Respuesta :
Answer:
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.
[tex]-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-[/tex].
Explanation:
Consider four possible cases.
Case A: 12.0 V.
[tex]-\begin{array}{c}-{\bf 2.0\;\mu\text{F}-}\\-1.5\;\mu\text{F}- \\-3.0\;\mu\text{F}-\end{array}-[/tex]
In case all three capacitors are connected in parallel, the [tex]2.0\;\mu\text{F}[/tex] capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.
Case B: 5.54 V.
[tex]-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-[/tex]
In case the [tex]2.0\;\mu\text{F}[/tex] capacitor is connected in parallel with the [tex]1.5\;\mu\text{F}[/tex] capacitor, and the two capacitors in parallel is connected to the [tex]3.0\;\mu\text{F}[/tex] 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,
[tex]\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_3}+ \dfrac{1}{C_1+C_2}} = \frac{1}{\dfrac{1}{3.0}+\dfrac{1}{2.0+1.5}} = 1.62\;\mu\text{F}[/tex].
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.
[tex]Q = C(\text{Effective}) \cdot V = 1.62\;\mu\text{F}\times 12\;\text{V} = 19.4\;\mu\text{C}[/tex].
Voltage is the same across two capacitors in parallel.As a result,
[tex]\displaystyle V_1 = V_2 = \frac{Q}{C_1+C_2} = \frac{19.4\;\mu\text{C}}{3.5\;\mu\text{F}} = 5.54\;\text{V}[/tex].
Case C: 2.76 V.
[tex]-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-[/tex].
Similarly,
- the effective capacitance of the two capacitors in parallel is 5.0 μF;
- the effective capacitance of the three capacitors, combined: [tex]\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_2}+ \dfrac{1}{C_1+C_3}} = \frac{1}{\dfrac{1}{1.5}+\dfrac{1}{2.0+3.0}} = 1.15\;\mu\text{F}[/tex].
Charge stored:
[tex]Q = C(\text{Effective}) \cdot V = 1.15\;\mu\text{F}\times 12\;\text{V} = 13.8\;\mu\text{C}[/tex].
Voltage:
[tex]\displaystyle V_1 = V_3 = \frac{Q}{C_1+C_3} = \frac{13.8\;\mu\text{C}}{5.0\;\mu\text{F}} = 2.76\;\text{V}[/tex].
Case D: 4.00 V
[tex]-2.0\;\mu\text{F}-1.5\;\mu\text{F}-3.0\;\mu\text{F}-[/tex].
Connect all three capacitors in series.
[tex]\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_1} + \dfrac{1}{C_2}+\dfrac{1}{C_3}} =\frac{1}{\dfrac{1}{2.0} + \dfrac{1}{1.5}+\dfrac{1}{3.0}} =0.667\;\mu\text{F}[/tex].
For each of the three capacitors:
[tex]Q = C(\text{Effective})\cdot V = 0.667\;\mu\text{F} \times 12\;\text{V} = 8.00\;\mu\text{C}[/tex].
For the [tex]2.0\;\mu\text{F}[/tex] capacitor:
[tex]\displaystyle V_1=\frac{Q}{C_1} = \frac{8.00\;\mu\text{C}}{2.0\;\mu\text{F}} = 4.0\;\text{V}[/tex].
