Chapter 18, Problem 91P

(a)

To determine

Find the value of current $I_1$  and $I_2$ and voltage $V_1$ and $V_2$ at $t=0$ time.

Answer to Problem 91P

The value of current $I_1$  is $0.30\text{ mA}$ and $I_2$ is $0.30\text{ mA}$ and voltage $V_1$ is $12\text{ V}$ and $V_2$ is $12\text{ V}$ at $t=0$ time.

Explanation of Solution

Initially the capacitor has zero voltage across it then the voltage $V_1$ and $V_2$ is equal to the voltage of the battery.

Write the equation for current.

$I_1=I_2=\frac{V}{R}$ (I)

Here, $V$ is the voltage and $R$ is the resistance. The current in $I_1$  and $I_2$ are same.

Conclusion:

Substitute $\text{12 V}$ for $V$ and $40.0\times {10}^3\Omega$ for $R$ in equation I.

$I_1=I_2=\frac{\text{12 V}}{40.0\times {10}^3\Omega }=0.30\text{ mA}$

Therefore, the value of current $I_1$  is $0.30\text{ mA}$ and $I_2$ is $0.30\text{ mA}$ and voltage $V_1$ is $12\text{ V}$ and $V_2$ is $12\text{ V}$ at $t=0$ time.

(b)

To determine

Find the value of current $I_1$  and $I_2$ and voltage $V_1$ and $V_2$ at $t=1.0\text{ ms}$ time.

Answer to Problem 91P

The value of current $I_1$  is $0.18\text{ mA}$ and $I_2$ is $0.18\text{ mA}$ and voltage $V_1$ is $12\text{ V}$ and $V_2$ is $7.3\text{ V}$ at $t=1.0\text{ ms}$ time.

Explanation of Solution

Write the equation for time constant,

$\tau =RC$ (II)

Here, $\tau$ is the time constant, $R$ is the resistance and $C$ is the current.

Write the equation for current.

$I_1=I_2=I_0{e}^{-t/\tau }$ (III)

Here, $I_1=I_2$ is the current, $I_0$ is the initial current and $t$ is the time.

Conclusion:

Substitute $40.0\times {10}^3\Omega$ for $R$  and $\text{5}\text{.0}\times {\text{10}}^{-6}\text{ F}$ for $C$ in equation II.

$\begin{array}{c}\tau =\left(40.0\times {10}^3\Omega \right)\left(\text{5}\text{.0}\times {\text{10}}^{-6}\text{ F}\right)\\ =2.0\text{ ms}\end{array}$

Substitute $3.0\times {10}^{-4}\text{ A}$ for $I_0$, $2.0\text{ ms}$ for $\tau$ and $1.0\text{ ms}$ for $t$ in equation III.

$I_1=I_2=\left(3.0\times {10}^{-4}\text{ A}\right)e^{-1.0\text{ ms}/2.0\text{ ms}}=0.18\text{ mA}$

Then the voltage,

${\text{V}}_1\text{=12 V}$ and $V_2=I_{2}R$

Substitute $1.82\times {10}^{-4}\text{ A}$ for $I_2$ and $40.0\times {10}^3\Omega$ for $R$  in above equation.

$\begin{array}{c}{V}_2=\left(1.82\times {10}^{-4}\text{ A}\right)\left(40.0\times {10}^3\Omega \right)\\ =7.3\text{ V}\end{array}$

Therefore, the value of current $I_1$  is $0.18\text{ mA}$ and $I_2$ is $0.18\text{ mA}$ and voltage $V_1$ is $12\text{ V}$ and $V_2$ is $7.3\text{ V}$ at $t=1.0\text{ ms}$ time.

(c)

To determine

Find the value of current $I_1$  and $I_2$ and voltage $V_1$ and $V_2$ at $t=5.0\text{ ms}$ time.

Answer to Problem 91P

The value of current $I_1$  is $\text{25 μA}$ and $I_2$ is $\text{25 μA}$ and voltage $V_1$ is $12\text{ V}$ and $V_2$ is $\text{0}\text{.99 V}$ at $t=5.0\text{ ms}$ time.

Explanation of Solution

Use the previous section equations to find current and voltage.

Conclusion:

Substitute $3.0\times {10}^{-4}\text{ A}$ for $I_0$, $2.0\text{ ms}$ for $\tau$ and $5.0\text{ ms}$ for $t$ in equation III.

$I_1=I_2=\left(3.0\times {10}^{-4}\text{ A}\right)e^{-1.0\text{ ms}/5.0\text{ ms}}=25\text{ μA}$

Then the voltage,

${\text{V}}_1\text{=12 V}$ and $V_2=I_{2}R$

Substitute $2.463\times {10}^{-5}\text{ A}$ for $I_2$ and $40.0\times {10}^3\Omega$ for $R$  in above equation.

$\begin{array}{c}{V}_2=\left(2.463\times {10}^{-5}\text{ A}\right)\left(40.0\times {10}^3\Omega \right)\\ =0.99\text{ V}\end{array}$

Therefore, the value of current $I_1$  is $\text{25 μA}$ and $I_2$ is $\text{25 μA}$ and voltage $V_1$ is $12\text{ V}$ and $V_2$ is $\text{0}\text{.99 V}$ at $t=5.0\text{ ms}$ time.