Electric circuitry, confusion about the electric charge

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Homework Help Overview

The discussion revolves around understanding electric charge in the context of a car's battery, specifically focusing on the voltage, internal resistance, and the implications of using different units of electric charge such as ampere-hours and coulombs. The original poster expresses confusion regarding the concepts and units involved in calculating the number of ignition attempts possible before the battery charge falls below a certain threshold.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between different units of electric charge, questioning the validity of using ampere-hours versus coulombs in calculations. There are discussions about the current draw of the car starter and the implications of the battery's specifications on the calculations. Some participants also reflect on the nature of the problem and the relevance of provided values like resistance and voltage.

Discussion Status

Participants have provided some guidance on unit conversions and the nature of electric charge, with some confirming the original poster's calculations. However, there remains a degree of confusion regarding the use of ampere-hours in the context of the problem, indicating that further clarification may be needed.

Contextual Notes

There is an ongoing discussion about the appropriateness of using different units of charge and the potential implications of the assumptions made in the problem statement. The original poster also notes a lack of clarity regarding the current draw of the starter motor and its comparison to household electrical systems.

late347
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Homework Statement


car's battery has voltage (E) 11,5 volts and internal resistance of 0,040 ohms.
battery's charge is 20 amperehours (20 Ah)

How many ignition attempts of the car can be made, when the charge can be allowed to fall to 15 amperehours.

One attempt of at engine ignition withdraws 150 amperes current from the battery during 6 seconds time.

Homework Equations


Q= I x t

U= R x I

The Attempt at a Solution



I'm pretty confused about the electric charge concept and the unit Coulomb.
Also, I'm not a big car guy, I'm afraid. I know how to drive them though!

I don't really see which unit is As (amperesecond) or Ah, or Coulomb.
Why are there so many units for electric charge is the thing I'm wondering.

Actually I was also wondering... Does the car starter engine really draw 150amps, from the battery? Regular house fuse, only holds up to about 15 amps, as I recall. Maybe it truly is so because the problem statement was that way.

I also wonder, what the current through the circuit would be under load... Perhaps there is not enough information given to ascertain what the current would be?

Also it seemed as though the starting values for resistance and voltage were basically redundant values as they were not used in the calculation process. At least the book answer at the end of the problems states. that it was enough to find out what amount of electric charge is reduced by one attempt at starting the car. Then find out how many attempts could be made, until the battery charge is reduced by the correct amount.Basically I know I need to calculate the charge, that occurs within one car start attempt. If that could be known, then simple division could be made, to see how many attempts it takes to reduce the electric charge until the threshold is reached (5 Ah)

I tried to convert the units at first to As , though it might not be necessary.

My book says that

I = Q/ t
where current is electric charge divided by, time taken for the charge to flow.
normally the unit of Q = 1 coulomb

I suppose that 1C= 1As
So one could surmise that
amperes= As/ seconds

I x t = Q

so one would end up with electric charge, as the product of current times the time.
I have some doubts mentally though. It seems to me that 150 amps is a lot of current to my mind. Could it possibly be that much? The problem statement clearly said though, that one attempt to start the engine takes 150amp current within time of 6 seconds.

I = Q/t ] both sides times t
I x t= Q

150A x 6s= 900 As

this much electric charge is taken in one attempt at engine starting it seems.

5 Ah= 18 000 As

18 000 As / 900 As = 20

20 attempts at ignition could be made such that the battery life falls down to that level.
 
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Your result is correct. As is really unit of charge and 1 C is equal to 1 As.
 
You seem to have arrived at the right answer.
Yes, a starter motor does take a very large current, maybe up to 1000 Amps. A household electricity supply uses higher voltage, which allows smaller currents to be used.
A Coulomb is an Amp-second, a small unit often used in electronics, and is much smaller than an Amp-Hour, which is used for the large amounts of charge found in heavy current engineering.
 
Is it allowed to cakculate this problem using ampere-hours instead of As ?Are you allowed to plug into the formula ampere hours, instead of the usual unit of charge the coulomb.

Q/t = I

Amperehours / hours = current (??now I am confused here??)

That seems like it would be that way.

When you calculate mathematically
(AxB) ÷ (B) = A

Both numerator and denominator are divisible by B.

Therefore that quotient would be A

It was very confusing to think about amperehours and amperes together initially, because coulombs are associated with ampereseconds.
 
The name of the unit As is Coulomb. You can use Ampere-hours to measure charge, but it is 3600 Coulomb. If you divide x Ah with the time in hours, you get current in amps.
Remember we use the unit kWh to measure how much electric energy we consumed. It is energy, and 1 kWh is equal to 3600000 joule.
 
late347 said:
Why are there so many units for electric charge is the thing I'm wondering.

Good question. A perfectly good unit exists for charge (coulomb), but when dealing with batteries, milliamp-hours or amp-hours are commonly used, which necessitates unnecessary conversion. I suppose amp-hour gives people an intuitive feel for how long and how much current a battery can supply.
 

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