Suppose, you are given three capacitors C 1 = 2.0 F, C 2 = 4.0 F, and C 3 = 6.0

Question

Suppose, you are given three capacitors C1 = 2.0F, C2 = 4.0F, and C3 = 6.0F.
1. What is the minimum and the maximum capacitance you can obtain using all three capacitors in a circuit?

2. If you apply a voltage Vab = 12.0V to the terminals of the series and parallel capacitor circuits of part [a] above to charge the capacitors, what is the ratio of the energy that can be stored in the 2.0F and the 4.0F capacitors when they are parts of

[a] the minimum capacitance?

[b] the maximum capacitance configuration?

Explanation / Answer

Minimum in series

=C1C2C3/(C1C2+C2C3+C3C1)

=1.091uF

Maximum capacitance in parallel

=C1+C2+C3=12uF

a)In min capacitance case,

energy stored=Q^2/2C

Ratio=C2/C1

=4/2=2

b)Ratio in maximum capacitance configuration

=0.5C1V^2/0.5C2V^2

=C1/C2=0.5

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