الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Lectures in General Physics
Solution
C 2 and C 3 are connected in parallel, therefore
C′ =C 2 +C 3 =4+12=16µF
Now C′ is connected in series with C 1 , therefore the equivalent capacitance
is
1
C
1 1
= +
C′
C
1
=
1
6
+
1
16
=
11
48
C = 4.36µF
The total charge Q =CV = 4.36x12 = 52.36µC
The charge will be equally distributed on the capacitor C 1 and C′
Q 1 =Q′ =Q=52.36µC
But Q′ = C′ V’, therefore
V ′ = 52.36/16=3.27 volts
The potential difference on C 1 is
V 1 =12-3.27=8.73volts
The potential difference on both C 2 and C 3 is equivalent to V ′ since they
are connected in parallel.
V 2 = V 3 =3.27volts
Q 2 = C 2 V 2 = 13.08µC
Q 3 = C 3 V 3 = 39.24µC
Dr. Hazem Falah Sakeek