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A 3.0-µF capacitor charged to 40 V and a 5.0-µF capacitor charged to 18 V are connected to each other, with the positive plate of each conne
Question
A 3.0-µF capacitor charged to 40 V and a 5.0-µF capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other. What is the final charge on the 3.0-µF capacitor? *
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Physics
3 years
2021-07-16T07:16:31+00:00
2021-07-16T07:16:31+00:00 1 Answers
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Answers ( )
Answer:
Explanation:
Given that,
Capacitor 1
Capacitance C1 = 3.0-µF
Voltage V1 = 40V
Capacitor 2
Capacitance C2 = 5.0-µF
Voltage V2 = 18V
The two capacitor are connected such that, the positive place is connected to the negative plate of the other, I.e series connection
Charge on each capacitor before connection
Q1 = C1V1 = 3.0-µF × 40
Q1 = 120 -µC
Q2 = C2V2 = 5.0-µF × 18
Q2 =90 µC
So, when they are connected together in series, the charge with combine together to have a single end equal charge since series connection have the same charge
Therefore,
1/Qeq = 1/Q1 + 1/Q2
1/Qeq = (Q1+Q2)/Q1Q2
Qeq = Q1Q2/(Q1+Q2)
Qeq = 51.43 -µC
So the charge on the will be
Q = Qeq/2
Q = 51.43/2
Q = 25.71
Q≈ 26 -µC