Calculating Capacitor Charge and Energy: A Quick Guide

  • Thread starter CharlieBewn
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    Capacitance
In summary, a fully charged capacitor with 4600 μF and a 12 V power supply has an initial charge of 55.2 mC and an initial energy of 331.2 mJ. After 49 seconds of discharging through a 50 kΩ resistor, the residual charge is 44.6 mC, the potential difference is 9.7 V, and the energy content is 216.3 mJ. The values are rounded to 2 significant figures.
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CharlieBewn
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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
 
Physics news on Phys.org

1. What is capacitance?

Capacitance is the ability of a system to store an electric charge. It is measured in farads (F) and is determined by the physical dimensions and materials used in a capacitor.

2. How is capacitance calculated?

The capacitance of a system can be calculated by dividing the charge stored in the system by the potential difference across it. In equation form, it is expressed as C = Q/V, where C is capacitance, Q is charge, and V is voltage.

3. What factors affect capacitance?

Capacitance is affected by the distance between conductors, the surface area of the conductors, and the type of material used as the dielectric between the conductors.

4. How does capacitance impact electronic circuits?

Capacitance plays a crucial role in electronic circuits as it can store and release electrical energy on demand, making it a key component in power supply circuits, timing circuits, and filters.

5. How can capacitance be increased?

Capacitance can be increased by using materials with higher dielectric constants, increasing the surface area of the conductors, and decreasing the distance between the conductors. Additionally, connecting capacitors in parallel can also increase the overall capacitance.

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