The capacitors in $FIGUREP28.74$are charged and the switch closes at$t=0s.$ At what time has the current in the $8\Omega$resistor decayed to half the value it had immediately after the switch was closed?

we have to find at what time has the current in the $8\Omega$resistor decayed to half the value it had immediately after the switch was closed.

We will find the equivalent RC circ0uit one given in the problem we start finding the equivalent capacitance in the first step we find the equivalent capacitance for $C_{1}and{C}_2$:

$\frac{1}{C_{12}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{2}{C_1}$\

where we have used the fact that $C_1=C_{2.}$this yields

$$\begin{gathered} C_{12}=\frac{C_1}{}=30\mu F. \\ {C}_{12}=30\mu F,C_{123}=50\mu F. \end{gathered}$$

in the next step,$C_{12}$is parallelly connected to $C_2$meaning that to equivalent capacitance is simply.

$C=C_{123}=C_{12}+C_3=50\mu F.$

Now we find the equivalent resistance in this circuit. The resistors $R_{1}and{R}_2$are connected parallelly. Therefore, their equivalent resistance satisfies

$$\begin{gathered} \frac{1}{R_{23}}=\frac{1}{R_2}+\frac{1}{R_3}=\frac{R_2+R_3}{R_2{R}_3}, \\ {R}_{23}=\frac{R_2{R}_3}{R_2+R_3}=12\Omega . \end{gathered}$$

further,$R_{23}$is connected serially to$R_1$so

$R=R_{123}=R_1+R_{23}=20\Omega .$

The equivalent RC circuit is given in the figure bellow. The$8\Omega$resistor $\left(R_1\right)$is in the main branch of our circuit so the current that flows through it is the same as the current through the equivalent resistor R in our equivalent circuit. The time dependent current is given by

$I\left(t\right)=Io{e}^{-\frac{t}{RC}}$,

where $I_0$is the initial current we need to find time t such that $I\left(t\right)=\frac{I_0}{2}$this yields

$\frac{I_0}{2}={I_{0}e}^{-\frac{t}{RC}}$.

i.e.

$2=e^{\frac{t}{RC}}$,

and finally

$t=RCin2=0.7ms.$