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Internal Resistance of a Battery problem

  • #1
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Homework Statement



A voltmeter whose resistance is 1000 ohms measures the voltage of a worn out 1.5V battery as .9V. What is the internal Resistance

Homework Equations





The Attempt at a Solution



In order to do this, I Solved for the Equiv Resistance of the circuit (leaving in the unknown Ri) by assuming they were two resistors in parallel. as:

Requiv=(1000+Ri)/(1000Ri)

Using ohms law I calculated the current through the circuit:

1.5=IRequiv

Plugging in my answer for Ri and simplifying I got:

I=(1500+1.5Ri)/1000Ri

Now assuming the resistors are in parallel, I would know that the current through each resistor is the same, meaning that I would get:

V=IRi

assuming the voltage drop across Ri as .6V (because 1.5-.9 is .6) I plugged that in and got:

.6=(1500+1.5Ri)/1000

However, when I solved for Ri, I got a negative number!

Help would be great!!

Next, thinking that they were in parallel, the current through both resistors was the same, so
 

Answers and Replies

  • #2
gneill
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Resistors in *series* share the same current. Resistors in parallel share the same voltage (across them).

If the internal resistance of a battery was in parallel with the cell, the battery charge wouldn't last very long -- it would drain away through the internal resistance, creating a lot of heat in the process. Battery internal resistance is modeled as being in series with the cell.
 
  • #3
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yeah I realized my series/parallel current mixup as soon as I posted.

I know that the internal resistance is in series with the battery, however, I am saying that the voltmeter is in parallel with the battery (or maybe I'm wrong) because a voltmeter is supposed to be in parallel with the resistor (or so I thought)
 
  • #4
gneill
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20,793
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Draw a circuit diagram. The battery is an ideal cell in series with its internal resistance (draw a box around them - that's the model for the "real" battery). Now connect your voltmeter, represented as a 1000 Ohm resistor, across the battery.
 
  • #5
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Alright so I think I got this:

If i consider the equiv Resistance as Requiv=(1000+Ri), I could then find the current by saying:

1.5=I(1000+Ri)
and furthermore I=1.5/(1000+Ri)

Using the fact that current is constant through SERIES :) I can say that the voltage drop (which should be .6V right?) is:

.6=(1.5/(Ri+1000))Ri
Cross multiplying I get
.6Ri+600=1.5Ri
and thus Ri=666.67 ohms

I am not sure whether I am supposed to use .6 or .9 V for the V though...
 
  • #6
gneill
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You did good.

You could also have determined the current directly knowing that the voltage drop across the voltmeter was 0.9V and that its resistance is 1000 Ohms.

You also know that the voltage drop across the unknown internal resistance must 1.5V - 0.9V (that is, what's left of the voltage from the battery to be dropped around the circuit).

Given the current and voltage drop, you can determine the value of the unknown resistance.
 
  • #7
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Thanks so much for your help :)
 

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