(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An ammeter whose internal resistance is 63 ohm reads 5.25 mA when connected in a circuit comtaining a battery and two resistors in series whose values are 750 ohm and 480 ohm. What is the actual current when the ammeter is absent?

2. Relevant equations

V = IR

R_{series}= R_{1}+ R_{2}

3. The attempt at a solution

First I found the voltage using the internal resistance of the ammeter and the current:

V = IR

V = 5.25 x 10^{-3}(63) = 0.33 V

Then I added the two resistances in series to find the total resistance

R = 750 + 480 = 1230 ohm

Then I tried to apply V = IR with the new resistance to find the current

0.33 V = I (1230 ohm)

And get a value of 2.68 x 10^{-4}A, which is not correct (correct answer: 5.52 x 10^{-3})

I'm not very good at these problems, so I have a feeling I'm doing something silly wrong, thanks for the input!

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# Current without ammeter

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