Ammeter Assignment Help

Electromagnetism - Ammeter


An ammeter is a low resistance galvanometer. It is used to measure the current in a circuit in amperes. A galvanometer can be converted into an ammeter by sing a low resistance wire in parallel with the galvanometer. The resistance of this wire (called the shunt wire) depends upon the range of the ammeter and can be calculated as follows.

Let G = resistance of galvanometer,

N = number of scale divisions in the galvanometer,

K = figure of merit or current for one scale deflection in the galvanometer.

Then current which produces full-scale deflection in the galvanometer I = nK

Let I be the maximum current to be measured by galvanometer.

To do so a shunt of resistance S is connected in parallel with the galvanometer so that out of the total current I a part I should pass through the galvanometer and the remaining part (I – I) flows through the shunt VA – VB = I G = (I – Ig) S

S = (Ig / I – Ig) G

Thus S can be calculated.

If this value of shunt resistance S is connected in parallel with galvanometer is works as an ammeter of the range O to I ampere. Not the same scale of the galvanometer which was recording the maximum current I before conversion into ammeter will record the maximum current I after conversion into ammeter. It means each division of the sale in ammeter will be showing higher* currnetthan that of galvanometer. 

The effective resistance Rp of ammeter (shunted galvanometer) will be 

1 / Rp = 1 / G + 1 / S = S + G / GS or Rp GS/ G + S

As the shunt resistance is small the combined resistance of the galvanometer and the shunt is very low and hence ammeter has a much lower resistance than galvanometer. An ideal ammeter has zero resistance. 

A galvanometer with a coil of resistance 12 OO shows a full scale deflection for a current of 25 mA. How it can be converted into an ammeter of range 7.5 A? What will be the resistance of ammeter formed?

A galvanometer can be converted into an ammeter by connecting a shunt resistance in parallel to the galvanometer given by

S = Ig G / I – I = (2.5 x 120-3) x 12 / 7.5 – 2.5 x 10 – 3 = 4 x 10-3 O

The effective resistance R of ammeter formed is 

R = GS / G + S = 12 x 4 x 10-3 / 12 +4 x 10-3 = 4 x 10-3 O. - Ammeter Assignment Help, Ammeter Homework Help, Ammeter Assignment Tutors, Ammeter Solutions, Ammeter Answers, Electromagnetism Assignment Tutors

Help with Assignments

Why Us ?

Online Instant Experts Tutors

~Experienced Tutors

~24x7 hrs Support

~Plagiarism Free

~Quality of Work

~Time on Delivery

~Privacy of Work