Findiing internal resitance trouble

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Homework Help Overview

The discussion revolves around a problem involving the internal resistance of a battery in a circuit with car lights and a starting motor. Participants are exploring the relationship between current, voltage, and resistance in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the emf of the battery using potential differences and current readings, but encounters confusion regarding the application of voltage values. Others suggest that the potential difference should be interpreted in a specific way, leading to different conclusions about the emf.

Discussion Status

Participants are actively engaging with the problem, offering various interpretations and calculations related to the emf and current through the starting motor. Some guidance has been provided regarding the relationship between the components in the circuit, but multiple interpretations are still being explored.

Contextual Notes

There is mention of specific values for internal resistance and current readings, as well as the implications of these values on the circuit's behavior. The discussion reflects uncertainty about how to apply these values correctly in the context of the problem.

mr_coffee
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Hello everyone, I keep missing his problem! and i only have 2 more trys...
When the lights of a car are switched on, an ammeter in series with them reads 14.0 A and a voltmeter connected across them reads 12.0 V (Figure 27-48). When the electric starting motor is turned on, the ammeter reading drops to 12.0 A and the lights dim somewhat.
Here is the image: http://www.webassign.net/hrw/hrw7_27-48.gif


Fig. 27-48

(a) If the internal resistance of the battery is 0.050 and that of the ammeter is negligible, what is the emf of the battery?
wrong check mark V
(b) What is the current through the starting motor when the lights are on?
A

I just thought i could find the EMF of the battery by finding the Potential difference between 2 points i made, a and b, i put a point a right before the resistor and right after the positve pole of the batery i put b. So here is what my equation looks like:
Va - RI + E = Vb;
now they say somthing about V being 12 V, does that mean
Va-Vb = 12?
I tried
E = RI - 12 = -11.6 which was wrong then i tried to just say screw the Va-Vb and say
E = rI
E = .7 also wrong! ahh!
 
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Vb-Va=12 because the current flows from b to a so the potential at b is higher than a. so emf =12.7 V
 
Awesome thank you, that wokred out great. Now if i wanted to find What is the current through the starting motor when the lights are on? I just drew a circuit and added up the following: EMF - rI = 0; I = 12.7/r; But this makes no sense because they give me the fact that, "When the electric starting motor is turned on, the ammeter reading drops to 12.0 A and the lights dim somewhat." So I'm confused on what they are hinting at.
 
from the previous part, find the resistance of the light,
r_light = v_light/i_light.
continue with the second part:
12 A is the current flowing through the light, you know its resistance from above. find the voltage across it.
now, this voltage must be the same as the voltage across the battery terminals since the are parallel.
so, emf - 0.050i = the voltage across the battery.(use the above value). i the current flowing through the battery.
the sum of the current flowing through the light and the starting motor must equal to this value. find the answer then.
 

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