stored by the group. _____________ mJ

Explanation / Answer

(a) Resultant of the first two capacitor connected in series -

Cs1 = (9*6) / (9+6) = 3.6 uF

and resultant of the lower two capacitors connected in series -

Cs2 = (2*2) / (2+2) = 1 uF

Now, these two capacitors are connected in parallel.

So, the resultant equivalent capacitance, Ceq = Cs1 + Cs2

= 3.6 + 1

= 4.6 uF

(b) Charge on capacitor C1 = 3.6 x 10^-6 x 90 = 324 x 10^-6 C = 324 uC

Charge of 6.0 uF capacitor = 324 uC

Charge on 2.0 uF capacitor = 1 x 10^-6 x 90 = 90 x 10^-6 C = 90 uC

Charge on capacitor C2 = 90 uC

(c) Potential difference across C1 = q1 / C1 = 324x10^-6 / (9x10^-6) = 36 V

Potential difference across C2 = q2 / C2 = 90x10^-6 / (2x10^-6) = 45 V

Potential difference across 6.0 uF capacitor = 324x10^-6 / (6x10^-6) = 54 V

Potential difference across 2.0 uF capacitor = 90x10^-6 / (2x10^-6) = 45 V

(d) Energy stored by the group = (1/2)*Ceq*v^2 = 0.5*4.6x10^-6*90^2 = 18630 x 10^-6 J = 18.63 x 10^-3 J

= 18.63 mJ

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