Its said that for an ammeter to give good reading, the full current in the circuit must pass through it. But if I am right, the ammeter is basically a galvanometer connected parallel to a very low resistance called a shunt. I am aware that connecting a low resistance in parallel will reduce effective resistance to a value lesser than the least resistance.

But in an ammeter, if the shunt is a low resistance (lesser than galvanometer's resistance), then most of the current would pass through the shunt than the galvanometer. Thus, the reading given by galvanometer would decrease (as its the component which gives deflection in an ammeter), which means that the reading of ammeter would decrease.

Is my interpretation correct? If its wrong please explain me where I have gone wrong.

Also, how will range and sensitivity of a an ammeter change if we increase or decrease shunt resistance?

  • $\begingroup$ Would Electrical Engineering be a better home for this question? $\endgroup$
    – Qmechanic
    Feb 25, 2016 at 16:34
  • $\begingroup$ @Qmechanic aah i had this topic in my physics text book.also since EMI comes in physics i thought physics would be better. $\endgroup$ Feb 25, 2016 at 16:50

2 Answers 2


In a practical ammeter there will be a number of fixed shunt resistances, selected by a switch. The galvanometer is acting as a high-resistance voltmeter, measuring the voltage across the shunt, and has little effect on the current through the circuit was a whole.

If we know the value of the shunt resistor, then the voltage read by the galvanometer is proportional to the current passing. I = V/R.

Changing the shunt resistor affects the reading greatly. It's up to the user to select the correct shunt for the current being measured.

  • $\begingroup$ So what you are saying is that an ammeter works like a voltmeter internally right?.Which means internally an ammeter measures the potential drop across a known resistance and is calibrated according to this resistance (shunt) . $\endgroup$ Feb 25, 2016 at 17:43
  • $\begingroup$ your answer is a bit concise and tricky to follow.(But its better than to read a whole lot of para with all sorts uncommon and hard english ). $\endgroup$ Feb 25, 2016 at 18:05
  • $\begingroup$ @AbhishekPallipparagopakumar Yes. The idea is just to put a known low-value resistor in series with the rest of the circuit. Then use a voltmeter to measure the voltage across it. That's all there is to it. $\endgroup$
    – Simon B
    Feb 25, 2016 at 22:06
  • $\begingroup$ would you please explain how the range/sensitivity gets affected on increasing or decreasing shunt resistance $\endgroup$ Feb 26, 2016 at 1:13
  • $\begingroup$ @AbhishekPallipparagopakumar It's just down to I = V/R again. If I put a 1 ohm resistor in series with the load, and the meter reads 1V, then 1A must be flowing. If I replaced it with a 0.1 ohm shunt, the voltmeter would now read 0.1V for 1A instead. Or to put it another way, if the meter reads 1V with a 0.1 ohm shunt, then the current must now be 10A. $\endgroup$
    – Simon B
    Feb 27, 2016 at 21:03

An Ammeter which is a current measuring device has some limitations. If the instrument is designed as a moving coil galvanometer -it can measure small currents and the maximum limit may be about 30 micro amperes. Therefore one uses a shunt resistor to share larger part of the current and limited(within range ) current is allowed to pass through the ammeter.

one can select a range of shunt resistance for utilizing them in measuring current of different ranges. the ammeter can be calibrated along with different shunts and the same instrument can have switching device for different ranges.

Regarding sensitivity of an ammeter usually its defined as amount of current required for a full scale deflection- and it varies from 10 micro-Ampere to 30 micro-Ampere .

  • $\begingroup$ original contents have been added and those references removed as objected to ; complied with the suggestions,thanks inconvenience is regretted-will learn slowly! $\endgroup$
    – drvrm
    Mar 4, 2016 at 4:40
  • $\begingroup$ I've deleted comments posted prior to the latest edit. $\endgroup$
    – David Z
    Mar 6, 2016 at 11:34

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