The circuit in the figure below contains two resistors, \( R_1 = 2.20 \, \text{k}\Omega \) and \( R_2 = 3.10 \, \text{k}\Omega \), and two capacitors, \( C_1 = 2.10 \, \mu\text{F} \) and \( C_2 = 3.10 \, \mu\text{F} \), connected to a battery with emf \( \mathcal{E} = 115 \, \text{V} \). There are no charges on the capacitors before switch \( S \) is closed.### Diagram Description:- The circuit diagram shows \( R_1 \) and \( R_2 \) in series, then in parallel with \( C_1 \) and \( C_2 \) also in series.- A switch \( S \) is shown open but is meant to be closed.- The battery provides an emf of 115 V.**(a)** Determine the charge on capacitor \( C_1 \) as a function of time (in ms), after the switch is closed. (Use the following as necessary: \( t \).)\[ q_1 = 598 \left( 1 - e^{\frac{-t}{5.13}} \right) \, \mu\text{C} \]**(b)** Determine the charge on capacitor \( C_2 \) as a function of time (in ms), after the switch is closed. (Use the following as necessary: \( t \).)\[ q_2 = \, \mu\text{C} \] (Note: The equation for \( q_2 \) is left unfilled in the image, indicating more information may be needed to complete this part of the problem.)

Given Resistor R1 = 2.20 kΩ Resistor R2 = 3.10 kΩ Capacitor C1 = 2.10 μF Capacitor C2 = 3.10…

Answered: The circuit in the figure below…

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