The switch in FIGURE EX28.36 has been in position a for a long time. It is changed to position b at$t=0$ . What are the charge $Q$on the capacitor and the current $I$through the resistor $\left(a\right)$immediately after the switch is closed? $\left(b\right)$at$t=50\mu s$ ? $\left(c\right)$at $t=200\mu s$?

We need to find the charge $Q$on the capacitor and the current I through the resistor immediately after the switch is closed?

At position $a$, the battery charges the capacitor after the position is changed to $b$the capacitor discharges due to the resistor. To discharge the capacitor, the charges flow through the circuit in time $t$. The initial charge $Q_o$is related to the final charge $Q$by the equation in the form

localid="1648963587476" $Q=Q_o{e}^{-t/RC}\left(1\right)$

$Q_o$is the charge on the capacitor immediately after the switch is off. This charge could be calculated using the potential difference between the capacitor which is the same for the battery. Immediately after the position is at $b$, the charge will be given as

$$\begin{gathered} Q_o=C△V \\ =\left(4\mu F\right)\left(9V\right) \\ =36\mu C \end{gathered}$$

The initial current immediately after the switch closes is

localid="1648964010113" $I_o=\frac{Q_o}{RC}{e}^{-t/RC}\left(2\right)$

Immediately means the time is zero $t=0.$after putting the values for $R,C,Q_o$and $t$into equation $\left(2\right)$to get the initial current $I_o$

$$\begin{gathered} I_o=\left(\frac{Q_o}{RC}\right)e^{-t/RC} \\ =\frac{36\times {10}^{-6}C}{\left(25\Omega \right)(4\times {10}^{-6}F)}{e}^0 \\ =0.360A \\ =360mA \end{gathered}$$

We need to find that resistor immediately after the switch is closed at$t=50\mu s$.

At time $t=50\mu s$equation $\left(1\right)$to get the final charge $Q$by

$$\begin{gathered} Q=Q_o{e}^{-t/RC} \\ =\left(36\mu C\right)e^{\left[-50\mu s/\left(25\Omega \right)\left(4\mu F\right)\right]} \\ =22\mu C \end{gathered}$$

Using equation $\left(2\right)$to get $I$by

$$\begin{gathered} I=\left(\frac{Q}{RC}\right)e^{-t/RC} \\ =\left(\frac{36\mu C}{\left(25\Omega \right)\left(4\mu F\right)}\right)e^{\left[-50\mu s/\left(25\Omega \right)\left(4\mu F\right)\right]} \\ =0.212A \\ =212mA \end{gathered}$$

We need to find that resistor immediately after the switch is closed at time $t=200\mu s$.

At time $t=200\mu s$equation $\left(1\right)$to get the final charge $Q$by

$$\begin{gathered} Q=Q_o{e}^{-t/RC} \\ =\left(36\mu C\right)e^{\left[-200\mu s/\left(25\Omega \right)\left(4\mu F\right)\right]} \\ =5\mu C \end{gathered}$$

Using equation $\left(2\right)$to get $I$by

$$\begin{gathered} I=\left(\frac{Q}{RC}\right)e^{-t/RC} \\ =\left(\frac{36\mu C}{\left(25\Omega \right)\left(4\mu F\right)}\right)e^{\left[-50\mu s/\left(25\Omega \right)\left(4\mu F\right)\right]} \\ =6.75\times {10}^{-3}A \\ =6.75mA \end{gathered}$$