Chapter 19, Problem 43P

To determine

The charge on each of the other two capacitors. Also, the voltage across each capacitor and the voltage across the entire combination.

Answer to Problem 43P

The charge on two other capacitors is,

  ${\mathrm{Q}}_1=48\mathrm{\mu }\text{C}$

  ${\mathrm{Q}}_3={\mathrm{Q}}_2=24\mathrm{\mu }\text{C}$

The voltage across each capacitor is,

  $\begin{array}{l}{\mathrm{V}}_2={\mathrm{V}}_3=1.5\text{V}\\ {\mathrm{V}}_1=3\text{V}\end{array}$

The voltage across the entire combination is $\mathrm{V}=3\text{V}$

Explanation of Solution

Given:

The given circuit is shown below.

  

The value of capacitors is, ${\mathrm{C}}_1={\mathrm{C}}_2={\mathrm{C}}_3=16\mathrm{\mu }\text{F}$

Charge across second capacitor is ${\mathrm{Q}}_2=24\mathrm{\mu }\text{C}$

Formula used:

The relation between charge Q , voltage V and capacitance C is,

  $\mathrm{Q}=\mathrm{C}\mathrm{V}$

Calculation:

The voltage across capacitor ${\mathrm{C}}_2$ is,

  $\begin{array}{c}{\mathrm{Q}}_2={\mathrm{C}}_2{\mathrm{V}}_2\\ {\mathrm{V}}_2=\frac{{\mathrm{Q}}_2}{{\mathrm{C}}_2}\\ =\frac{24\times {10}^{-6}}{16\times {10}^{-6}}\\ =1.5\text{V}\end{array}$

Capacitor ${\mathrm{C}}_2$ and ${\mathrm{C}}_3$ are connected in series, so the charge on both the capacitors will be same.

So,

  ${\mathrm{Q}}_3={\mathrm{Q}}_2=24\mathrm{\mu }\text{C}$

Also, the voltage across capacitor ${\mathrm{C}}_3$ is,

  $\begin{array}{c}{\mathrm{Q}}_3={\mathrm{C}}_3{\mathrm{V}}_3\\ {\mathrm{V}}_3=\frac{{\mathrm{Q}}_3}{{\mathrm{C}}_3}\\ =\frac{24\times {10}^{-6}}{16\times {10}^{-6}}\\ =1.5\text{V}\end{array}$

Capacitor ${\mathrm{C}}_1$ is connected in parallel with ${\mathrm{C}}_2$ and ${\mathrm{C}}_3$ . So, the voltage across capacitor ${\mathrm{C}}_1$ will be equal to the total voltage across the branch containing capacitor ${\mathrm{C}}_2$ and ${\mathrm{C}}_3$

  $\begin{array}{c}{\mathrm{V}}_1={\mathrm{V}}_2+{\mathrm{V}}_3\\ =1.5+1.5\\ =3\text{V}\end{array}$

Also, the total voltage across the combination will be equal to the voltage across capacitor ${\mathrm{C}}_1$ and the branch containing capacitor ${\mathrm{C}}_2$ and ${\mathrm{C}}_3$

So,

  $\mathrm{V}={\mathrm{V}}_1=3\text{V}$

The charge on capacitor ${\mathrm{C}}_1$ is,

  $\begin{array}{c}{\mathrm{Q}}_1={\mathrm{C}}_1\times {\mathrm{V}}_1\\ =16\times {10}^{-6}\times 3\\ =48\times {10}^{-6}\\ =48\mathrm{\mu }\text{C}\end{array}$

Conclusion:

Therefore, the charge on two other capacitors is,

  ${\mathrm{Q}}_1=48\mathrm{\mu }\text{C}$

  ${\mathrm{Q}}_3={\mathrm{Q}}_2=24\mathrm{\mu }\text{C}$

The voltage across each capacitor is,

  $\begin{array}{l}{\mathrm{V}}_2={\mathrm{V}}_3=1.5\text{V}\\ {\mathrm{V}}_1=3\text{V}\end{array}$

The voltage across the entire combination is $\mathrm{V}=3\text{V}$