In the circuit depicted in the figure, the capacitors are initially uncharged. If the switch is closed at $t = 0$ , determine the charge on the capacitor $C_1$ as a function of time $t$ .

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In the circuit depicted in the figure, the capacitors are initially uncharged. If the switch is closed at t = 0 , determine the charge on the capacitor C_1 as a function of time t .

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Correct answer: A. E C_1 C_2 C_1 + C_2 ( 1 - e^ -(t(C_1 + C_2))/(r C_1 C_2) )

The given circuit is an RC circuit with two capacitors C_1 and C_2 connected in series with a resistor r . The equivalent capacitance of the series combination is: C_ eq = (C_1 C_2)/(C_1 + C_2) The time constant of the circuit is: = r C_ eq = (r C_1 C_2)/(C_1 + C_2) The maximum steady-state charge on the equivalent capacitor is: q_0 = E C_ eq = E C_1 C_2 C_1 + C_2 The charge on the equivalent capacitor as a function of time is given by the standard charging equation: q(t) = q_0 ( 1 - e^ -(t)/( ) ) Since the capacitors are in series, the charge on each individual capacitor is the same as the charge on the equivalent capacitor. Thus, the charge on C_1 is: q_1(t) = E C_1 C_2 C_1 + C_2 ( 1 - e^ -(t(C_1 + C_2))/(r C_1 C_2) )

In the circuit depicted in the figure, the capacitors are initially...

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