A 0.1 shunt is connected to a 100-uA, 600-ohm meter movement. What is the range of the ammeter?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

When connecting a shunt in parallel to the internal resistance of an ammeter the maximum range of current intensity that can be indicated is increased. Because the internal resistance is in parallel with shunt resistance the voltage drop is the same on both (U).

with the shunt connected we have

`I_("total") = I_("shunt") + I_("meter")`

The ratio of meter current to shunt current is

`n = I_(meter)/I_(shunt) = (U/R_(meter))*(R_(shunt)/U) = R_(shunt)/R_(meter) =0.1/600 =1.667*10^-4`

which means

`I_(shunt) =I_(meter)/n = (100*10^-6)/(1.667*10^-4) =0.6 A`

This means the total scale of the ammeter with shunt is

`I_("total") = 0.6 +0.0001 = 0.6001 A ~~0.6 A`

Approved by eNotes Editorial Team

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial