yalgaar ho . A 4 µF capacitor is charged to a voltage of 100 V and a 6 µF capacitor is charged to a voltage of 200 V. They are now joined with plates of like charges connected together. Find (a) potential difference across each after joining, (b) the total energy stored before joining, and (c) the total energy stored after joining.
Add an Answer
More Questions From This User See All
Answers & Comments
Answer:
Before disconnection , as the both capacitor are in parallel so potential across them will be same i.e 660V.
Now using Q=CV, the charge on 4μF is Q
1
=(4×10
−6
)×660=2.64×10
−3
C and the charge on 6μF is Q
2
=(6×10
−6
)×660=3.96×10
−3
C
When disconnected and again connected each other with terminals of unlike sign together, the total charge is Q
t
=Q
2
+(−Q
1
)=(3.96−2.64)10
−3
=1.32×10
−3
C and total capacitance C
t
=C
1
+C
2
=4+6=10μF
After re-connection, they are in parallel so the potential across will be same.
Thus, common potential is V
c
=
C
t
Q
t
=
10×10
−6
1.32×10
−3
=132V
Now charges become , Q
1
′
=(4×10
−6
)×132=5.28×10
−4
C and Q
2
′
=(6×10
−6
)×132=7.92×10
−4
C