Question

Three Capacitor of capacitance 2pF, 3pF and 4pF are connected in parallel. What is the total

capacitanceofthecombination? (1pF= 10-12F)​

Answer 1

Answer:

As Q=CV

∴ Initially charge on each capacitor is

Q  

1

​  

=C  

1

​  

V  

1

​  

=(3μF)(6V)=18μC and Q  

2

​  

=C  

2

​  

V  

2

​  

=(4μF)(6V)=24μC

When two capacitors are joined to each other such that negative plate of one is attached with the positive plate of the other. The charges Q  

1

​  

 and Q  

2

​  

 are redistributed till they attain the common potential which is given by

Common potential V=  

Totalcapacitiance

Totalcharge

​  

 

=  

3μF+4μF

24μC−18μC

​  

=  

7

6

​  

V

Final energy stored

U  

f

​  

=  

2

1

​  

(C  

1

​  

+C  

2

​  

)V  

2

=  

2

1

​  

[3×10  

−6

+4×10  

−6

]×(  

7

6

​  

)  

2

=  

2

1

​  

×7×10  

−6

×  

49

36

​  

=2.57×10  

−6

J