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Given three capacitors, C1C1C_1 = 2.0 μFμF, C2C2C_2 = 1.5 μFμF, and C3C3C_3 = 3.0 μFμF, what arrangement of parallel and series connections
Question
Given three capacitors, C1C1C_1 = 2.0 μFμF, C2C2C_2 = 1.5 μFμF, and C3C3C_3 = 3.0 μFμF, what arrangement of parallel and series connections with a 12-VV battery will give the minimum voltage drop across the 2.0- μFμF capacitor?
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Physics
3 years
2021-08-26T08:20:00+00:00
2021-08-26T08:20:00+00:00 1 Answers
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Answers ( )
Answer:
Explanation:
If joined in parallel ,each will have potential of 12 volt
so on 2 μF capacitor , voltage will be 12 V
When joined in series , if C be equivalent capacitor
1/C = 1/2 + 1/1.5 + 1/3
= .5 + .667 + .333
= 1.5
C = .666 μF
Charge on each capacitor
= CV
= .666 x 12
= 8 μC
Potential on 2 μF capacitor
= charge / capacitance
= 8 μC / 2 x μF
= 4 V
So in series , potential at 2μF capacitor is 4 V which is less than 12 V in parallel combination.