Electric circuitry, confusion about the electric charge

In summary, the car's battery has a voltage of 11.5 volts and an internal resistance of 0.040 ohms. Its charge is 20 ampere-hours (20 Ah). One attempt at engine ignition draws 150 amperes of current from the battery for 6 seconds. Using the formula Q=I x t, it can be calculated that one attempt at starting the car withdraws 900 As of electric charge. If the threshold is 5 Ah, then approximately 20 attempts can be made before the charge falls to that level. This is due to the fact that 1 C is equal to 1 As, and the formula I=Q/t can be used to find the current. The starter motor
  • #1

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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  • #2
Your result is correct. As is really unit of charge and 1 C is equal to 1 As.
  • #3
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.
  • #4
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.
  • #5
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.
  • #6
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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