Chapter 16, Problem 48P

Four capacitors are connected as shown in Figure P16.48.

(a) Find the equivalent capacitance between points a and b.

(b) Calculate the charge on each capacitor, taking ΔV_ab = 15.0 V.

Figure P16.48

To determine

The equivalent capacitance.

Answer to Problem 48P

The equivalent capacitance is $5.96\text{μF}$.

Explanation of Solution

The capacitors $15.0\text{μF}$ and $3.00\text{μF}$ are connected in series combination. The equivalent capacitance is,

$C_s=\frac{C_{15.0\text{μF}}{C}_{3.00\text{μF}}}{C_{15.0\text{μF}}+C_{3.00\text{μF}}}$

The capacitors $C_s$ and $6.00\text{μF}$ are connected in parallel combination. The equivalent capacitance is,

$C_p=C_s+\left(6.00\text{μF}\right)$

The capacitance $C_p$ and $20.0\text{μF}$ are connected in series combination. The total equivalent capacitance is,

$C_{eq}=\frac{\left(\frac{C_{15.0\text{μF}}{C}_{3.00\text{μF}}}{C_{15.0\text{μF}}+C_{3.00\text{μF}}}+C_{6.00\text{μF}}\right)\left(C_{20.0\text{μF}}\right)}{\left(\frac{C_{15.0\text{μF}}{C}_{3.00\text{μF}}}{C_{15.0\text{μF}}+C_{3.00\text{μF}}}+C_{6.00\text{μF}}\right)+\left(C_{20.0\text{μF}}\right)}$

Substitute $15.0\text{μF}$ for $C_{15.0\text{μF}}$, $3.00\text{μF}$ for $C_{3.00\text{μF}}$, $6.00\text{μF}$ for $C_{6.00\text{μF}}$ and $20.0\text{μF}$ for $C_{20.0\text{μF}}$

$\begin{array}{c}{C}_{eq}=\frac{\left(\frac{\left(15.0\text{μF}\right)\left(3.00\text{μF}\right)}{\left(15.0\text{μF}\right)+\left(3.00\text{μF}\right)}+6.00\text{μF}\right)\left(20.0\text{μF}\right)}{\left(\frac{\left(15.0\text{μF}\right)\left(3.00\text{μF}\right)}{\left(15.0\text{μF}\right)+\left(3.00\text{μF}\right)}+6.00\text{μF}\right)+\left(20.0\text{μF}\right)}\\ =5.96\text{μF}\end{array}$

Conclusion:

The equivalent capacitance is $5.96\text{μF}$.

To determine

The charge on each capacitor.

Answer to Problem 48P

The charge on $20.0\text{μF}$ capacitor is $89.4\text{μC}$

The charge on $6.00\text{μF}$ capacitor is $63.0\text{μC}$

The charge on $15.0\text{μF}$ and $3.00\text{μF}$ capacitors is $26.3\text{μC}$

Explanation of Solution

Formula to calculate the charge on $20.0\text{μF}$ capacitor is,

$Q_{20}=C_{eq}\left(\Delta V\right)$

Substitute $5.96\text{μF}$ for $C_{eq}$ and 15.0 V for V.

$\begin{array}{c}{Q}_{20}=\left(5.96\text{μF}\right)\left(15.0\text{V}\right)\\ =89.4\text{μC}\end{array}$

Formula to calculate the charge on $20.0\text{μF}$ capacitor is,

$Q_6=Q_{20}\left(\frac{C_{6.00\text{μF}}}{C_s+C_{6.00\text{μF}}}\right)$

Therefore,

$Q_6=Q_{20}\left(\frac{C_{6.00\text{μF}}}{\frac{C_{15.0\text{μF}}{C}_{3.00\text{μF}}}{C_{15.0\text{μF}}+C_{3.00\text{μF}}}+C_{6.00\text{μF}}}\right)$

Substitute15.0 V for V, $3.00\text{μF}$ for $C_{3.00\text{μF}}$, $6.00\text{μF}$ for $C_{6.00\text{μF}}$ and $20.0\text{μF}$ for $C_{20.0\text{μF}}$

$\begin{array}{l}{Q}_6=\left(89.4\text{μC}\right)\left(\frac{6.00\text{μF}}{\frac{\left(15.0\text{μF}\right)\left(3.00\text{μF}\right)}{\left(15.0\text{μF}\right)+\left(3.00\text{μF}\right)}+6.00\text{μF}}\right)\\ =63.0\text{μC}\end{array}$

Explanation:

Formula to calculate the charge on charge on $15.0\text{μF}$ and $3.00\text{μF}$ capacitors is,

$Q=Q_{20}\left(\frac{C_s}{C_s+C_{6.00\text{μF}}}\right)$

Therefore,

$Q_6=Q_{20}\left(\frac{\frac{C_{15.0\text{μF}}{C}_{3.00\text{μF}}}{C_{15.0\text{μF}}+C_{3.00\text{μF}}}}{\frac{C_{15.0\text{μF}}{C}_{3.00\text{μF}}}{C_{15.0\text{μF}}+C_{3.00\text{μF}}}+C_{6.00\text{μF}}}\right)$

Substitute15.0 V for V, $3.00\text{μF}$ for $C_{3.00\text{μF}}$, $6.00\text{μF}$ for $C_{6.00\text{μF}}$ and $20.0\text{μF}$ for $C_{20.0\text{μF}}$

$\begin{array}{l}{Q}_6=\left(89.4\text{μC}\right)\left(\frac{\frac{\left(15.0\text{μF}\right)\left(3.00\text{μF}\right)}{\left(15.0\text{μF}\right)+\left(3.00\text{μF}\right)}}{\frac{\left(15.0\text{μF}\right)\left(3.00\text{μF}\right)}{\left(15.0\text{μF}\right)+\left(3.00\text{μF}\right)}+6.00\text{μF}}\right)\\ =26.3\text{μC}\end{array}$

Conclusion:

The charge on $20.0\text{μF}$ capacitor is $89.4\text{μC}$.

The charge on $6.00\text{μF}$ capacitor is $63.0\text{μC}$.

The charge on $15.0\text{μF}$ and $3.00\text{μF}$ capacitors is $26.3\text{μC}$.