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Calculate current in shunt resistor

Ok using the following:

Ish/Im = Rm/Rsh

Rsh = shunt resistance
Rm = meter resistance
Ish = shunt current
Im = meter current

A moving coil ammeter has a coil resistance of 28.5 ohms and a shunt resistor of 1.6 ohms. When the 0-740 mA display reads it maximum, how much current flows through the shunt resistor?

I need help on this. I have tried with calculating the voltage and power ... help :confused:
 
This is not as much a mathematics problem as it is science problem. This is exactly why you should not be in trapped by memorizing formulas. You should be deriving all your formulas from definitions and scientific facts. If you cannot derive a formula you shouldn't be using it. Without looking at a formula tell me what happens when you apply a current to two resistances in parallel.
 
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Hello


This is a math problem and it needed to be rearranged. I did it!:D

I(sh) = (( Rm / Rs + Rm)). IT

IT= Total current
 
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Sorry, I think you're missing the point. If all is you want is the answer you can do it using a formula. There are virtually an infinite number of combinations of parts and many types of parts involving inductors, capacitors, transistors, chemical batteries, and so forth, not just resistors. So are you going to memorize different formulas for all these different situations? Memorizing formulas is backwards. You must first understand the physics or the science and then wrap formulas around them. And if you do not understand how a resistor works get a resistor, a battery, and a multimeter and experiment with them. I design circuits for a living. I only rely on what I know about parts and data sheets. I see formulas on data sheets and I do not use them. I wrap my own numbers around the circuits based on what i know. If i relied only on formulas i never would be able to do my job. When I looked at your problem I realized instantly that you lacked understanding of the parts. So I say again that if you don't understand formula and you cannot derive if based on scientific fasts you should not be using it.
 
Edited because I was being daft.

It's not obvious if the meter is 740mA with the shunt in place or if that's the 'standalone' reading without a shunt.
 
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