Chapter 27, Problem 75PQ

Figure P27.75 shows four capacitors with C_A = 4.00 μF, C_B = 8.00 μF. C_C = 6.00 μF. and C_D = 5.00 μF connected across points a and b, which have potential difference ∆V_ab = 12.0 V. a. What is the equivalent capacitance of the four capacitors? b. What is the charge on each of the four capacitors?

To determine

The equivalent capacitance of the four capacitors $\left(C_{\text{A}},C_{\text{B}},C_{\text{C}}\text{and}{C}_{\text{D}}\right)$ .

Answer to Problem 75PQ

The equivalent capacitance of the four capacitors $\left(C_{\text{A}},C_{\text{B}},C_{\text{C}}\text{and}{C}_{\text{D}}\right)$ is $2.91\times {10}^{-6}\text{F}$ .

Explanation of Solution

From figure 1,

Capacitors C and D are connected in series.

Write the expression for the equivalent capacitance CD in series.

    $\frac{1}{C_{\text{eqCD}}}=\frac{1}{C_{\text{C}}}+\frac{1}{C_{\text{D}}}$                                         (I)

Here, $C_{\text{C}}$ is the capacitor C and $C_{\text{D}}$ is the capacitor D.

The equivalent capacitance CD and B are connected in parallel.

Write the expression for the equivalent capacitance BCD in parallel.

    $C_{\text{eqBCD}}=C_{\text{B}}+C_{\text{eqCD}}$                                      (II)

Here, $C_{\text{B}}$ is the capacitor B.

The equivalent capacitance BCD and capacitor A are connected in series.

Write the expression for the equivalent capacitance ABCD in series.

    $\frac{1}{C_{\text{eqABCD}}}=\frac{1}{C_{\text{A}}}+\frac{1}{C_{\text{eqBCD}}}$                                        (III)

Conclusion:

Substitute $6.00\mu \text{F}$ for $C_{\text{C}}$ , $5.00\mu \text{F}$ for $C_{\text{D}}$ in (I) to find $C_{\text{eqCD}}$.

    $\begin{array}{c}\frac{1}{C_{\text{eqCD}}}=\frac{1}{6.00\mu \text{F}}+\frac{1}{5.00\mu \text{F}}\\ =\frac{30.0}{11.0}\mu \text{F}\end{array}$

Substitute $8.00\mu \text{F}$ for $C_{\text{B}}$ , $\frac{30.0}{11.0}\mu \text{F}$ for $C_{\text{eqCD}}$ in (II) to find $C_{\text{eqBCD}}$.

    $\begin{array}{c}{C}_{\text{eqBCD}}=8.00\mu \text{F+}\frac{30.0}{11.0}\mu \text{F}\\ \text{=}\frac{118}{11.0}\mu \text{F}\end{array}$

Substitute $4.00\mu \text{F}$ for $C_{\text{A}}$ , $\frac{118.0}{11.0}\mu \text{F}$ for $C_{\text{eqBCD}}$ in (III) to find $C_{\text{eqABCD}}$.

    $\begin{array}{c}\frac{1}{C_{\text{eqABCD}}}=\frac{1}{4.0\mu \text{F}}+\frac{11.0}{118.0\mu \text{F}}\\ =\frac{472}{162}\mu \text{F}\\ \text{=}\text{2}\text{.91}\mu \text{F}\\ =2.91\times {10}^{-6}\text{F}\end{array}$

Therefore, the equivalent capacitance of the four capacitors $\left(C_{\text{A}},C_{\text{B}},C_{\text{C}}\text{and}{C}_{\text{D}}\right)$ is $2.91\times {10}^{-6}\text{F}$ .

To determine

Find the charge on each of the four capacitors

Answer to Problem 75PQ

The charge on each of the four capacitors are $3.50\times {10}^{-5}C$ on $C_{\text{A}}$, $2.61\times {10}^{-5}C$ on $C_{\text{B}}$ and $8.93\times {10}^{-6}C$ on $C_{\text{C}}\text{ and }{C}_{\text{D}}$.

Explanation of Solution

Write the expression for the charge on the equivalent capacitance.

    $Q=C\Delta V$                                                        (IV)

Here, $Q$ is the charge on the equivalent capacitance, $C$ is the capacitance and $\Delta V$ is the potential difference.

From (IV),

Write the expression for the potential difference.

    $\Delta V=\frac{Q}{C}$                                                             (V)

From (V), we can find the voltage across A and then the voltage across the BCD combination.

    $\Delta V_{\text{BCD}}=\Delta V-\Delta V_{\text{A}}$                                               (VI)

Here, $\Delta V_{\text{BCD}}$ is the voltage across the BCD and $\Delta V_{\text{A}}$ is the voltage across A.

Conclusion:

The charge on A.

Substitute $\frac{472}{162}\mu \text{F}$ for $C_{\text{eq}}$ , $12.0\text{V}$ for $\Delta V$ in (IV) to find $Q_{\text{A}}$.

    $\begin{array}{c}{Q}_{\text{A}}=\left(\frac{472}{162}\mu \text{F}\right)\left(1.0\times {10}^{-6}\text{F/μF}\right)\left(12.0\text{V}\right)\\ =3.50\times {10}^{-5}\text{C}\end{array}$

Substitute $\left(\frac{472}{162}\mu \text{F}\right)\left(1.0\times {10}^{-6}\text{F/μF}\right)\left(12.0\text{V}\right)$ for Q and $4.0\times {10}^{-6}\text{F}$ for $C_{\text{A}}$ in (V) to get $\Delta V_{\text{A}}$ value.

    $\Delta V_{\text{A}}=\frac{\left(\frac{472}{162}\mu \text{F}\right)\left(1.0\times {10}^{-6}\text{F/μF}\right)\left(12.0\text{V}\right)}{4.0\times {10}^{-6}\text{F}}$             (VII)

Substitute $\frac{\left(\frac{472}{162}\mu \text{F}\right)\left(1.0\times {10}^{-6}\text{F/μF}\right)\left(12.0\text{V}\right)}{4.0\times {10}^{-6}\text{F}}$ for $\Delta V_{\text{A}}$  and $12.0\text{V}$ for $\Delta V$ in (VI) to get $\Delta V_{\text{BCD}}$ value.

    $\Delta V_{\text{BCD}}=\left(12.0\text{V}\right)-\frac{\left(\frac{472}{162}\mu \text{F}\right)\left(1.0\times {10}^{-6}\text{F/μF}\right)\left(12.0\text{V}\right)}{4.0\times {10}^{-6}\text{F}}$

The charge on B.

Substitute $\left(12.0\text{V}\right)-\frac{\left(\frac{472}{162}\mu \text{F}\right)\left(1.0\times {10}^{-6}\text{F/μF}\right)\left(12.0\text{V}\right)}{4.0\times {10}^{-6}\text{F}}$ for $\Delta V_{\text{BCD}}$ and $8.0\mu \text{F}$ for $C_{\text{B}}$ in (IV) to find $Q_{\text{B}}$ .

    $\begin{array}{c}{Q}_{\text{B}}=\left(8.0\mu \text{F}\right)\left(\left(12.0\text{V}\right)-\frac{\left(\frac{472}{162}\mu \text{F}\right)\left(1.0\times {10}^{-6}\text{F/μF}\right)\left(12.0\text{V}\right)}{4.0\times {10}^{-6}\text{F}}\right)\\ =2.61\times {10}^{-5}\text{C}\end{array}$

The charge on C and D.

Here, C and D are in series, so the charge will be same. And the charge of B and C or D must sum to that on A.

    $Q_{\text{C}\text{or}\text{D}}=Q_{\text{A}}-Q_{\text{B}}$

Substitute $35\mu \text{C}$ for $Q_{\text{A}}$ and $\left(8.0\mu \text{F}\right)\left(\left(12.0\text{V}\right)-\frac{\left(\frac{472}{162}\mu \text{F}\right)\left(1.0\times {10}^{-6}\text{F/μF}\right)\left(12.0\text{V}\right)}{4.0\times {10}^{-6}\text{F}}\right)$ for $Q_{\text{B}}$ in the above equation to find $Q_{\text{C}\text{or}\text{D}}$.

    $\begin{array}{c}{Q}_{\text{C}\text{or}\text{D}}=\left(35\mu \text{C}\right)-\left(8.0\mu \text{F}\right)\left(\left(12.0\text{V}\right)-\frac{\left(\frac{472}{162}\mu \text{F}\right)\left(1.0\times {10}^{-6}\text{F/μF}\right)\left(12.0\text{V}\right)}{4.0\times {10}^{-6}\text{F}}\right)\\ =8.93\times {10}^{-6}\text{C}\end{array}$

Therefore, the charge on each of the four capacitors are $3.50\times {10}^{-5}C$ on $C_{\text{A}}$, $2.61\times {10}^{-5}C$ on $C_{\text{B}}$ and $8.93\times {10}^{-6}C$ on $C_{\text{C}}\text{ and }{C}_{\text{D}}$.