How long will it take for the current in a circuit to drop from its initial value to \(1.50 \mathrm{~mA}\) if the circuit contains two \(3.8-\mu \mathrm{F}\) capacitors that are initially uncharged, two \(2.2-\mathrm{k} \Omega\) resistors, and a \(12.0-\mathrm{V}\) battery all connected in series?

Since the two capacitors and the two resistors are connected in series, we will calculate the equivalent resistance (R) and capacitance (C) of the circuit. For resistors connected in series, we can simply add their resistance values together: \(R_{eq} = R_1 + R_2 = 2.2\,\text{k}\Omega + 2.2\,\text{k}\Omega = 4.4\,\text{k}\Omega\) For capacitors connected in series, the inverse of the equivalent capacitance is equal to the sum of the inverses of the two capacitances: \(\frac{1}{C_{eq}} = \frac{1}{C_1}+\frac{1}{C_2}\) Insert the values of the given capacitances: \(\frac{1}{C_{eq}} = \frac{1}{3.8\,\mu\text{F}}+\frac{1}{3.8\,\mu\text{F}} = \frac{2}{3.8\,\mu\text{F}}\) Now, we can find the equivalent capacitance (C) as follows: \(C_{eq} = \frac{1}{\frac{2}{3.8\,\mu\text{F}}} = 1.9\,\mu\text{F}\)

The time constant (τ) of an RC circuit can be calculated using the formula: \(\tau = RC\) Insert the values of the equivalent resistance and capacitance: \(\tau = (4.4\,\text{k}\Omega)(1.9\,\mu\text{F}) = 8.36\,\text{ms}\)

The charging current (I) in the circuit after a time t can be given by the formula: \(I(t) = I_0 e^{-t/\tau}\) Where \(I_0\) is the initial current, and t is the time. The initial current in the circuit can be found using Ohm's law: \(I_0 = \frac{V}{R_{eq}}=\frac{12.0\,\text{V}}{4.4\,\text{k}\Omega}=2.73\,\text{mA}\) We will now insert the given final current value, \(1.50\,\text{mA} \), and solve for t: \(1.50\,\text{mA} = 2.73\,\text{mA} e^{-t/8.36\,\text{ms}}\)

First, divide both sides by \(2.73\,\text{mA}\): \(\frac{1.50\,\text{mA}}{2.73\,\text{mA}} = e^{-t/8.36\,\text{ms}}\) Now, take the natural logarithm of both sides: \(-t/8.36\,\text{ms} = \ln\left(\frac{1.50\,\text{mA}}{2.73\,\text{mA}}\right)\) Finally, solve for t: \(t = -8.36\,\text{ms} \times \ln\left(\frac{1.50\,\text{mA}}{2.73\,\text{mA}}\right) \approx 4.58\,\text{ms}\) Therefore, it will take approximately 4.58 ms for the current in the circuit to drop from its initial value to 1.50 mA.