Question

What is the charge on each capacitor in the figure, if V = 9.0V ?(Figure 1)

Problem 29.55 Part A what is the charge on each capacitor in the figure, if V = 9.0 V? Q1,Q2, Q3 =Part B What is the potential difference across each capacitor in the figure ? delta V1, delta V2, delta V3 =

Answer 1

Concepts and reason

The concepts required to solve this problem are capacitance, voltage and the charge.

Initially, calculate the charge in each capacitor using the relation of charge with the capacitance and the voltage. Then, calculate the voltage in each capacitor using the relation of voltage with the charge and the capacitance.

Fundamentals

The relation of the capacitor with the charge and the voltage is,

$Q = CV$

Here, Q is the charge, C is the capacitance, and V is the voltage.

The equivalent capacitance of series combination of two capacitors is,

${C_{{\rm{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}$

Here, ${C_{{\rm{eq}}}}$ is the equivalent capacitance, ${C_1}$ and ${C_2}$ are the capacitors connected in series.

(A)

Refer the figure given in the question.

The charge ${Q_1}$ on the capacitor ${C_1}$ is,

${Q_1} = {C_1}V$

Substitute ${\rm{5 }}\mu {\rm{F}}$ for ${C_1}$ and 9.0 V for V.

$\begin{array}{c}\\{Q_1} = \left( {{\rm{5 }}\mu {\rm{F}}} \right)\left( {\frac{{{{10}^{ - 6}}{\rm{ F}}}}{{{\rm{1 }}\mu {\rm{F}}}}} \right)\left( {9.0{\rm{ V}}} \right)\\\\ = 45 \times {10^{ - 6}}{\rm{ C}}\\\\{Q_1}{\rm{ = 45 }}\mu {\rm{C}}\\\end{array}$

The equivalent capacitance of series combination of two capacitors is,

$\frac{1}{{{C_{{\rm{eq}}}}}} = \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}$

Substitute $4{\rm{ }}\mu {\rm{F}}$ for ${C_2}$ and ${\rm{6 }}\mu {\rm{F}}$ for ${C_3}$ .

$\begin{array}{c}\\\frac{1}{{{C_{{\rm{eq}}}}}} = \frac{1}{{4{\rm{ }}\mu {\rm{F}}}} + \frac{1}{{{\rm{6 }}\mu {\rm{F}}}}\\\\\frac{1}{{{C_{{\rm{eq}}}}}} = \frac{{{\rm{6 }}\mu {\rm{F}} + 4{\rm{ }}\mu {\rm{F}}}}{{\left( {{\rm{6 }}\mu {\rm{F}}} \right)\left( {4{\rm{ }}\mu {\rm{F}}} \right)}}\\\\{C_{{\rm{eq}}}} = \frac{{\left( {{\rm{6 }}\mu {\rm{F}}} \right)\left( {4{\rm{ }}\mu {\rm{F}}} \right)}}{{\left( {{\rm{6 }}\mu {\rm{F}}} \right) + \left( {4{\rm{ }}\mu {\rm{F}}} \right)}}\\\\ = 2.4{\rm{ }}\mu {\rm{F}}\\\end{array}$

The charge ${Q_{23}}$ on the capacitor ${C_{{\rm{eq}}}}$ is,

${Q_{23}} = {C_{{\rm{eq}}}}V$

Substitute ${\rm{2}}{\rm{.4 }}\mu {\rm{F}}$ for ${C_{{\rm{eq}}}}$ and 9.0 V for V.

$\begin{array}{c}\\{Q_{23}} = \left( {{\rm{2}}{\rm{.4 }}\mu {\rm{F}}} \right)\left( {\frac{{{{10}^{ - 6}}{\rm{ F}}}}{{{\rm{1 }}\mu {\rm{F}}}}} \right)\left( {9.0{\rm{ V}}} \right)\\\\ = 21.6 \times {10^{ - 6}}{\rm{ C}}\\\\{Q_{23}}{\rm{ = 21}}{\rm{.6 }}\mu {\rm{C}}\\\end{array}$

The charge remains the same in the series combination.

The charge ${Q_2}$ on the capacitor ${C_2}$ is,

${Q_2} = {\rm{21}}{\rm{.6 }}\mu {\rm{C}}$

The charge ${Q_3}$ on the capacitor ${C_3}$ is,

${Q_3} = {\rm{21}}{\rm{.6 }}\mu {\rm{C}}$

(B)

The voltage $\Delta {V_1}$ across the capacitor ${C_1}$ is,

$\Delta {V_1} = 9.0{\rm{ V}}$

The voltage $\Delta {V_2}$ across ${C_2}$ is

$\Delta {V_2} = \frac{{{Q_2}}}{{{C_2}}}$

Substitute ${\rm{21}}{\rm{.6 }}\mu {\rm{C}}$ for ${Q_2}$ and $4{\rm{ }}\mu {\rm{F}}$ for ${C_2}$ .

$\begin{array}{c}\\\Delta {V_2} = \frac{{{\rm{21}}{\rm{.6 }}\mu {\rm{C}}}}{{4{\rm{ }}\mu {\rm{F}}}}\\\\ = 5.4{\rm{ V}}\\\end{array}$

The voltage $\Delta {V_3}$ across ${C_3}$ is

$\Delta {V_3} = \frac{{{Q_3}}}{{{C_3}}}$

Substitute ${\rm{21}}{\rm{.6 }}\mu {\rm{C}}$ for ${Q_3}$ and ${\rm{6 }}\mu {\rm{F}}$ for ${C_3}$ .

$\begin{array}{c}\\\Delta {V_3} = \frac{{{\rm{21}}{\rm{.6 }}\mu {\rm{C}}}}{{{\rm{6 }}\mu {\rm{F}}}}\\\\ = 3.6{\rm{ V}}\\\end{array}$

Ans: Part A

The value of the charges ${Q_1}$ , ${Q_2}$ , and ${Q_3}$ are ${\rm{45}}{\rm{.0 }}\mu {\rm{C}}$ , ${\rm{21}}{\rm{.6 }}\mu {\rm{C}}$ , and ${\rm{21}}{\rm{.6 }}\mu {\rm{C}}$ respectively.

Part B

The value of the voltages $\Delta {V_1}$ , $\Delta {V_2}$ , and $\Delta {V_3}$ are $9.0{\rm{ V}}$ , $5.4{\rm{ V}}$ , and $3.6{\rm{ V}}$ respectively.