Chapter 25, Problem 103P

(a)

To determine

The battery current just after closing switch S.

Answer to Problem 103P

The battery current just after closing the switch S is $0.250\text{A}$ .

Explanation of Solution

Given:

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

The value of capacitances are $C=5.00\mu \text{F}$ .

Formula used:

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

  $\epsilon -I_0\left(200\Omega \right)-V_C=0$ ..... (1)

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

Calculation:

Initially, the capacitor is uncharged so,

  $V_C=0$

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

  $\begin{array}{c}\left(50.0\text{V}\right)-I_0\left(200\Omega \right)-0=0\\ I_0\left(200\Omega \right)=50.0\text{V}\\ I_0=0.250\text{A}\end{array}$

Conclusion:

Therefore, the battery current just after closing the switch S is $0.250\text{A}$ .

(b)

To determine

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

Answer to Problem 103P

The battery current a long time after closing the switch S is $62.5\text{mA}$ .

Explanation of Solution

Formula used:

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

  $\epsilon -I_{\infty }\left(200\Omega \right)-I_{\infty }\left(600\Omega \right)=0$ ..... (2)

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

Calculation:

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

  $\begin{array}{c}\left(50.0\text{V}\right)-I_{\infty }\left(200\Omega \right)-I_{\infty }\left(600\Omega \right)=0\\ I_{\infty }\left(800\Omega \right)=50.0\text{V}\\ I_{\infty }=62.5\times {10}^{-3}\text{A}\\ I_{\infty }=62.5\text{mA}\end{array}$

Conclusion:

Therefore, the battery current a long time after closing the switch S is $62.5\text{mA}$ .

(c)

To determine

The current in $600\Omega$ resistor as a function of time.

Answer to Problem 103P

The current in $600\Omega$ resistor as a function of time is $\left(62.5\text{mA}\right)\left(1-e^{-t/0.500\text{ms}}\right)$ .

Explanation of Solution

Formula used:

Apply Kirchhoff’s rule at j unction of resistor $200\Omega$ and capacitors $5.00\mu \text{F}$ ,

  $I_1=I_2+I_3$ ..... (3)

Apply Kirchhoff’s rule in loop 1,

  $\epsilon -\frac{Q}{C}-R_1{I}_1=0$ ..... (4)

Apply Kirchhoff’s rule in loop containing resistor $600\Omega$ and capacitors $5.00\mu \text{F}$ ,

  $\frac{Q}{C}-R_2{I}_2=0$ ..... (5)

Calculation:

Differentiate equation (4) with respect to time,

  $\begin{array}{c}\frac{d}{dt}\left(\epsilon -\frac{Q}{C}-R_1{I}_1\right)=0\\ 0-\frac{1}{C}\frac{dQ}{dt}-R_1\frac{d{I}_1}{dt}=0\\ -\frac{1}{C}{I}_3-R_1\frac{d{I}_1}{dt}=0\\ R_1\frac{d{I}_1}{dt}=-\frac{I_3}{C}\end{array}$ ..... (6)

Differentiate equation (5) with respect to time,

  $\begin{array}{c}\frac{d}{dt}\left(\frac{Q}{C}-R_2{I}_2\right)=0\\ \frac{1}{C}\frac{dQ}{dt}-R_2\frac{d{I}_2}{dt}=0\\ R_2\frac{d{I}_2}{dt}=\frac{I_3}{C}\end{array}$ ..... (7)

From equation (3) and (7),

  $\begin{array}{c}{R}_2\frac{d{I}_2}{dt}=\frac{\left(I_1-I_2\right)}{C}\\ \frac{d{I}_2}{dt}=\frac{\left(I_1-I_2\right)}{R_{2}C}\end{array}$ ..... (8)

From equation (4),

  $\begin{array}{c}{I}_1=\frac{\epsilon -Q/C}{R_1}\\ =\frac{\epsilon -R_2{I}_2}{R_1}\end{array}$ ..... (9)

From equation (8) and (9),

  $\begin{array}{c}\frac{d{I}_2}{dt}=\frac{1}{R_{2}C}\left(\frac{\epsilon -R_2{I}_2}{R_1}-I_2\right)\\ =\frac{\epsilon }{R_1{R}_{2}C}-\frac{R_1+R_2}{R_1{R}_{2}C}{I}_2\end{array}$

Let the solution of above differential equation is,

  $I_2=a+b{e}^{-t/\tau }$ ..... (10)

Differentiate equation (10) with respect to time,

  $\frac{d{I}_2}{dt}=-\frac{b}{\tau }{e}^{-t/\tau }$ ..... (11)

From equation (8) and (11),

  $-\frac{b}{\tau }{e}^{-t/\tau }=\frac{\epsilon }{R_1{R}_{2}C}-\left(\frac{R_1+R_2}{R_1{R}_{2}C}\right)\left(a+b{e}^{-t/\tau }\right)$

Equate coefficient of $e^{-t/\tau }$ of above equation on both sides,

  $\begin{array}{c}-\frac{b}{\tau }=-\left(\frac{R_1+R_2}{R_1{R}_{2}C}\right)b\\ \tau =\frac{R_1{R}_{2}C}{R_1+R_2}\end{array}$ ..... (12)

And,

  $a=\frac{\epsilon }{R_1+R_2}$

At $t=0$ , $I_2=0$ ,

  $\begin{array}{c}a+b=0\\ b=-a\\ b=-\frac{\epsilon }{R_1+R_2}\end{array}$

From equation (10),

  $I_2=\frac{\epsilon }{R_1+R_2}-\frac{\epsilon }{R_1+R_2}{e}^{-t/\tau }$ ..... (13)

From equation (10),

  $\begin{array}{c}\tau =\frac{\left(200\Omega \right)\left(600\Omega \right)\left(5.00\mu \text{F}\right)}{\left(200\Omega \right)\left(600\Omega \right)}\\ =5\times {10}^{-6}\text{s}\\ =0.500\text{ms}\end{array}$

Substitute values in equation (13),

  $\begin{array}{c}{I}_2=\frac{50.0\text{V}}{200\Omega +600\Omega }\left(1-e^{-t/0.500\text{ms}}\right)\\ =\left(62.5\text{mA}\right)\left(1-e^{-t/0.500\text{ms}}\right)\end{array}$

Conclusion:

Therefore, the current in $600\Omega$ resistor as a function of time is $\left(62.5\text{mA}\right)\left(1-e^{-t/0.500\text{ms}}\right)$ .

(d)

To determine

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

Answer to Problem 103P

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.