Question

A 24-V dc voltage source was connected in series with a resistor and capacitor. Calculate the current after

0.00, 0.010, 0.10, 1.0, and 10 s if the resistance was 25 M$\Omega$and the capacitance 0.20 μF.

Short answer

The current after 0.00s, 0.010s, 0.10s, 1s, and 10s is$0.96\mu A,0.958\mu A,0.941\mu A,0.786\mu A,and0.130\mu Arespectively.$

Step-by-step solution

Given information:

A 24V battery connected to a resistor of 25M$\Omega$and a capacitor of 0.20$\mu$F.

Explanation:

Clearly we can see that the given circuit is an RC circuit. The capacitor will be charged and the current will slowly decrease until it becomes zero when the capacitor is completely charged. As the time continues the current flow reduces as the capacitor starts to store the current as electric energy. This reduction in current is given by:

$$\begin{gathered} I\left(t\right)=I_0{e}^{-\frac{t}{RC}} \\ hereRC=25M\Omega \times 0.2\mu F \\ RC=5\Omega F \\ Nowtheinitialcurrent{I}_{0}whenthecapacitorisnotactive: \\ {I}_0=\frac{V}{R}=\frac{24V}{25M\Omega }=0.96\mu A \\ Now,calculatingthecurrentI\left(t\right)atthegiventimeweget: \\ I\left(t\right)=0.96e^{-\frac{t}{5}} \\ so, \\ I(0.00)=0.96e^{-\frac{0.00}{5}}=0.96\mu A \\ I(0.010)=0.96e^{-\frac{0.010}{5}}=0.958\mu A \\ I(0.10)=0.96e^{-\frac{0.10}{5}}=0.941\mu A \\ I\left(1\right)=0.96e^{-\frac{1}{5}}=0.786\mu A \\ I\left(10\right)=0.96e^{-\frac{10}{5}}=0.130\mu A \end{gathered}$$