In Fig. 27-66 ${\mathit{R}}_{\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathit{K}\mathit{\Omega }\mathbf{,}{\mathit{R}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathit{K}\mathit{\Omega }\mathbf{,}\mathit{C}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{400}\mathit{\mu }\mathit{F}\mathbf{}$and the ideal battery has emf $\mathit{\epsilon }\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{\text{V}}$. First, the switch is closed a longtime so that the steady state is reached. Then the switch is opened at time. What is the current in resistor 2 at t =$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\text{ms}}$?

$R_1=10.0K\Omega ,R_2=15.0K\Omega ,C=0.400\mu F$

b. The emf of the given ideal battery of the circuit,$\epsilon =20.0\text{V}$

c. The switch is opened at time.t = 0

d. The given time, t - 4 ms

At a steady current situation, the voltage is divided among the resistors in the given resistance ratio values. Thus, the voltage in the steady situation can be calculated and can be used for further knowing the voltage at the given time. Now, using Ohm's law, the current value can be calculated.

Formulae:

The charge equation of a RC circuit, $\mathit{q}\mathbf{=}{\mathit{q}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{R}\mathbf{C}}$ (i)

The charge stored within the plates of a capacitor, q = CV (ii)

The voltage equation using Ohm’s law, V = IR (iii)

In a steady current situation, the capacitor voltage will equal the voltage across resistor 2 and thus can be given using equation (iii) and the given figure as follows:

$$\begin{gathered} V_0=R_2\frac{\epsilon }{R_1+R_2} \\ =\left(15.0\text{kΩ}\right)\left(\frac{20\text{V}}{10.0\text{kΩ}+15.0\text{kΩ}}\right) \\ =12.0\text{V} \end{gathered}$$

Now, using equations (i) and (ii), the voltage across resistor 2 at$t=4\times {10}^{-3}\text{s}$ can be given using the above voltage value as follows:

$$\begin{gathered} V=V_0{e}^{-t/RC} \\ =\left(12.0\text{V}\right)e^{-4\times {10}^{-3}\text{s}/\left(15\times {10}^3\Omega \right)\left(0.4\times {10}^6\text{C}\right)} \\ =6.16\text{V} \end{gathered}$$

Thus, the current value at resistor 2 branch can be given using the above voltage value in equation (iii) as follows:

$$\begin{gathered} I_2=\frac{6.16\text{V}}{15\times {10}^3\Omega } \\ =4.11\times {10}^{-4}\text{A} \end{gathered}$$

Hence, the value of the current is $4.11\times {10}^{-4}\text{A}$.