Question

找到如图所示连接的电容器组A和B之间的等效电容。以C1 = 5µF，C2 = 10µF、C3 = 2µF？

（a）找到如图所示连接的电容器组A和B之间的等效电容。以C1 = 5µF，C2 = 10µF、C3 = 2µF（B）是电荷存储在C3如果A点和B点之间的电位差为60 V？(a) Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in the right Figure. Take C1 = 5.00µF, C2 = 10.0µF, and C3 = 2.00µF. (b) What charge is stored on C3 if the potential difference between points a and b is 60.0 V?

Answer 1

(a)， ab之间的等效电容C：
设，上方框与下方框的连接点为d，则ab间的等效电容Cab等于Cad与Cdb的串联；
Cad=三个支路电容的并联。
支路1的电容=C1串联C2=支路3=（C1xC2)/(C1+C2)=10/3uF;
支路2的电容=C3=2； 所以Cad=10/3+2+10/3=26/3uF;
Cdb=C2+C2=20uF
所以Cab=Cad x Cdb/(Cad+Cdb)=260/43uF =6.05uF
(b), 求C3带的电荷量
由于Uab=60V, 所以电容C3两端的等效电压
Uad=Uab x Cdb/(Cad+Cdb) =60V x 20/(20+26/3)=1800/43V=41.86V
则，C3上的电荷量q=C3 x Uad=2uF x 41.86=83.79库伦

Answer 2

电容网络的解法跟电阻网络其实是一样的，只不过是按照1/C来处理就行了……
（1）a到中间点(2×5×10)/15+2 = 26/3 μF = 8.67 μF，中间点到b是10×2 = 20 μF，最后a到b的电容是26×20÷86 = 260/43 μF = 6.05 μF
（2）当Uab = 60V时，C3两端电压为60×(1/8.67)÷(1/8.67 + 1/20) = 41.86V
套公式Q = UC = 41.86×8.67 = 362.93 C