Question

2-21 (a) The circuit shown next is a network of four capacitors connected in parallel. Show that the parallel capacitance $C_{\text{p}}$ is given by .

$C_{\text{p}}=C_1+C_2+C_3+C_4$

(b) If $V=5.00\text{V},C_1=0.050\mu \text{F},C_2=0.010\mu \text{F},C_3=0.075\mu \text{F}$, and , find the parallel capacitance $C_4=0.020\mu \text{F}$, the charge on each capacitor, and the total charge $Q_p$.

(c) A series combination of capacitors is shown in the next figure. Show that the series capacitance is given by

$\frac{1}{C_S}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

(d) $V=3.0\text{V},C1=1.00\mu \text{F},C2=0.75\mu \text{F}$

find the series capacitance Cs and the voltage drops across each capacitor.

(e) For the series circuit of part (d) suppose that there were only two capacitors,

and . Show that the series capacitance in this case is the product of the two capacitances divided by the sum of the two.

(f) The complex capacitive network shown next is wired. Find the capacitance of the network, the voltage across each capacitor, and the charge on each capacitor.

Short answer

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Step-by-step solution

Part (a)

In a capacitor, the charge Q is directly proportional to the applied voltage.

Q = CV

The total charge Q_T stored in the all capacitors equal to the summation of the individual charges stored on each capacitor;

${\text{Q}}_{\text{T}}{\text{=Q}}_{\text{1}}{\text{+Q}}_{\text{2}}{\text{+Q}}_{\text{3}}{\text{+Q}}_{\text{4}}$

But, Q = CV

${\text{Q}}_{\text{T}}{\text{=C}}_p{\text{V}}_{\text{T}}{\text{=C}}_{\text{1}}{\text{V}}_{\text{1}}{\text{+C}}_{\text{2}}{\text{V}}_{\text{2}}{\text{+C}}_{\text{3}}{\text{V}}_{\text{3}}{\text{+C}}_{\text{4}}{\text{V}}_4$

Voltage (V) common for all parallel connected capacitors. Therefore, divide both side of the equation by voltage.

$\begin{array}{l}{\text{C}}_p{\text{V}}_{\text{T}}{\text{=C}}_{\text{1}}{\text{V}}_{\text{1}}{\text{+C}}_{\text{2}}{\text{V}}_{\text{2}}{\text{+C}}_{\text{3}}{\text{V}}_{\text{3}}{\text{+C}}_{\text{4}}{\text{V}}_4\\ {\text{C}}_p{\text{=C}}_{\text{1}}{\text{+C}}_{\text{2}}{\text{+C}}_{\text{3}}{\text{+C}}_{\text{4}}\end{array}$

Part (b). 

All the capacitor ahve 5.0 V

The charge on each is as follow

$\begin{array}{l}\text{Q}=\text{CV}\\ {\text{Q}}_1=0.050\times {10}^{-6}\text{F}\times 5.00\text{V}=2.5\times {10}^{-7}\text{C}\\ {\text{Q}}_2=0.010\times {10}^{-6}\text{F}\times 5.00\text{V}=5\times {10}^{-8}\text{C}\\ {\text{Q}}_3=0.075\times {10}^{-6}\text{F}\times 5.00\text{V}=3.75\times {10}^{-7}\text{C}\\ {\text{Q}}_4=0.020\times {10}^{-6}\text{F}\times 5.00\text{V}=1\times {10}^{-7}\text{C}\end{array}$

$\begin{array}{l}{\text{Q}}_p={\text{C}}_p{\text{V}}_T\\ {\text{Q}}_p=0.155\times {10}^{-6}\text{F}\times 5.00\text{V}=7.75\times {10}^{-7}\text{C}\end{array}$

Part (c)

In a capacitor, the charge Q is directly proportional to the applied voltage.

Q = CV

When capacitors are in series, all the capacitors have same current flow.

Therefore, same amount of charge store in all the capacitors.

Q_p = Q₁ = Q₂ = Q₃

When capacitors are in series manner, voltage is not equal in each capacitor.

V_T = V_C1 + V_C2 + V_c3

$\begin{array}{l}{\text{V}}_{\text{c1}}\text{=}\frac{{\text{Q}}_{\text{1}}}{{\text{C}}_{\text{1}}}\\ {\text{V}}_{\text{c2}}\text{=}\frac{{\text{Q}}_{\text{2}}}{{\text{C}}_{\text{2}}}\\ {\text{V}}_{\text{C3}}\text{=}\frac{{\text{Q}}_{\text{3}}}{{\text{C}}_{\text{3}}}\\ \\ {\text{V}}_T=\frac{{\text{Q}}_{\text{1}}}{{\text{C}}_{\text{1}}}+\frac{{\text{Q}}_{\text{2}}}{{\text{C}}_{\text{2}}}+\frac{{\text{Q}}_{\text{3}}}{{\text{C}}_{\text{3}}}\\ {\text{V}}_T=\frac{Q_p}{C_s}\\ \frac{Q_p}{C_s}=\frac{{\text{Q}}_{\text{1}}}{{\text{C}}_{\text{1}}}+\frac{{\text{Q}}_{\text{2}}}{{\text{C}}_{\text{2}}}+\frac{{\text{Q}}_{\text{3}}}{{\text{C}}_{\text{3}}}\\ \sin ce,Q_p=Q_1=Q_2=Q_3\\ so\\ \frac{1}{C_s}=\frac{1}{{\text{C}}_{\text{1}}}+\frac{1}{{\text{C}}_{\text{2}}}+\frac{1}{{\text{C}}_{\text{3}}}\end{array}$

part (d)

In a capacitor, the charge Q is directly proportional to the applied voltage.

Q = CV

$\begin{array}{l}\frac{1}{C_s}=\frac{1}{{\text{C}}_{\text{1}}}+\frac{1}{{\text{C}}_{\text{2}}}+\frac{1}{{\text{C}}_{\text{3}}}\\ \frac{1}{C_s}=\frac{1}{1\times {10}^{-6}}+\frac{1}{0.75\times {10}^{-6}}+\frac{1}{0.5\times {10}^{-6}}\\ \frac{1}{C_s}={10}^6+\frac{1}{0.75}\times {10}^6+2\times {10}^6=4.33\times {10}^6\\ C_s=\frac{1}{4.33\times {10}^6}=2.31\times {10}^{-7}\text{F}\end{array}$

Q = CV

$\text{Q=2.31}\times {\text{10}}^{-7}\text{F}\times \text{3.00V=6.93}\times {\text{10}}^{-7}\text{C}$

When capacitors are in series, all the capacitors have same current flow.

Therefore, same amount of charge store in all the capacitors.

Q_p = Q₁ = Q₂ = Q₃

When capacitors are in series manner, voltage is not equal in each capacitor.

V_T = V_C1 + V_C2 + V_c3

$\begin{array}{l}{\text{V}}_{\text{c1}}\text{=}\frac{{\text{Q}}_{\text{1}}}{{\text{C}}_{\text{1}}}=\frac{6.93\times {10}^{-7}\text{C}}{1\times {10}^{-6}\text{F}}=0.693\text{V}\\ \\ {\text{V}}_{\text{c2}}\text{=}\frac{{\text{Q}}_{\text{2}}}{{\text{C}}_{\text{2}}}=\frac{6.93\times {10}^{-7}\text{C}}{0.75\times {10}^{-6}\text{F}}=0.924\text{V}\\ \\ {\text{V}}_{\text{C3}}\text{=}\frac{{\text{Q}}_{\text{3}}}{{\text{C}}_{\text{3}}}\frac{6.93\times {10}^{-7}\text{C}}{0.5\times {10}^{-6}\text{F}}=1.386\text{V}\end{array}$

Part (e)

In a capacitor, the charge Q is directly proportional to the applied voltage.

Q = CV

When capacitors are in series, all the capacitors have same current flow.

Therefore, same amount of charge store in all the capacitors.

Q_p = Q₁ = Q₂

When capacitors are in series manner, voltage is not equal in each capacitor.

V_T = V_C1 + V_C2

_{$\begin{array}{l}{\text{V}}_{\text{c1}}\text{=}\frac{{\text{Q}}_{\text{1}}}{{\text{C}}_{\text{1}}}\\ {\text{V}}_{\text{c2}}\text{=}\frac{{\text{Q}}_{\text{2}}}{{\text{C}}_{\text{2}}}\\ \\ {\text{V}}_T=\frac{{\text{Q}}_{\text{1}}}{{\text{C}}_{\text{1}}}+\frac{{\text{Q}}_{\text{2}}}{{\text{C}}_{\text{2}}}\\ {\text{V}}_T=\frac{Q_p}{C_s}\end{array}$}

$\begin{array}{l}\\ \frac{Q_p}{C_s}=\frac{{\text{Q}}_{\text{1}}}{{\text{C}}_{\text{1}}}+\frac{{\text{Q}}_{\text{2}}}{{\text{C}}_{\text{2}}}\\ Q_p=Q_1=Q_2\\ \\ \frac{1}{C_s}=\frac{1}{{\text{C}}_{\text{1}}}+\frac{1}{{\text{C}}_{\text{2}}}\\ \frac{1}{C_s}=\frac{{\text{C}}_{\text{1}}+{\text{C}}_{\text{2}}}{{\text{C}}_{\text{1}}{\text{C}}_{\text{2}}}\\ C_s=\frac{{\text{C}}_{\text{1}}{\text{C}}_{\text{2}}}{{\text{C}}_{\text{1}}+{\text{C}}_{\text{2}}}\end{array}$