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Findiing internal resitance trouble!

by mr_coffee
Tags: findiing, internal, resitance, trouble
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mr_coffee
#1
Oct26-05, 12:34 PM
P: 1,629
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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Leong
#2
Oct26-05, 08:10 PM
P: 377
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
mr_coffee
#3
Oct27-05, 09:48 AM
P: 1,629
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.

Leong
#4
Oct28-05, 12:44 AM
P: 377
Findiing internal resitance trouble!

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 throught 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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