Consider the system of capacitors shown in the figure below (C1 = 8.00 µF, C2 =

Question

Consider the system of capacitors shown in the figure below (C1 = 8.00 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

Given that

C1 = 8.00 µF, C2 = 6.00 µF

Now from the above figure

C1 and 6uF are in series than the equivalent capacitance is

1/Ceq1 =1/8+1/6

Ceq1 =(8)(6)/(8+6) =48/14=3.428uF

Now 2uF and C2 are in series than the equivalent capacitance is

1/Ceq2 =1/2+1/6

Ceq2 =(2)(6)/(2+6) =12/8 =1.5uF

Now 3.428uF and 1.5uF are in parallel than the total equivalent capacitance is
Ceqtotal =Ceq1+Ceq2 =3.428+1.5 =4.928uF

The charge is given by

Weknow that Q =CV =Ceqtotal*V =4.928*10-6F*90V =444.52*uC

Now in parallel combination the voltage remains the same, but the charge will be different

Then the charge across the C1 then Q(6) =Q (3.428/3.428+1.5) =0.695Q =0.695*444.53uC =309.06uC

The same will be manitained across the C1 then Q1 =309.06uC

Now Q(C2) =Q*(1.5/3.428+1.5) =Q(1.5/4.928) =0.304*444.53uC =135.307uC

The same charge will be across the 2 and 6 uF =135.307uF

Now for the potenital difference we can find Q =CV

V1 =Q1/C1 =

V2 =Q2/C2 =

similalry we can find for the rest of the capacitors

Now the total energy stored is given by

U =(1/2)CeqtotalV2 =0.5*4.928*10-6*(90V)2 =19.958mJ

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