Chapter 16, Problem 43P

Find the charge on each of the capacitors in Figure P16.43.

Figure P16.43

To determine

The charge on each capacitor.

Answer to Problem 43P

The charge on $1.00\text{μF}$ capacitor is $16.0\text{μC}$

The charge on $5.00\text{μF}$ capacitor is $80.0\text{μC}$

The charge on $8.00\text{μF}$ capacitor is $64.0\text{μC}$

The charge on $4.00\text{μF}$ capacitor is $32.0\text{μC}$

Explanation of Solution

The capacitors $1.00\text{μF}$ and $5.00\text{μF}$ are connected in series combination. The equivalent capacitance is,

$C_p^t=\frac{C_{1.00\text{μF}}{C}_{5.00\text{μF}}}{C_{1.00\text{μF}}+C_{5.00\text{μF}}}$

The capacitors $8.00\text{μF}$ and $4.00\text{μF}$ are connected in series combination. The equivalent capacitance is,

$C_p^b=\frac{C_{8.00\text{μF}}{C}_{4.00\text{μF}}}{C_{8.00\text{μF}}+C_{4.00\text{μF}}}$

The capacitances $C_p^t$ and $C_p^b$ are in parallel combination. The total equivalent capacitance is,

$C_{eq}=C_p^t+C_p^b$

Therefore,

$C_{eq}=\left(\frac{C_{1.00\text{μF}}{C}_{5.00\text{μF}}}{C_{1.00\text{μF}}+C_{5.00\text{μF}}}\right)+\left(\frac{C_{8.00\text{μF}}{C}_{4.00\text{μF}}}{C_{8.00\text{μF}}+C_{4.00\text{μF}}}\right)$

Substitute $1.00\text{μF}$ for $C_{1.00\text{μF}}$, $5.00\text{μF}$ for $C_{5.00\text{μF}}$, $8.00\text{μF}$ for $C_{8.00\text{μF}}$ and $4.00\text{μF}$ for $C_{4.00\text{μF}}$

$\begin{array}{c}{C}_{eq}=\left[\frac{\left(1.00\text{μF}\right)\left(5.00\text{μF}\right)}{\left(1.00\text{μF}\right)+\left(5.00\text{μF}\right)}\right]+\left[\frac{\left(8.00\text{μF}\right)\left(4.00\text{μF}\right)}{\left(8.00\text{μF}\right)+\left(4.00\text{μF}\right)}\right]\\ =4.00\text{μF}\end{array}$

Formula to calculate the total charge is,

$Q_{total}=C_{eq}V$

Substitute $4.00\text{μF}$ for $C_{eq}$ and 24.0 V for V.

$\begin{array}{c}{Q}_{total}=\left(4.00\text{μF}\right)\left(24.0\text{V}\right)\\ =96.0\text{μC}\end{array}$

Formula to calculate the charge on $1.00\text{μF}$ capacitor is,

$Q_1=C_{1.00\text{μF}}\left(\frac{Q_{total}}{C_{1.00\text{μF}}+C_{5.00\text{μF}}}\right)$

Substitute $96.0\text{μC}$ for $Q_{total}$, $1.00\text{μF}$ for $C_{1.00\text{μF}}$ and $5.00\text{μF}$ for $C_{5.00\text{μF}}$

$\begin{array}{c}{Q}_1=\left(1.00\text{μF}\right)\left(\frac{96.0\text{μC}}{\left(1.00\text{μF}\right)+\left(5.00\text{μF}\right)}\right)\\ =16.0\text{μC}\end{array}$

Formula to calculate the charge on $5.00\text{μF}$ capacitor is,

$Q_2=C_{5.00\text{μF}}\left(\frac{Q_{total}}{C_{1.00\text{μF}}+C_{5.00\text{μF}}}\right)$

Substitute $96.0\text{μC}$ for $Q_{total}$, $1.00\text{μF}$ for $C_{1.00\text{μF}}$ and $5.00\text{μF}$ for $C_{5.00\text{μF}}$

$\begin{array}{c}{Q}_2=\left(5.00\text{μF}\right)\left(\frac{96.0\text{μC}}{\left(1.00\text{μF}\right)+\left(5.00\text{μF}\right)}\right)\\ =80.0\text{μC}\end{array}$

Formula to calculate the charge on $8.00\text{μF}$ capacitor is,

$Q_3=C_{8.00\text{μF}}\left(\frac{Q_{total}}{C_{8.00\text{μF}}+C_{4.00\text{μF}}}\right)$

Substitute $96.0\text{μC}$ for $Q_{total}$, $8.00\text{μF}$ for $C_{8.00\text{μF}}$ and $4.00\text{μF}$ for $C_{4.00\text{μF}}$

$\begin{array}{c}{Q}_3=\left(8.00\text{μF}\right)\left(\frac{96.0\text{μC}}{\left(8.00\text{μF}\right)+\left(4.00\text{μF}\right)}\right)\\ =64.0\text{μC}\end{array}$

Formula to calculate the charge on $4.00\text{μF}$ capacitor is,

$Q_4=C_{4.00\text{μF}}\left(\frac{Q_{total}}{C_{8.00\text{μF}}+C_{4.00\text{μF}}}\right)$

Substitute $96.0\text{μC}$ for $Q_{total}$, $8.00\text{μF}$ for $C_{8.00\text{μF}}$ and $4.00\text{μF}$ for $C_{4.00\text{μF}}$

$\begin{array}{c}{Q}_4=\left(4.00\text{μF}\right)\left(\frac{96.0\text{μC}}{\left(8.00\text{μF}\right)+\left(4.00\text{μF}\right)}\right)\\ =32.0\text{μC}\end{array}$

Conclusion:

The charge on $1.00\text{μF}$ capacitor is $16.0\text{μC}$

The charge on $5.00\text{μF}$ capacitor is $80.0\text{μC}$

The charge on $8.00\text{μF}$ capacitor is $64.0\text{μC}$

The charge on $4.00\text{μF}$ capacitor is $32.0\text{μC}$