EQUIVALENT CAPACITANCE GOAL Solve a complex combination of series and parallel capacitors. PROBLEM (a) Calculate the equivalent capacitance between a and b for the combination of capacitors shown in Figure 16.22a. All capacitances are in microfarads. (b) If a 12−V battery is connected across the

Question

EQUIVALENT CAPACITANCE

GOAL Solve a complex combination of series and parallel capacitors.

PROBLEM (a) Calculate the equivalent capacitance between a and [latex]b[/latex] for the combination of capacitors shown in Figure 16.22a. All capacitances are in microfarads. (b) If a [latex]12-\mathrm{V}[/latex] battery is connected across the system between points [latex]a[/latex] and [latex]b[/latex], find the charge on the [latex]4.0-\mu \mathrm{F}[/latex] capacitor in the first diagram and the voltage drop across it.

STRATEGY For part (a), use Equations [latex]16.12[/latex]

[latex]C_{eq} = C_1 + C_2 + C_3 + \cdots[/latex] [16.12]

and [latex]16.15[/latex]

[latex]PE = k_e \frac{q_1 q_2}{r}[/latex] [16.15]

to reduce the combination step by step, as indicated in the figure. For part (b), to find the charge on the [latex]4.0-\mu \mathrm{F}[/latex] capacitor, start with Figure [latex]16.22 \mathrm{c}[/latex], finding the charge on the [latex]2.0-\mu \mathrm{F}[/latex] capacitor. This same charge is on each of the [latex]4.0-\mu \mathrm{F}[/latex] capacitors in the second diagram, by fact [latex]5 \mathrm{E}[/latex] of the Problem-Solving Strategy. One of these [latex]4.0-\mu \mathrm{F}[/latex] capacitors in the second diagram is simply the original [latex]4.0-\mu \mathrm{F}[/latex] capacitor in the first diagram.

Accepted Answer

(a) Calculate the equivalent capacitance.

Find the equivalent capacitance of the parallel [latex]1.0-\mu \mathrm{F}[/latex] and 3.0- [latex]\mu \mathrm{F}[/latex] capacitors in Figure 16.22a:

[latex] C_{\mathrm{eq}} =C_{1}+C_{2}=1.0 \mu \mathrm{F}+3.0 \mu \mathrm{F}=4.0 \mu \mathrm{F} [/latex]

Find the equivalent capacitance of the parallel [latex]2.0-\mu \mathrm{F}[/latex] and [latex]6.0-\mu \mathrm{F}[/latex] capacitors in Figure 16.22a:

[latex]C_{\mathrm{eq}} =C_{1}+C_{2}=2.0 \mu \mathrm{F}+6.0 \mu \mathrm{F}=8.0 \mu \mathrm{F}[/latex]

Combine the two series [latex]4.0-\mu \mathrm{F}[/latex] capacitors in Figure [latex]16.22 \mathrm{~b}[/latex] :

[latex]\begin{aligned}\frac{1}{C_{\mathrm{eq}}} &=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{4.0 \mu \mathrm{F}}+\frac{1}{4.0 \mu \mathrm{F}} \ &=\frac{1}{2.0 \mu \mathrm{F}} ightarrow C_{\mathrm{eq}}=2.0 \mu \mathrm{F}\end{aligned}[/latex]

Combine the two series [latex]8.0-\mu \mathrm{F}[/latex] capacitors in Figure [latex]16.22 \mathrm{~b}[/latex] :

[latex]\begin{aligned}\frac{1}{C_{\mathrm{eq}}} &=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{8.0 \mu \mathrm{F}}+\frac{1}{8.0 \mu \mathrm{F}} \&=\frac{1}{4.0 \mu \mathrm{F}} \quad ightarrow \quad C_{\mathrm{eq}}=4.0 \mu \mathrm{F}\end{aligned}[/latex]

Finally, combine the two parallel capacitors in Figure 16.22c

[latex]C_{\mathrm{eq}} =C_{1}+C_{2}=2.0 \mu \mathrm{F}+4.0 \mu \mathrm{F}=6.0 \mu \mathrm{F}[/latex]

(b) Find the charge on the [latex]4.0-\mu \mathrm{F}[/latex] capacitor and the voltage drop across it.

Compute the charge on the [latex]2.0-\mu \mathrm{F}[/latex] capacitor in Figure 16.22c, which is the same as the charge on the [latex]4.0-\mu \mathrm{F}[/latex] capacitor in Figure 16.22a:

[latex]\begin{gathered}C=\frac{Q}{\Delta V} \quad ightarrow \quad Q=C \Delta V=(2.0 \mu \mathrm{F})(12 \mathrm{~V})=24 \mu \mathrm{C} \end{gathered}[/latex]

Use the basic capacitance equation to find the voltage drop across the [latex]4.0-\mu \mathrm{F}[/latex] capacitor in Figure [latex]16.22 \mathrm{a}[/latex] :

[latex]\begin{gathered}C=\frac{Q}{\Delta V} \quad ightarrow \quad \Delta V=\frac{Q}{C}=\frac{24 \mu \mathrm{C}}{4.0 \mu \mathrm{F}}=6.0 \mathrm{~V}\end{gathered}[/latex]

REMARKS To find the rest of the charges and voltage drops, it's just a matter of using [latex]C=Q / \Delta V[/latex] repeatedly, together with facts [latex]5 \mathrm{C}[/latex] and [latex]5 \mathrm{E}[/latex] in the Problem-Solving Strategy. The voltage drop across the [latex]4.0-\mu \mathrm{F}[/latex] capacitor could also have been found by noticing, in Figure 16.22b, that both capacitors had the same value and so by symmetry would split the total drop of 12 volts between them.

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