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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:)