Calculating Capacitor Charge and Energy: A Quick Guide

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SUMMARY

The discussion focuses on calculating the charge and energy of a capacitor with a capacitance of 4600 μF charged to 12 V and discharging through a 50 kΩ resistor. The maximum charge (Q max) is calculated as 55.2 mC, and the maximum energy (E max) stored is 331.2 mJ. After 49 seconds, the residual charge is determined to be 44.6 mC, the voltage across the capacitor is 9.697 V, and the energy content is 216.3 mJ. The calculations utilize the formulas Q = CV, E = 1/2 CV², and the time constant τ = RC.

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Homework Statement


A capacitor with C = 4600 μ F is fully charged using a 12 V power supply.
At time t = 0, you start discharging the capacitor via a 50 k Ω resistor.
Calculate the initial charge Q max (in mC) and the initial energy E max (in mJ) stored in the capacitor.

Then, calculate the residual charge Q, the potential difference V across the capacitor and its energy E content after 49 seconds.
Q=mC
V=V
E=mJ

Have to answer in decimal numbers to 2 sig fig.

Homework Equations





The Attempt at a Solution


Here’s my attempt:
Q max = CV = 4600 uF * 12 V = 55200 u coulomb = 55.2 mC
E max = 1/2 CV^2 = 1/2 4600 uF (12 V)^2 = 331.2 mJ

tau = RC = 50e3*4600e-6 = 230 S.

Q = Q max exp(-t/tau) = 55.2 mC * exp(-49/230) = 55.2 mC * 0.808121 = 44.6 mC.
V = V power supply * 0.808121 = 12 * 0.808121 = 9.697 V
E = 1/2 CV^2 = 1/2 4600 uF (9.697 V)^2 = 216.3 mJ


Is this looking correct?
The only thing i can't do (stupid i know) is convert these to 2 sig fig whilst also in decimal form. For example V = 9.697 is 9.7
Is it 55.2 = 50.2
331.2 = 330.0
44.6 = 45.0
216.3 = 210.0
 
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