Chapter 26, Problem 26.23P

Four capacitors are connected as shown in Figure P25.11. (a) Find the equivalent capacitance between points a and b. (b) Calculate the charge on each capacitor, taking ΔV_ab = 15.0 V.

Figure P25.11

To determine

The equivalent capacitance between $a$ and $b$.

Answer to Problem 26.23P

The equivalent capacitance between $a$ and $b$ is $5.96\text{μF}$.

Explanation of Solution

Given information:

The four capacitors are connected in figure given below:

Figure (1)

Explanation:

The capacitors $15.0\text{μF}\text{and}3.00\text{μF}$ are in series.

Formula to calculate the equivalent capacitance when they are connected in series.

${C^{\prime }}_S=\frac{C_{15.0\text{μF}}\times C_{3.00\text{μF}}}{C_{15.0\text{μF}}+C_{3.00\text{μF}}}$ (1)

Here,

${C^{\prime }}_S$ is the equivalent capacitance when they are connected in series.

Substitute $15.0\text{μF}$ for $C_{15.0\text{μF}}$, $3.00\text{μF}$ for $C_{3.00\text{μF}}$ in equation (1) to find ${C^{\prime }}_S$,

$\begin{array}{c}{C^{\prime }}_S=\frac{15.0\text{μF}\times 3.00\text{μF}}{15.0\text{μF}+3.00\text{μF}}\\ =2.50\text{μF}\end{array}$

The capacitors ${C^{\prime }}_S$ and $6.00\text{μF}$ are in parallel.

Formula to calculate the equivalent capacitance when they are connected in parallel.

${C^{\prime }}_P={C^{\prime }}_S+C_{6.00\text{μF}}$ (2)

Here,

${C^{\prime }}_P$ is the equivalent capacitance when they are connected in parallel.

Substitute $2.50\text{μF}$ for ${C^{\prime }}_S$, $6.00\text{μF}$ for $C_{6.00\text{μF}}$ in equation (1) to find ${C^{\prime }}_P$,

$\begin{array}{c}{C^{\prime }}_P=2.50\text{μF}+6.00\text{μF}\\ \text{=8}.50\text{μF}\end{array}$

The capacitors ${C^{\prime }}_P$ and $20.0\text{μF}$ are in series.

Formula to calculate the equivalent capacitance when they are connected in series.

$C_{\text{eq}}=\frac{{C^{\prime }}_P\times C_{20.0\text{μF}}}{{C^{\prime }}_P+C_{20.0\text{μF}}}$ (3)

Here,

$C_{\text{eq}}$ is the equivalent capacitance between a and b.

Substitute $8.50\text{μF}$ for ${C^{\prime }}_P$, $20.0\text{μF}$ for $C_{20.0\text{μF}}$ in equation (1) to find $C_{\text{eq}}$,

$\begin{array}{c}{C}_{\text{eq}}=\frac{8.50\text{μF}\times 20.0\text{μF}}{8.50\text{μF}+20.0\text{μF}}\\ =5.96\text{μF}\end{array}$

Thus, the equivalent capacitance between $a$ and $b$ is $5.96\text{μF}$.

Conclusion:

Therefore, the equivalent capacitance between $a$ and $b$ is $5.96\text{μF}$.

To determine

The charge on each capacitor.

Answer to Problem 26.23P

The charge across capacitors $15.0\text{μF}\text{and}3.00\text{μF}$ is $C_2$ is $\text{26}\text{.3}\text{μC}$, the charge through $20.0\text{μF}$ is $\text{89}\text{.5}\text{μC}$, the charge across $6.00\text{μF}$ capacitor is $\text{63}\text{.2}\text{μC}$.

Explanation of Solution

Given information:

The voltage across $a$ and $b$ is $15.0\text{V}$, the four capacitors are connected in figure given below:

Explanation:

Formula to calculate the total charge in the circuit.

$Q_{\text{total}}=C_{\text{eq}}\times \Delta V_{\text{ab}}$ (4)

Here,

$Q_{\text{total}}$ is the total charge in the circuit.

$\Delta V_{\text{ab}}$ is the voltage across a and b.

Substitute $15.0\text{V}$ for $\Delta V_{\text{ab}}$, $5.965\text{μF}$ for $C_{\text{eq}}$, in equation (3) to find $V_{{{\text{C}}^{\prime }}_{\text{P}}}$,

$\begin{array}{c}{Q}_{\text{total}}=5.965\text{μF}\times 15.0\text{V}\\ =\text{89}\text{.475}\text{μC}\\ Q_{20.0\text{μF}}\approx \text{89}\text{.5}\text{μC}\end{array}$

Thus, the total charge in the circuit and the charge through $20.0\text{μF}$ is $\text{89}\text{.5}\text{μC}$.

Formula to calculate the potential drop across $20.0\text{μF}$ capacitor.

$\Delta V_{20.0\text{μF}}=\frac{Q_{20.0\text{μF}}}{C_{20.0\text{μF}}}$ (5)

Substitute $20.0\text{μF}$ for $C_{20.0\text{μF}}$, $\text{89}\text{.5}\text{μC}$ for $Q_{20.0\text{μF}}$, in equation (5) to find $\Delta V_{20.0\text{μF}}$,

$\begin{array}{c}\Delta V_{20.0\text{μF}}=\frac{\text{89}\text{.5}\text{μC}}{20.0\text{μF}}\\ =4.475\text{V}\end{array}$

Thus, the potential drop across $20.0\text{μF}$ capacitor is $4.475\text{V}$.

Formula to calculate the potential drop across $6.00\text{μF}$ capacitor.

$\Delta V_{6.00\text{μF}}=\Delta V_{ab}-\Delta V_{20.0\text{μF}}$ (6)

Substitute $15.0\text{V}$ for $\Delta V_{\text{ab}}$, $4.475\text{V}$ for $\Delta V_{20.0\text{μF}}$, in equation (6) to find $\Delta V_{6.00\text{μF}}$,

$\begin{array}{c}\Delta V_{6.00\text{μF}}=15.0\text{V}-4.475\text{V}\\ =10.525\text{V}\end{array}$

Thus, the potential drop across $6.00\text{μF}$ capacitor is $10.525\text{V}$.

Formula to calculate the charge across $6.00\text{μF}$ capacitor.

$Q_{6.00\text{μF}}=C_{6.00\text{μF}}\times \Delta V_{6.00\text{μF}}$ (7)

Substitute $6.00\text{μF}$ for $C_{6.00\text{μF}}$, $10.525\text{V}$ for $\Delta V_{6.00\text{μF}}$ in equation (7) to find $Q_{6.00\text{μF}}$,

$\begin{array}{c}{Q}_{6.00\text{μF}}=6.00\text{μF}\times 10.525\text{V}\\ \text{=63}\text{.15}\text{μC}\\ \approx \text{63}\text{.2}\text{μC}\end{array}$

Thus, the charge across $6.00\text{μF}$ capacitor is $\text{63}\text{.2}\text{μC}$.

The charge across capacitors $15.0\text{μF}\text{and}3.00\text{μF}$ will be same.

Calculate the charge for the capacitor $15.0\text{μF}\text{and}3.00\text{μF}$.

$Q_{15.0\text{μF}}=Q_{3.00\text{μF}}=Q_{20.0\text{μF}}-Q_{6.00\text{μF}}$ (8)

Substitute $\text{89}\text{.5}\text{μC}$ for $Q_{20.0\text{μF}}$, $\text{63}\text{.2}\text{μC}$ for $Q_{6.00\text{μF}}$ in equation (8) to find $Q_{15.0\text{μF}}\text{and}{Q}_{3.00\text{μF}}$,

$\begin{array}{c}{Q}_{15.0\text{μF}\text{and}3.00\text{μF}}=\text{89}\text{.5}\text{μC}-\text{63}\text{.2}\text{μC}\\ =\text{26}\text{.3}\text{μC}\end{array}$

Thus, the charge across capacitors $15.0\text{μF}\text{and}3.00\text{μF}$ is $C_2$ is $\text{26}\text{.3}\text{μC}$.

Conclusion:

Therefore, the charge across capacitors $15.0\text{μF}\text{and}3.00\text{μF}$ is $C_2$ is $\text{26}\text{.3}\text{μC}$, the charge through $20.0\text{μF}$ is $\text{89}\text{.5}\text{μC}$, the charge across $6.00\text{μF}$ capacitor is $\text{63}\text{.2}\text{μC}$.