الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Lectures in General Physics
Therefore for C 2 , V 2 =24V and Q 2 =C 2 V 2 =48µC
Also in circuit (ii) the potential difference
V de =V gh =24V
And
Q de =C de V de =2×24=48µC
The two capacitors shown in circuit (i) between points d and a are in series,
and therefore the charge on each is equal to Q de .
Therefore for C 6 , Q 6 =48µC
Q6
V6 = = 8V
C
6
For C 3 , Q 3 =48µC and V 3 =Q 3 /C 3 =16V
The results can be summarized as follow:
Capacitor
Potential Charge
Difference (V) (µC)
C 1 48 48
C 2 24 48
C 3 16 48
C 4 24 96
C 5 48 240
C 6 8 48
C eq 48 384
Dr. Hazem Falah Sakeek