Suppose in the figure (Figure 1) that C1=C2=C3= 18.0 microF and C4 = 28.2microF
ID: 2237664 • Letter: S
Question
Suppose in the figure (Figure 1) that C1=C2=C3= 18.0 microF and C4 = 28.2microF If the charge C2 is Q2 = 37.4 microCoulombs, determine the charge on each of the other capacitors. Figure 1:
The attempt at the solution:
The rules my teacher gave me for this stuff:
Parallel:
QT = Q1 + Q2 +Q3...
Ceq = C1 + C2 + C3...
Voltage is same in parallel
Series:
Q or charge is same in series
1/Ceq = 1/C1 + 1/C2...
VT = V1 + V2 + V3...
My answer...
Let 1 and 2 be in Series. Then Capacitance can be added as reciprocals
1/Ceq = 1/C1 + 1/C2 or 1/18 + 1/18...Ceq = 9microF for 1 and 2
Q1=Q2...because they are in series, Q is same. Q1 = Q2 = 37.4 microC
Q = CV. So, 37.4 = 9*V
V = 4.155 V
I figured that because Voltage is the same in parallel, then this voltage is the same across the board for 3 and 4.
So I found C for 3 and 4
3 and 4 are in series with each other, so 1/Ct = 1/18 + 1/28.2
End up with Ct = 10.987...
Then Q = CtV
= 10.987*4.155 = 45.65 microC
Q3 = Q4 because Q's are equal in series...final answer I put down was Q1=Q2 = 37.4 and Q3 = Q4 = 45.65 all in microColoumbs
Where did I go wrong? Please help, I did my best and honestly tried.
Explanation / Answer
you did correct till
v= 4.155 V
then now the combination of C1 C2 and C3 are in parallel
Ceq = 9 +18 = 27 microF
charge on C3 = 4.155 * 18 = 74.8 microC
now C4 and Equivalent of C1,C2 and C3 are in series
therefore
Cnet = C4 * Ceq/(C4 + Ceq)
= 28.2 *27/55.2
= 13.793 microF
Qnet = 4.155 * 13.793
= 57.311 micro C
charge on C4 = 57.311/2 =28.65 micro C