Two capacitor of capacitance 4 micro farad and 6 micro farad are connected in series across a potential difference of 250v. the capacitance are disconnected from the supply and are reconnected in parallel with each other. calculate the new potential difference and the charge on each capacitor.
Answers
Answer:
Let C1 = 3 μ F , C2 = 6 μ F C1 and C2 are connected in series with potential difference V = 900 V acros the combinationThus for the series combination ..
Answer:
Hey mate, here is your answer.
Explanation:
Two capacitors of 3μF and 6μF are connected in series and a potential difference of 900V is applied across the combination. They are then disconnected and reconnected in parallel. The potential difference across the combination is
December 26, 2019
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Shaurya Dan
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ANSWER
Let C1=3 μF,C2=6 μF
C1 and C2 are connected in series with potential difference V=900 V acros the combination
Thus for the series combination Charge on both capacitor is same
∴C
1
V
1
=C 2
V
2
=Q ...1
∴
V
2
V
1
=
C
1
C
2
=2 μC ...2
also V
1
+V
2
=900 V ...3
From 2 and 3
V
2
=300 V and V
1
=600 V
Therefore charge on each capacitor Q=300×6=600×3=1800 μC
When these capacitors are disconnected and reconnected in parallel the charge is distributed and the potential difference is same
∴
C
1
Q
1
=
C
2
Q
2
=V ...4
∴
Q
2
Q
1
=
C
2
C
1
=
2
1
...5
and Q
1
+Q
2
=Q=1800 μC ...6
From 5 and 6
Q
1
=600 μC and Q
2
=1200 μC
Therefore the potential difference across the combination is V=
3
600
=
6
1200
=200 V
Ans is 200 V
Thank you
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