Find the resistance that must be placed in parallel with a 25.0Ω25.0 \Omega galvanometer having a 50.0μΩ50.0 \mu \Omega sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a 300-mA full-scale reading.
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Final Answer

4.17 mΩ4.17 \textrm{ m}\Omega

Solution video

OpenStax College Physics for AP® Courses, Chapter 21, Problem 47 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. When this ammeter is placed in this circuit we want full scale deflection of the galvanometer to occur when the current, the total current through the circuit is 300 milliamps. Now we're told that the galvanometer has a sensitivity of 50 microamps which means this current I subscript capital G which is the current through the galvanometer has to be 50 microamps when this I t for total is 300 milliamps. We're told that the internal resistance of the galvanometer is 25 ohms. So the voltage across this shunt resistor which is the name for this resistor here that we have to calculate what it should be, the voltage across that is going to be whatever current is leftover after some of it is distributed through the galvanometer. This current here is going to be the total minus whatever went up through the galvanometer multiplied by the shunt resistance capital R, and this voltage also is going to be the current through the galvanometer multiplied by the galvanometer's internal resistance because since the shunt resistor and the galvanometer are connected in parallel, the voltage across each would be the same. So we can equate the two. Now we solve this for the shunt resistance by dividing both sides by I t minus I G and we get that the shunt resistance must be 50 microamps multiplied by 25 ohms divided by 300 milliamps minus 50 microamps. That gives 4.17 milliohms which you could also write as 4.17 times ten to the minus three ohms.