In the figure below, battery B is a 24 Volt battery. In this circuit C 1 = 2 . 0
ID: 1536102 • Letter: I
Question
In the figure below, battery B is a 24 Volt battery. In this circuit C1 = 2.0 µF, C2 = 4.0 µF, C3 = 6.0 µF and C4 = 8.0 µF.
Find the charge on each capacitor when the switch S1 is closed (S2 remains open).
Find the charge on each capacitor when later S2 is also closed.
Find the energy stored in each capacitor in a) and in b)
Find the equivalent capacitance of the circuit for parts a) and b). Confirmthat energy conservation works in this situation by finding the energy stored in each equivalent capacitance and comparing with your answers to c).
Explanation / Answer
Given circuit diagram consisting of
battery B supplies 24 V. Take C1 = 2.0 µF, C2 = 4.0 µF, C3 = 6.0 µF and C4 = 8.0 µF.
when s1 only closed, S2 is insignificant
the capacitors C2,C4 and C1,C3 are in series with each other
so C2,4 = C2*C4/(C2+C4) = 4*8 /(4+8) F = 2.67F
C1,3 = C1*C3/(C1+C3) = 2*6 /(2+6) F = 1.5 F
so charge is Q2,4 = C24*V = 2.67*24 = 64.08 C = Q2 = Q4 and
charge on C1,C3 is Q1,3= C13*V = 1.5*24 = 36 C = Q1 = Q3 and
when s2 alos closed
C4,C3 and C1,C2 are in parallel combination
C4,3 = C4+C3 = 8+6 F = 14 F,
C1,2 = C1+C2 = 2+4 F= 6 F
these are agin in series combination so the net capacitance is C = (C4,3 *C1,2)/(C4,3 +C1,2) = (14*6)/(14+6 )= 4.2 F
so total cahrge is Q = C*V = 4.2*24 = 100.8 C
so all the capacitors are in series then the charge will be same across each capacitor that is
Q12 = Q34 = Q1234 = 100.8 C
charge on C1 , is Q1 = C1*V/(C1+C2) = 2*100.8/(2+4) = 33.6 c
charge on C2 , is Q2 = C2*V/(C1+C2) = 4*100.8/(2+4) = 67.2 c
charge on C3 , is Q3 = C3*V/(C3+C4) = 6*100.8/(6+8) = 43.2 c
charge on C4 , is Q4 = C4*V/(C1+C2) = 8*100.8/(2+4) = 57.6 c
energy stored in the capacitor is U = 0.5 c*V^2 = Q^2/2C
U1 = Q1^2/2C1 = 33.6^2/(2*2) J = 282.24 J
U2 = Q2^2/2C2 = 67.2^2/(2*4) J = 564.48 J
U3 = Q3^2/2C3 = 43.2^2/(2*6) J = 155.52 J
U4 = Q4^2/2C4 = 57.6^2/(2*8) J = 207.36 J