Question

Consider the situation shown in figure (32−E36). The switch is closed at t = 0 when the capacitors are uncharged. Find the charge on the capacitor C1 as a function of time t.
Figure

Answer 1

Answer:

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Answer 2

The charge on the capacitor C1 is $Ceqv \varepsilon 1 -e-1rCeqv$ as a function of time t.

Explanation:

• At t = 0, the switch is closed when there is no charge in the capacitors. It is to be noted that the two capacitors are linked in a series. Their equivalent capacitance is denoted as  

       $\text { Ceqv }=\mathrm{C} 1 \mathrm{C} 2 \mathrm{C} 1+\mathrm{C} 2$

• In the capacitors, the charge’s growth is q = $\text { Ceqv }=\mathrm{C} 1 \mathrm{C} 2 \mathrm{C} 1+\mathrm{C} 2$. The equation is:

      $Q=C_{\text {eff }} E\left(1-e^{-t / R C}\right)=\frac{C_{1} C_{2}}{C_{1}+C_{2}} E\left(1-e^{-t / R C}\right)$