P28_085 - 1 + R 2 + R A = R A r + R 1 + R 2 + R A = . 10 2...

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85. The current in the ammeter is given by i A = E / ( r + R 1 + R 2 + R A ). The current in R 1 and R 2 without the ammeter is i = E / ( r + R 1 + R 2 ). The percent error is then i i = i i A i =1 r + R 1 + R 2 r + R
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Unformatted text preview: 1 + R 2 + R A = R A r + R 1 + R 2 + R A = . 10 2 . 0 + 5 . 0 + 4 . 0 + 0 . 10 = 0 . 90% ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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