الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Lectures in General Physics
Example 6.8
1 2
Consider the circuit shown in figure
S
6.11 where C 1 =4µF, C 2 =6µF, C 3 =2µF,
and V=35V. C 1 is first charged by
closing switch S to point 1. S is then
connected to point 2 in the circuit.
(a) Calculate the initial charge
Figure 6.11
acquired by C 1 ,
(b) Calculate the final charge on each of the three capacitors.
(c) Calculate the potential difference across each capacitor after
the switch is connected to point 2.
Solution
When switch S is connected to point 1, the potential difference on C 1 is
35V. Hence the charge Q 1 is given by
Q 1 = C 1 xV=4x35 =140µC
When switch S is connected to point 2, the charge on C 1 will be distributed
among the three capacitors. Notice that C 2 and C 3 are connected in series,
therefore
1 1 1
= +
C′ C C
2
C ′ = 1.5µF
3
=
1
6
+
1
2
=
4
6
We know that the charges are distributed equally on capacitor connected in
series, but the charges are distributed with respect to their capacitance when
they are connected in parallel. Therefore,
140
Q
1
= × 4 = 101.8µC
4 + 1.5
But the charge Q′ on the capacitor C′ is
Q ′ = 140 −101.8
= 38.2µC
Dr. Hazem Falah Sakeek