Final answer:
The question asks for the time needed for the current to drop to 1.52 mA in an RC circuit consisting of two capacitors and two resistors in series with a voltage source. The procedure involves finding the initial current, calculating the time constant, and then determining the time for the current to decay according to the exponential decay formula for RC circuits.
Step-by-step explanation:
The given physics question involves an RC (resistor-capacitor) circuit and requires determining the time it takes for the current to decrease to a specific value from its initial state in a series configuration. In such circuitry, the time constant τ (tau) is fundamental, which is defined as τ = R·C, where R is the total resistance and C is the total capacitance of the circuit. Upon connecting two capacitors in series, the equivalent capacitance (Ceq) is given by 1/Ceq = 1/C1 + 1/C2, and the total resistance is simply the sum of the two resistors since they are in series.
Initially, when the switch is closed, the current (I0) can be found using Ohm's Law, I0 = V/R, where V is the source voltage. Over time, the current in an RC circuit decays exponentially according to I(t) = I0 · e^(-t/τ). To find the time t when the current has dropped to 1.52 mA, we solve this equation with the known values of initial current, τ, and the target current value.
Example calculation based on the provided example 68: For a 500-ohm resistor and a 1.50-μF capacitor with a 6.16-V source, the initial current I0 would be 6.16 V / 500 Ω = 12.32 mA. The RC time constant τ = RC = 500 Ω · 1.50 µF. Current after one time constant I(τ) = I0 · e^(-1) and voltage on the capacitor V(τ) = V(1 - e^(-1)).