wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

Open in App
Solution

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
C1V1=C2V2=Q ...1
V1V2=C2C1=2 μC ...2
also V1+V2=900 V ...3
From 2 and 3
V2=300 V and V1=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
Q1C1=Q2C2=V ...4
Q1Q2=C1C2=12 ...5
and Q1+Q2=Q=1800 μC ...6
From 5 and 6
Q1=600 μC and Q2=1200 μC
Therefore the potential difference across the combination is V=6003=12006=200 V
Ans is 200 V

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Kirchoff's Voltage Law
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon