Consider the system of capacitors shown in the figure below ( C 1 = 6.00 µF, C 2

Question

Consider the system of capacitors shown in the figure below (C1 = 6.00 µF, C2 = 6.00 µF).

(a) Find the equivalent capacitance of the system.
__________ µF

(b) Find the charge on each capacitor.
__________ µC (on C1)
__________ µC (on C2)
__________ µC (on the 6.00 µF capacitor)
__________ µC (on the 2.00 µF capacitor)

(c) Find the potential difference across each capacitor.
__________ V (across C1)
__________ V (across C2)
__________ V (across the 6.00 µF capacitor)
__________ V (across the 2.00 µF capacitor)

(d) Find the total energy stored by the group.
__________ mJ

Explanation / Answer

a)Ceq = 6*6/(6+6) + 6*2/(6+2)

=3 + 1.5 =4.5 uF

B) charge on c1= Capacitance in that branch * potential

=3uF * 90V = 270uC

charge on c2= 1.5uF *90v = 135uC

charge on the 6.00 µF capacitor =3uF * 90V = 270uC

charge on the 2.00 µF capacitor = 1.5uF *90v = 135uC

c) V (across C1) = Q/C1= 270uC /6uF = 45V

V (across C2) = 135 uC/6uF= 22.5V

V (across the 6.00 µF capacitor)= 270uC /6uF = 45V

V (across the 2.00 µF capacitor) =  135 uC/2uF= 67.5V

(d) total energy stored by the group = 0.5C V2 = 0.5 x 4.5 x 90 x 90 = 18225 uJ = 18.225 mJ

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