two capacitor of 10microfarad and 20microfarad are connected across batteries of 600volt and 1000 volt respectively.and then disconnected.the capacitor are then joined in parallel.what is the charge of each capacitor
Answers
Answer:
ANSWER
We know that half the energy supplied by the battery is lost as heat while charging the capacitor it means if a battery supplies the energy E
B
, then the energy store in the capacitor is
2
1
E
B
.
Now, from the question
equivalent capacitance =
C
1
+C
2
C
1
C
2
=
10+20
10×20
=200/30=
3
20
Energy stored in this capacitors =
2
1
CV
2
=
2
1
×
30
20
×(6)
2
=10×12=120
Total charge =CV=
3
20
×6=40μC
Thus potential drop across the capacitor (C
1
) is given by V
1
=
10
40
=4
and potential drop across the capacitor (C
2
) is given by V
2
=
20
40
=2
Energy stored in C
1
=
2
1
×40×4=80
Energy stored in C
2
=
2
1
×40×2=40=E (given)
Thus energy stored in capacitor C
1
=2E
Thus total energy in the capacitors =120=3E
Hence energy supplied by battery =2×3E=6E
How satisfied are you with the answer?
Explanation:
hope this helps you