**Problem 5** A 6.00 µF capacitor with an initial charge of 1200 µC is discharged through a 4.0 MΩ resistor. a) **Calculate the magnitude of the current in the resistor 12.0 s after the resistor is connected across the terminals of the capacitor.** - Formula: \( V_C = \frac{V_0}{R}(1 - e^{-t/\tau}) \) - Substituting given values: \[ V_C = \frac{1200 \times 10^{-6}}{4 \times 10^6}(1 - e^{-12/24}) \] - Simplification: \[ V_C = 1.2 \times 10^{-10} \] b) **What is the charge that remains on the capacitor after 12.0 s?** - Formula: \( q = Q (1 - e^{-t/\tau}) \) - Substituting given values: \[ q = [1200 \times 10^{-6}](1 - e^{-12/24}) \] - Result: \[ q = 4.7 \times 10^{-4} \] Where: - \(\tau = (4)(6) \, \text{s}\) - \(\tau = 24 \, \text{seconds}\) c) **How much energy has the capacitor dissipated through the resistor in those 12 s?** - Hint: Energy is the integral of the power with respect to time. \[ \int_0^{12} Q (1 - e^{-t/\tau}) \] - Calculation: \[ \int_0^{12} (1200 \times 10^{-6})(1 - e^{-t/24}) \] - Solution: \[ 1200 \times 10^{-6} \ln(1 - t/24) \] The boxed result gives the dissipated energy: \[ Q = 8.31 \times 10^{-4} \]

Problem 5 A 6.00 µF capacitor with an initial charge of 1200 µC is discharged through a 4.0 MΩ resistor. a) Calculate the magnitude of the current in the resistor 12.0 s after the resistor is connected across the terminals of the capacitor. - Formula: \( V_C = \frac{V_0}{R}(1 - e^{-t/\tau}) \) - Substituting given values: \[ V_C = \frac{1200 \times 10^{-6}}{4 \times 10^6}(1 - e^{-12/24}) \] - Simplification: \[ V_C = 1.2 \times 10^{-10} \] b) What is the charge that remains on the capacitor after 12.0 s? - Formula: \( q = Q (1 - e^{-t/\tau}) \) - Substituting given values: \[ q = [1200 \times 10^{-6}](1 - e^{-12/24}) \] - Result: \[ q = 4.7 \times 10^{-4} \] Where: - \(\tau = (4)(6) \, \text{s}\) - \(\tau = 24 \, \text{seconds}\) c) How much energy has the capacitor dissipated through the resistor in those 12 s? - Hint: Energy is the integral of the power with respect to time. \[ \int_0^{12} Q (1 - e^{-t/\tau}) \] - Calculation: \[ \int_0^{12} (1200 \times 10^{-6})(1 - e^{-t/24}) \] - Solution: \[ 1200 \times 10^{-6} \ln(1 - t/24) \] The boxed result gives the dissipated energy: \[ Q = 8.31 \times 10^{-4} \]

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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