Science, asked by skfurkan24, 8 months ago

266.67 V
С
6 x 10-6
F :C=4 UF: C. = 3 uF and C = 6 UF
Problem 5.25. Four capacitors C,C,,Cand C, are connected to a battery of 12 V as shown in
ig. 5.48. Find the potential difference between points A and B.​

Answers

Answered by Sabavatlinisha
0

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

Explanation:

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