Question

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?

Answer 1

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.

`-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-`.

Step-by-step explanation:

Consider four possible cases.

Case A: 12.0 V.

`-\begin{array}{c}-{\bf 2.0\;\mu\text{F}-}\\-1.5\;\mu\text{F}- \\-3.0\;\mu\text{F}-\end{array}-`

In case all three capacitors are connected in parallel, the
`2.0\;\mu\text{F}` capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.

Case B: 5.54 V.

`-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-`

In case the
`2.0\;\mu\text{F}` capacitor is connected in parallel with the
`1.5\;\mu\text{F}` capacitor, and the two capacitors in parallel is connected to the
`3.0\;\mu\text{F}` 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,

`\displaystyle C(\text{Effective}) = (1)/((1)/(C_3)+ (1)/(C_1+C_2)) = (1)/((1)/(3.0)+(1)/(2.0+1.5)) = 1.62\;\mu\text{F}`.

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.

`Q = C(\text{Effective}) \cdot V = 1.62\;\mu\text{F}* 12\;\text{V} = 19.4\;\mu\text{C}`.

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

`\displaystyle V_1 = V_2 = (Q)/(C_1+C_2) = \frac{19.4\;\mu\text{C}}{3.5\;\mu\text{F}} = 5.54\;\text{V}`.

Case C: 2.76 V.

`-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-`.

Similarly,

• the effective capacitance of the two capacitors in parallel is 5.0 μF;
• the effective capacitance of the three capacitors, combined:
  `\displaystyle C(\text{Effective}) = (1)/((1)/(C_2)+ (1)/(C_1+C_3)) = (1)/((1)/(1.5)+(1)/(2.0+3.0)) = 1.15\;\mu\text{F}`.

Charge stored:

`Q = C(\text{Effective}) \cdot V = 1.15\;\mu\text{F}* 12\;\text{V} = 13.8\;\mu\text{C}`.

Voltage:

`\displaystyle V_1 = V_3 = (Q)/(C_1+C_3) = \frac{13.8\;\mu\text{C}}{5.0\;\mu\text{F}} = 2.76\;\text{V}`.

Case D: 4.00 V

`-2.0\;\mu\text{F}-1.5\;\mu\text{F}-3.0\;\mu\text{F}-`.

Connect all three capacitors in series.

`\displaystyle C(\text{Effective}) = (1)/((1)/(C_1) + (1)/(C_2)+(1)/(C_3)) =(1)/((1)/(2.0) + (1)/(1.5)+(1)/(3.0)) =0.667\;\mu\text{F}`.

For each of the three capacitors:

`Q = C(\text{Effective})\cdot V = 0.667\;\mu\text{F} * 12\;\text{V} = 8.00\;\mu\text{C}`.

For the
`2.0\;\mu\text{F}` capacitor:

`\displaystyle V_1=(Q)/(C_1) = \frac{8.00\;\mu\text{C}}{2.0\;\mu\text{F}} = 4.0\;\text{V}`.