Find the following. (In the figure, use C 1 = 37.20 F and C 2 = 31.20 F.) (a) th
ID: 1446960 • Letter: F
Question
Find the following. (In the figure, use
C1 = 37.20 F
and
C2 = 31.20 F.)
(a) the equivalent capacitance of the capacitors in the figure above
(b) the charge on each capacitor
on the right 37.20-F capacitor
on the left 37.20-F capacitor
on the 31.20-F capacitor
on the 6.00-F capacitor
(c) the potential difference across each capacitor
on the right 37.20-F capacitor
on the left 37.20-F capacitor
on the 31.20-F capacitor
on the 6.00-F capacitor
on the right 37.20-F capacitor
on the left 37.20-F capacitor
on the 31.20-F capacitor
on the 6.00-F capacitor
Explanation / Answer
Capacitors in parallel are added
Capacitor in series are added like thi, Ceq = (1/C1 + 1/C2)^-1
1. Eq capacitance = (1/C1 + 1/(6+C2) + 1/C1)^-1 micro F = (2/37.2 + 1/(6+31.2) )^-1= 12.4 micro F
2. Due to symmetry, C1, C2+6 , C1 will have same charge deposited, let this be Q
Q = Ceq*V = 12.4*10^-6 * 9 = 111.6 micro C
Now, C2 and 6 mu F cap will have diff charge stored on them
Potential diff across both = Q/Ceq = 111.6*10^-6/37.2*10^-6 = 3 V
charge on 6 micro F = CV = 3*6 = 18 micro C
charge on C2 = C2V = 93.6 micro C
charge on C1 (both) = 111.6 micro C
3. potential diff across C1 (both) = 3V
potential diff across 6 micro F and C2 = 3V
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