Direct Current Circuits and automobile battery

[blakmamba619]

1.An automobile battery has an emf of 12.6 V and an internal resistance of 0.0620 . The headlights together present equivalent resistance 6.00 (assumed constant). What is the potential difference across the headlight bulbs when they are the only load on the battery?
What is the potential difference across the headlight bulbs when the starter motor is operated, taking an additional 35.0 A from the battery?




2. V = IR


3. For the first part i just added up the resistance R+r = Req then dived the emf by the Req to get I = 2.07 A. Then i found the potential difference across the 56ohm resistor to be IR = 2.07A*6ohms = 12.42V. Which i am sure is correct. The second part is hard for me to understand the concept, what does it mean when it says taking an additional 35.0 A from this battery? ( I tried R = V/ R = 12.6/35A = 0.36 ohms then finding a new current but it doesn't work)

 

[Philosophaie]

Try subtracting 0.36ohms by 0.0620ohms and drawing a new circuit for startup conditions.

 

[blakmamba619]

naw, i tried that, i get too high of a potential across the 5.00 ohms. i got 12.003 v too high

 

[Redbelly98]

What is the voltage across the 0.0620 Ω resistor, given that the current is 35.0 A plus the unknown headlight current?

 

[blakmamba619]

that doesn't make sense, they want the voltage drop across the 6.00 ohm resistor. But i tried your method, and i get I = 37.07 (too high??) then V = IR = 37.07*6= 222.42 tooo high. Am i doing the calculations right? or is this method wrong?

 

[blakmamba619]

i just don't understand the wording of the problem, i am usually good at physics lol (i received an A in motion and mechanics) Can someone describe what the problem means when it states, it takes 35 A from the battery? is it possible to draw more current than what's provided? does

 

[Redbelly98]

that doesn't make sense, they want the voltage drop across the 6.00 ohm resistor. But i tried your method, and i get I = 37.07 (too high??)

Pretty close. We don't actually know the headlight current yet, but it should be close to the 2.07 A you got for the 1st part of the problem.

Using 37.07 A, I will repeat my question from before:

What is the voltage across the 0.0620 Ω resistor?

... then V = IR = 37.07*6= 222.42 tooo high. Am i doing the calculations right? or is this method wrong?

Wrong, because not all of that current goes through the headlight.

It might help if you draw a circuit diagram, including:

The battery, including the internal 0.0620 resistor.
The headlights, a 6 ohm resistor.

Then, how would you include the starter in the circuit, given that it must be connected to the battery?

 

[blakmamba619]

[Wrong, because not all of that current goes through the headlight.][/QUOTE]

current is the same in all resistors in series right? but your potential difference is not. So
V = I(R+r) is that correct?

 

[blakmamba619]

i got it, thx.