الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Lectures in General Physics
(b) We find the charge and the voltage across each capacitor by working
backwards through solution figures (c) through (a).
For the 5.96µF capacitor we have
Q = CV = 5 .96 × 15 = 89.5µ
C
In figure (b) we have, for the 8.5µF capacitor,
V Q 89.5
∆ = = = 10. V
ac
C 8.5
5
and for the 20µF in figure (b) and (a)
Q 89.5µ
C
20
=
V Q 89.5
∆ = = = 4. V
cb
C 20
47
Next (a) is equivalent to (b), so
∆ V cb
= 4. 47V
and ∆ V ac
= 10. 5V
Thus for the 2.5µF and 6µF capacitors
∆ V = 10. 5V
Therefore
Q = CV = 2.5×
10.5 26.3µ
C
2 .5
=
Q = CV = 6 × 10.5 63.2µ
C
6
=
Q
15
= 26.3µC
Q
3
= 26.3µ
C
For the potential difference across the capacitors C 15 and C 3 are
Q 26.3
∆ V15 = = = 1. 75V
C 15
Q 26.3
∆ V3 = = = 8. 77V
C 3
Dr. Hazem Falah Sakeek