Chapter 17, Problem 106QAP

To determine

(a)

• Charge on the capacitor
• Energy stored in the capacitor

Answer to Problem 106QAP

• Charge on the capacitor is $75\text{μC}$.
• Energy stored in the capacitor is $112.5\times {10}^{-6}\text{J}$.

Explanation of Solution

Given:

Circuit diagram

Figure.1

Formula used:

Capacitance is given as

  $C=\frac{q}{V}$

Here,

  $q$ is the charge

  $V$ is the voltage

Energy stored in the capacitor is given as

  $U=\frac{1}{2}C{V}^2$

Here,

  $C$ is the capacitance

  $V$ is the voltage

Calculation:

Consider the given circuit diagram

Figure.1

When the switch S is connected to the point A, then only capacitor $25\mu \text{F}$ is connected to the battery, charge stored in the capacitor is calculated as

  $C=\frac{q}{V}$

Plugging the values in the above equation

  $\begin{array}{c}25=\frac{q}{3}\\ q=75\text{μC}\end{array}$

Energy stored in the capacitor is given as

  $U=\frac{1}{2}C{V}^2$

Plugging the values in the above equation

  $\begin{array}{c}U=\frac{1}{2}\times 25\times {10}^{-6}\times 3^2\\ =112.5\times {10}^{-6}\text{J}\end{array}$

Conclusion:

• Charge on the capacitor is $75\text{μC}$.
• Energy stored in the capacitor is $112.5\times {10}^{-6}\text{J}$.

To determine

(b)

• Charge and energy stored in the $25\text{μF}$ capacitor is
• Charge and energy stored in the $20\text{μF}$ capacitor is

Answer to Problem 106QAP

• Charge and energy stored in the $25\text{μF}$ capacitor is $75\text{μC}$ and $112.5\times {10}^{-6}\text{J}$ respectively.
• Charge and energy stored in the $20\text{μF}$ capacitor is $75\text{μC}$ and $140.6\times {10}^{-6}\text{J}$respectively.

Explanation of Solution

Given:

Circuit diagram

Figure.1

Formula used:

Capacitance is given as

  $C=\frac{q}{V}$

Here,

  $q$ is the charge

  $V$ is the voltage

Energy stored in the capacitor is given as

  $U=\frac{1}{2}C{V}^2$

Here,

  $C$ is the capacitance

  $V$ is the voltage

Calculation:

Consider the given circuit diagram

Figure.1

When the switch S is connected to the point b, than none of the capacitor is connected to the battery and both the capacitor are connected in series.

In series connection both the capacitor has same charge, therefore the charge on both the capacitor is $75\text{μC}$

Energy stored in the capacitor is given as

  $U=\frac{1}{2}\times \frac{q^2}{C}$

Energy on the $25\text{μF}$ capacitor is calculated as

  $\begin{array}{c}U=\frac{1}{2}\times \frac{{75}^2}{25}\\ =112.5\times {10}^{-6}\text{J}\end{array}$

Energy on the $20\text{μF}$ capacitor is calculated as

  $\begin{array}{c}U=\frac{1}{2}\times \frac{{75}^2}{20}\\ =140.6\times {10}^{-6}\text{J}\end{array}$

Conclusion:

• Charge and energy stored in the $25\text{μF}$ capacitor is $75\text{μC}$ and $112.5\times {10}^{-6}\text{J}$ respectively.
• Charge and energy stored in the $20\text{μF}$ capacitor is $75\text{μC}$ and $140.6\times {10}^{-6}\text{J}$respectively.