Chapter 25, Problem 102P

(a)

To determine

The battery current just after closing switch S.

Answer to Problem 102P

The battery current just after closing the switch S is $3.42\text{A}$ .

Explanation of Solution

Given:

The value of emf is $\epsilon =50.0\text{V}$ .

The value of capacitances are $C_1=10.0\mu \text{F}$ and $C_2=5.0\mu \text{F}$ .

Formula used:

Apply Kirchhoff’s rule in circuit just after switch is closed,

  $\epsilon -I_0\left(10.0\Omega \right)-I_0{R}_{eq}=0$ ..... (1)

Here, $I_0$ is initial current in battery just after closing the switch.

Calculation:

The equivalent resistance of parallel resistors $15.0\Omega ,12.0\Omega$ and $15.0\Omega$ is calculated as,

  $\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{15.0\Omega }+\frac{1}{12.0\Omega }+\frac{1}{15.0\Omega }\\ \frac{1}{R_{eq}}=\frac{4+5+4}{60}\\ R_{eq}=\frac{60}{13}\\ R_{eq}=4.615\Omega \end{array}$

From equation (1), the battery current just after closing switch S is calculated as,

  $\begin{array}{c}\left(50.0\text{V}\right)-I_0\left(10.0\Omega \right)-I_0\left(4.615\Omega \right)=0\\ I_0\left(14.615\Omega \right)=50.0\text{V}\\ I_0=3.42\text{A}\end{array}$

Conclusion:

Therefore, the battery current just after closing the switch S is $3.42\text{A}$ .

(b)

To determine

The battery current a long time after closing the switch S.

Answer to Problem 102P

The battery current a long time after closing the switch S is $0.962\text{A}$ .

Explanation of Solution

Formula used:

Apply Kirchhoff’s rule in circuit a long time after switch is closed,

  $\epsilon -I_{\infty }\left(10.0\Omega \right)-I_{\infty }{R^{\prime }}_{eq}=0$ ..... (2)

Here, $I_{\infty }$ is initial current in battery a long time after closing the switch.

Calculation:

The equivalent resistance of series resistors $15.0\Omega ,12.0\Omega$ and $15.0\Omega$ is calculated as,

  $\begin{array}{c}{R^{\prime }}_{eq}=\left(15.0\Omega \right)+\left(12.0\Omega \right)+\left(15.0\Omega \right)\\ =42.0\Omega \end{array}$

From equation (2), the battery current a long time after closing switch S is calculated as,

  $\begin{array}{c}\left(50.0\text{V}\right)-I_{\infty }\left(10.0\Omega \right)-I_{\infty }\left(42.0\Omega \right)=0\\ I_{\infty }\left(52.0\Omega \right)=50.0\text{V}\\ I_{\infty }=0.962\text{A}\end{array}$

Conclusion:

Therefore, the battery current a long time after closing the switch S is $0.962\text{A}$ .

(c)

To determine

The charges on capacitors plates a long time after closing the switch S.

Answer to Problem 102P

The charges on capacitors plates a long time after closing the switch S are $260\mu \text{C}$ and $130\mu \text{C}$ respectively.

Explanation of Solution

Formula used:

Apply Kirchhoff’s rule in circuit containing resistors $15.0\Omega$ , $12.0\Omega$ and capacitors $10.0\mu \text{F}$ ,

  $V_{10.0\mu \text{F}}-I_{\infty }\left(15.0\Omega \right)-I_{\infty }\left(12.0\Omega \right)=0$ ..... (3)

Apply Kirchhoff’s rule in circuit containing resistors $15.0\Omega$ , $12.0\Omega$ and capacitors $5.00\mu \text{F}$ ,

  $V_{5.00\mu \text{F}}-I_{\infty }\left(15.0\Omega \right)-I_{\infty }\left(12.0\Omega \right)=0$ ..... (4)

The expression for charge on capacitor $10.0\mu \text{F}$ is,

  $Q_{10.0\mu \text{F}}=C_1\cdot V_{10.0\mu \text{F}}$

The expression for charge on capacitor $5.00\mu \text{F}$ is,

  $Q_{5.00\mu \text{F}}=C_2\cdot V_{5.00\mu \text{F}}$

Calculation:

From equation (3), the potential difference $V_{10.0\mu \text{F}}$ is calculated as,

  $\begin{array}{l}{V}_{10.0\mu \text{F}}-\left(0.962\text{A}\right)\left(15.0\Omega \right)-\left(0.962\text{A}\right)\left(12.0\Omega \right)=0\\ V_{10.0\mu \text{F}}=\left(27.0\text{A}\right)\left(0.962\text{A}\right)\\ V_{10.0\mu \text{F}}=25.974\text{V}\end{array}$

From equation (4), the potential difference $V_{5.00\mu \text{F}}$ is calculated as,

  $\begin{array}{l}{V}_{5.00\mu \text{F}}-\left(0.962\text{A}\right)\left(15.0\Omega \right)-\left(0.962\text{A}\right)\left(12.0\Omega \right)=0\\ V_{5.00\mu \text{F}}=\left(27.0\text{A}\right)\left(0.962\text{A}\right)\\ V_{5.00\mu \text{F}}=25.974\text{V}\end{array}$

The charge on charge on capacitor $10.0\mu \text{F}$ is calculated as,

  $\begin{array}{c}{Q}_{10.0\mu \text{F}}=\left(10.0\mu \text{F}\right)\left(25.974\text{V}\right)\\ =259.74\mu \text{C}\\ \approx 260\mu \text{C}\end{array}$

The charge on charge on capacitor $5.00\mu \text{F}$ is calculated as,

  $\begin{array}{c}{Q}_{5.00\mu \text{F}}=\left(5.00\mu \text{F}\right)\left(25.974\text{V}\right)\\ =129.87\mu \text{C}\\ \approx 130\mu \text{C}\end{array}$

Conclusion:

Therefore, the charges on capacitors plates a long time after closing the switch S are $260\mu \text{C}$ and $130\mu \text{C}$ respectively.

(d)

To determine

The charges on capacitors plates a long time after reopening the switch S.

Answer to Problem 102P

The charges on capacitors plates a long time after reopening the switch S is zero.

Explanation of Solution

Calculation:

If the switch S is reopened, then after long time there will not be any flow of current in the circuit. Thus,

  $I_{\infty }=0$

The potential difference across $10.0\mu \text{F}$ and $5.00\mu \text{F}$ capacitors will be zero,

  $V_{10.0\mu \text{F}}=V_{5.00\mu \text{F}}=0$

The charges on $10.0\mu \text{F}$ and $5.00\mu \text{F}$ capacitors will be zero,

  $Q_{10.0\mu \text{F}}=Q_{5.00\mu \text{F}}=0$

Conclusion:

Therefore, the charges on capacitors plates a long time after reopening the switch S is zero.