Consider the system of capacitors shown in the figure below ( C 1 = 6.00 µF, C 2
ID: 1592669 • Letter: C
Question
Consider the system of capacitors shown in the figure below (C1 = 6.00 µF, C2 = 4.00 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 = [ Cseries of 6,6] + [ Cseries of 4,2]
=[6*6/(6+6) + 4*2/(4+2)]
= 3+ 8/6= 4.33 uF
B) Charge on C1= [ Cseries of 6,6] *V= 3*90=270 uC
Charge on C2= [ Cseries of 4,2] *V=1.33*90=120 uC
Charge (on the 6.00 µF capacitor) =[ Cseries of 6,6] *V= 3*90=270 uC
Charge (on the 2.00 µF capacitor)= [ Cseries of 4,2] *V=1.33*90=120 uC
C) Voltage across c1= Charge/capacitance= 270/6= 45V
Voltage across C2=120/4=30V
Voltage across 6uF=270/6= 45V
Voltage across 2uF=120/2=60V
d) Total energy = 0.5 CV2 = 0.5*4.3333*90*90= 17550 uJ = 0.01755 mJ
It is the unit. You had to convert it into mJ. No problem, I have done.