Chapter 16, Problem 41P

For the system of capacitors shown in Figure P16.41, find (a) the equivalent capacitance of the system, (b) the charge on each capacitor, and (c) the potential difference across each capacitor.

Figure P16.41 Problems 41 and 60.

To determine

The equivalent capacitance.

Answer to Problem 41P

The equivalent capacitance is $3.33\text{μF}$.

Explanation of Solution

The capacitors $6.00\text{μF}$ and $3.00\text{μF}$ are connected in series combination. The equivalent capacitance is,

$C_p^t=\frac{C_{6.00\text{μF}}{C}_{3.00\text{μF}}}{C_{6.00\text{μF}}+C_{3.00\text{μF}}}$

The capacitors $2.00\text{μF}$ and $4.00\text{μF}$ are connected in series combination. The equivalent capacitance is,

$C_p^b=\frac{C_{2.00\text{μF}}{C}_{4.00\text{μF}}}{C_{2.00\text{μF}}+C_{4.00\text{μF}}}$

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,

$C_{eq}=\left(\frac{C_{6.00\text{μF}}{C}_{3.00\text{μF}}}{C_{6.00\text{μF}}+C_{3.00\text{μF}}}\right)+\left(\frac{C_{2.00\text{μF}}{C}_{4.00\text{μF}}}{C_{2.00\text{μF}}+C_{4.00\text{μF}}}\right)$

Substitute $6.00\text{μF}$ for $C_{6.00\text{μF}}$,$3.00\text{μF}$ for $C_{3.00\text{μF}}$,$2.00\text{μF}$ for $C_{2.00\text{μF}}$ and $4.00\text{μF}$ for $C_{4.00\text{μF}}$

$\begin{array}{c}{C}_{eq}=\left[\frac{\left(6.00\text{μF}\right)\left(3.00\text{μF}\right)}{\left(6.00\text{μF}\right)+\left(3.00\text{μF}\right)}\right]+\left[\frac{\left(2.00\text{μF}\right)\left(4.00\text{μF}\right)}{\left(2.00\text{μF}\right)+\left(4.00\text{μF}\right)}\right]\\ =3.33\text{μF}\end{array}$

On Re-arranging,

$\begin{array}{c}{C}_{eq}=\frac{8.00\text{μF}}{3}\\ =2.67\text{μF}\end{array}$

Conclusion:

The equivalent capacitance is $3.33\text{μF}$.

To determine

The charge on each capacitor.

Answer to Problem 41P

The charge on $6.00\text{μF}$ and $3.00\text{μF}$ capacitors is $180\text{μC}$

The charge on $2.00\text{μF}$ and $4.00\text{μF}$ capacitors is $120\text{μC}$

Explanation of Solution

Formula to calculate the charge on $6.00\text{μF}$ and $3.00\text{μF}$ capacitors is,

$Q_1=C_p^{t}V$

Therefore,

$Q_1=\left(\frac{C_{6.00\text{μF}}{C}_{3.00\text{μF}}}{C_{6.00\text{μF}}+C_{3.00\text{μF}}}\right)V$

Substitute $6.00\text{μF}$ for $C_{6.00\text{μF}}$,$3.00\text{μF}$ for $C_{3.00\text{μF}}$ and 90.0 V for V.

$\begin{array}{c}{Q}_1=\left(\frac{\left(6.00\text{μF}\right)\left(3.00\text{μF}\right)}{\left(6.00\text{μF}\right)+\left(3.00\text{μF}\right)}\right)\left(90.0\text{V}\right)\\ =180\text{μC}\end{array}$

Formula to calculate the charge on $2.00\text{μF}$ and $4.00\text{μF}$ capacitors is,

$Q_2=C_p^{b}V$

Therefore,

$Q_2=\left(\frac{C_{2.00\text{μF}}{C}_{4.00\text{μF}}}{C_{2.00\text{μF}}+C_{4.00\text{μF}}}\right)V$

Substitute $2.00\text{μF}$ for $C_{2.00\text{μF}}$,$4.00\text{μF}$ for $C_{4.00\text{μF}}$$C_{3.00\text{μF}}$ and 90.0 V for V.

$\begin{array}{c}{Q}_2=\left(\frac{\left(2.00\text{μF}\right)\left(4.00\text{μF}\right)}{\left(2.00\text{μF}\right)+\left(4.00\text{μF}\right)}\right)\left(90.0\text{V}\right)\\ =120\text{μC}\end{array}$

Conclusion:

The charge on $6.00\text{μF}$ and $3.00\text{μF}$ capacitors is $180\text{μC}$

The charge on $2.00\text{μF}$ and $4.00\text{μF}$ capacitors is $120\text{μC}$

To determine

The potential difference on each capacitor.

Explanation of Solution

The potential difference on $2.00\text{μF}$ and $3.00\text{μF}$ capacitors is 60 V

The potential difference on $6.00\text{μF}$ and $4.00\text{μF}$ capacitors is 30 V

Formula to calculate the potential difference on $2.00\text{μF}$ and $3.00\text{μF}$ capacitors is,

$V=\frac{Q_2}{C_{2.00\text{μF}}}$

Substitute $120\text{μC}$ for $Q_2$ and $2.00\text{μF}$ for $C_{2.00\text{μF}}$

$\begin{array}{c}V=\frac{120\text{μC}}{2.00\text{μF}}\\ =60\text{V}\end{array}$

Formula to calculate the potential difference on $6.00\text{μF}$ and $4.00\text{μF}$ capacitors is,

$V=\frac{Q_2}{C_{6.00\text{μF}}}$

Substitute $180\text{μC}$ for $Q_1$ and $6.00\text{μF}$ for $C_{6.00\text{μF}}$

$\begin{array}{c}V=\frac{180\text{μC}}{6.00\text{μF}}\\ =30\text{V}\end{array}$

Conclusion:

The potential difference on $2.00\text{μF}$ and $3.00\text{μF}$ capacitors is 60 V

The potential difference on $6.00\text{μF}$ and $4.00\text{μF}$ capacitors is 30 V