Direct Current Circuits

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Homework Statement


10) A voltmeter whose resistance is 1000Ω measures the voltage of a worn-out 1.5V flashlight batter as 0.9V. What is the internal resistance?

11) If the flashlight battery of the preceding problem had been measured with a voltmeter with a resistance of 10MΩ what voltage would have been read?

[Questions From CHP 1 of Principles of Electronics Instrumentation by Diefenderfer and Holton 3rd edition]

Homework Equations


V=IR


The Attempt at a Solution



10)
1.5V-.9V=.6V
.9V/1000Ω= 9x10^-4A
R=.6V/9x10^-4A=666.67Ω as an internal resistance

*this answer seems to large to me for the given physical situation. Am I going in the right direction with my math or have I totally diverged from what the true answer should be?

11)
.9V/10MΩ= 9x10^-8A
V=(9x10^-8A)(666.67Ω)=6x10^-5V

*I believe that my answer in number 10 is incorrect and if it is...obviously my answer for number 11 would be affected. Any and all help is greatly appreciated! thank you to all in PF:)
 

Answers and Replies

  • #2
ehild
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Homework Statement


10) A voltmeter whose resistance is 1000Ω measures the voltage of a worn-out 1.5V flashlight batter as 0.9V. What is the internal resistance?

11) If the flashlight battery of the preceding problem had been measured with a voltmeter with a resistance of 10MΩ what voltage would have been read?

[Questions From CHP 1 of Principles of Electronics Instrumentation by Diefenderfer and Holton 3rd edition]

Homework Equations


V=IR


The Attempt at a Solution



10)
1.5V-.9V=.6V
.9V/1000Ω= 9x10^-4A
R=.6V/9x10^-4A=666.67Ω as an internal resistance


*this answer seems to large to me for the given physical situation. Am I going in the right direction with my math or have I totally diverged from what the true answer should be?
It is correct.

11)
.9V/10MΩ= 9x10^-8A
V=(9x10^-8A)(666.67Ω)=6x10^-5V

*I believe that my answer in number 10 is incorrect and if it is...obviously my answer for number 11 would be affected. Any and all help is greatly appreciated! thank you to all in PF:)
That is wrong.
The emf of the battery is 1.5 V, but the current flowing through the internal resistance Ri=667 Ω causes a voltage drop inside the battery. The terminal voltage, you measure between the terminals is equal to V=EMF-IRi. At the same time, it is equal to the resistance of the load Rv (it is the voltmeter now) multiplied by the current V=IRv.
When you connect the voltmeter to the battery, the total resistance in the loop is Ri+Rv. Calculate the current from this equation.

ehild
 
  • #3
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im not sure I completely follow....so are you saying to use V=I(Ri+Rv), solve for I and the plug that current value in the equation V=emf-IRi? What values of V do I use in the first and in the second equations?
 
  • #4
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I have tried the problem again and I still don't quite follow...
 
  • #5
ehild
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im not sure I completely follow....so are you saying to use V=I(Ri+Rv), ?
It is emf=(Ri+Rv)I, emf=1.5 V.

ehild
 
  • #6
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Thank you for your help ehild!
 
  • #7
ehild
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  • #8
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okay heres my math

emf=(Ri+Rv)I
1.5V=(666.67Ω+10MΩ)I
I=1.5x10^-7A
V=IRi
V=(1.5x10^-7A)(666.67Ω)
V=1x10^-4V is what is read on the voltmeter
 
  • #9
ehild
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okay heres my math

emf=(Ri+Rv)I
1.5V=(666.67Ω+10MΩ)I
I=1.5x10^-7A
V=IRi
V=(1.5x10^-7A)(666.67Ω)
V=1x10^-4V is what is read on the voltmeter
IRi is the potential drop across the internal resistance. You have to subtract it from the emf to get the terminal voltage, measured by the voltmeter. V=emf-IRi.

(A voltmeter measures the voltage across its own internal resistance, it is 10 MΩ now. So also V=IRv)

ehild
 
  • #10
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Okay :) got it!
 
  • #11
ehild
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Okay :) got it!
You see the voltmeter measures the true voltage of the cell if its resistance is much higher than the internal resistance of he cell. It is very essential to use high resistance voltmeters.

ehild
 

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