Given three capacitors, c1 = 2.0 μf, c2 = 1.5 μf, and c3 = 3.0 μf, what arrangement of parallel and series connections with a 12-v battery will give the minimum voltage drop across the 2.0-μf capacitor?

Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.

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Explanation:

Consider four possible cases.

In case all three capacitors are connected in parallel, the capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.

In case the capacitor is connected in parallel with the capacitor, and the two capacitors in parallel is connected to the capacitor in series.

The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.

The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,

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What will be the voltage across the 2.0 μF capacitor?

The charge stored in two capacitors in series is the same as the charge in each capacitor.

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Voltage is the same across two capacitors in parallel.As a result,

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Similarly,

Charge stored:

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Voltage:

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Connect all three capacitors in series.

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For each of the three capacitors:

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For the capacitor:

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see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west