Help with relating Current and Charge

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SUMMARY

The discussion centers on two key physics problems involving electric charge and potential difference. The first question confirms that a car battery rated at 90 A hr can supply a total charge of 324,000 coulombs (C) when fully discharged, calculated using the formula Q = It, where I is 90 A and t is 3600 seconds. The second question explores the potential difference across a 20-meter long, 2.0 mm diameter copper wire carrying an 8.0 A current, utilizing the relationship V = (I*p*L)/A, where the resistivity (p) of copper is needed for the calculation. Participants validate the approach and provide insights on the necessary parameters.

PREREQUISITES
  • Understanding of electric charge and current (Q = It)
  • Familiarity with Ohm's Law and resistivity concepts
  • Knowledge of basic electrical properties of materials, specifically copper
  • Ability to perform unit conversions and calculations involving length and area
NEXT STEPS
  • Research the resistivity of copper and how it varies with temperature
  • Learn about the relationship between current, voltage, and resistance in circuits
  • Explore practical applications of electric charge calculations in automotive batteries
  • Study the effects of wire diameter on resistance and potential difference
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Students and professionals in physics, electrical engineering, and automotive technology who are looking to deepen their understanding of electric charge, current flow, and circuit analysis.

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1. First Question

A car battery is rated at 90 A hr, meaning that it can supply a current of 90 A current for 1 hr before being completely discharged. If you leave your headlights on until the battery is completely dead, how much charge leaves the battery?

Can charge be related to the current as well? Since current is the flow of charge, can it just be said that Q = It, where t is 3600 seconds, and I is 90 A?

so then Q = (90)(3600) = 3.24 * 10^5 C

Is this correct?

Second Question

What is the potential difference across a 20-m long, 2.0 mm diameter copper wire carrying an 8.0A current?

We know that E = V/L, and that I = (A/p) * E, so then E = (I*p)/A

then we can relate them as V = (I*p*L)/A

We have I, L and A, but how would I get p from the information given?

Thanks for the help guys!
 
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RafaFutbol said:
1. First Question

A car battery is rated at 90 A hr, meaning that it can supply a current of 90 A current for 1 hr before being completely discharged. If you leave your headlights on until the battery is completely dead, how much charge leaves the battery?

Can charge be related to the current as well? Since current is the flow of charge, can it just be said that Q = It, where t is 3600 seconds, and I is 90 A?

so then Q = (90)(3600) = 3.24 * 10^5 C

Is this correct?

Second Question

What is the potential difference across a 20-m long, 2.0 mm diameter copper wire carrying an 8.0A current?

We know that E = V/L, and that I = (A/p) * E, so then E = (I*p)/A

then we can relate them as V = (I*p*L)/A

We have I, L and A, but how would I get p from the information given?

Thanks for the help guys!

Looks ok to me.
 

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