Potential Difference ( Battery + Motor )

[Siune]

Homework Statement

Hey, I know this is exteremely basic problem and the answer is "easy", but I can't
just get the exact idea why it's so.

"The starting motor of particular car has a resistance of 0,15 Ω and is powered by 12 V battery through a cable which has a resistance of 30 mΩ .

a) What current flows in the starting motor when the voltage is applied?
b) The owner of the vechile finds that it fails to start even after he has replaced the battery. He measures the potential difference across the motor when the current to it is switched on and he finds that this is 11.4 V. Should he replace the motor or check the cable connections?

Homework Equations

V = RI

The Attempt at a Solution

a) I just calculated the full resistance and then got current I = 66,7 A.

b) Now here I'm not sure anymore. I know potential difference across any component is V = RI. And I know that if the cables didn't have resistance the potential difference across the motor should be equal to the 12 V.

I think the potential difference as "how much" the components between the two points use the "energy" that battery or emf provides.

Now the motor uses 11.4 V but how I know how much it should be using if it was working properly? I might have lack of understanding about the potential difference also.

Right answer is that he should replace the motor.

 

[rude man]

If the motor resistance was OK and the cable resistance was OK, what voltage should he have measured across the motor?

 

[Siune]

Okey, so I understand your question so that resistance is ok means that it has the resistance it's given ( for example in this case for motor 0,15 Ω ).

So voltage, potential difference across the motor should be with the current I = 66,7 A.

V = RI = 0,15 Ω * 66,7 A = 10,0 V

and in the cable

V = RI = 0,030 Ω * 66,7 A = 2,0 V

So the motor takes more of the power of the battery as it should? Because it should take 10,0V but now it uses 11,4 V?

 

[rude man]

Okey, so I understand your question so that resistance is ok means that it has the resistance it's given ( for example in this case for motor 0,15 Ω ).

So voltage, potential difference across the motor should be with the current I = 66,7 A.

V = RI = 0,15 Ω * 66,7 A = 10,0 V

and in the cable

V = RI = 0,030 Ω * 66,7 A = 2,0 V

So the motor takes more of the power of the battery as it should? Because it should take 10,0V but now it uses 11,4 V?

There are two ways the motor voltage could have increased.

One way is for the cable resistance to change.
The other is for the motor resistance to change.

In order for the motor voltage to go up, which of the two choices is possible and in which direction?

 

[Siune]

For motor voltage ( Potential difference ) to increase:

1) Motor resistance can go up ( V = R(motor) * I ), so if R increase -> V increase.

but couldn't equally

2) cable resistance to go down?

P.S Deeply sorry for my bad understanding about this subject or more especially how to use the potential difference.
And thanks for your patience rude man. :)

 

[rude man]

For motor voltage ( Potential difference ) to increase:

1) Motor resistance can go up ( V = R(motor) * I ), so if R increase -> V increase.

but couldn't equally

2) cable resistance to go down?

P.S Deeply sorry for my bad understanding about this subject or more especially how to use the potential difference.
And thanks for your patience rude man. :)

No, cable resistance can't go down so you may assume the motor resistance went up.

So, what would that mean?

 

[Siune]

Resistance is the reason why conductor/components produces heat/power by equation:

P = R I^2, so atleast power/heat the starting motor procudes, goes up.

But I don't see that as problem, so hmm..

Current decrease when resistance goes up as the current in circuit is E / R ( of all components ) .

As the emf of the battery is constant of 12V, so current must go down. And doesn't starting motors require a lot of current through them?

 

[rude man]

Current decrease when resistance goes up as the current in circuit is E / R ( of all components ) .

As the emf of the battery is constant of 12V, so current must go down. And doesn't starting motors require a lot of current through them?

You have got the right idea now!

 

[Siune]

Yeah, I see it now. :)

Thank you so much for your patience!