Question

A \(15.19-\mathrm{mF}\) capacitor is fully charged using a battery that supplies \(V_{\text {emf }}=131.1 \mathrm{~V}\). The battery is disconnected, and a \(616.5-\Omega\) resistor is connected across the capacitor. What current will be flowing through the resistor after 3.871 s?

Short answer

Answer: The current flowing through the resistor after 3.871 seconds is 0.1403 A.

Step-by-step solution

Identify the key parameters

We are given the capacitance (C), the resistance (R), the initial voltage (Vemf), and the time (t). Let's list the values: - Capacitance (C) = 15.19 mF - Resistance (R) = 616.5 Ω - Initial Voltage (Vemf) = 131.1 V - Time (t) = 3.871 s

Find the time constant 𝜏

The time constant (𝜏) of an RC circuit can be found using the formula: 𝜏 = RC. Plugging in the given resistance and capacitance values, we can find 𝜏. 𝜏 = RC = (616.5 Ω)(15.19 * 10^{-3} F) = 9.368895 s

Find the voltage across the capacitor at time t

The voltage across the capacitor at any time t can be found using the following formula: V(t) = Vemf * exp(-t/𝜏). Plugging in the given initial voltage, time and the calculated time constant, we can find V(t). V(t) = 131.1 V * exp(-3.871 s / 9.368895 s) = 131.1 V * exp(-0.413) = 86.531 V

Find the current through the resistor at time t

Now that we know the voltage V(t) across the capacitor and the given resistor, we can use Ohm's law to find the current flowing through the resistor at the given time. Ohm's law states that I = V/R. I(t) = V(t) / R = 86.531 V / 616.5 Ω = 0.1403 A The current flowing through the resistor after 3.871 s is 0.1403 A.