What is the effect of a shunt on the sensitivity of a moving coil galvanometer?

[palkia]

Homework Statement

If a shunt of 1/10th of the coil resistance is applied to a moving coil galvanometer,then it's sensitivity becomes

(a)10 fold
(b)11 fold
(c)1/10 fold
(d)1/11 fold

Homework Equations

ϕ / I = NAB/L

The Attempt at a Solution

[/B]
So sensitivity gets increased by only these factors
(A) Increasing number of turns of coil
(B) Increasing the magnetic field

I don't know which one is happening here. Usually shunt and galvanometer are used to convert it into ammeter but I don't find it's significance here

 

[gleem]

First forget the equation and consider what happens when you place a shunt across the galvanometer.

 

[palkia]

It starts working as a ammeter

 

[gleem]

Yes but what is happening to the current through the galvanometer and consequently the affect on the deflection of the needle.

 

[palkia]

The current through the galvanometer increases and it's value is 11ig(full scale current) tho idk why it gets increases ...?

The deflection will be more in the second case

 

[gleem]

Think again the shunt has a smaller resistance than the galvanometer's coil meaning that...?

 

[gleem]

Keep in mind that a galvanometer is just a very sensitive ammeter.

 

[palkia]

All the major current will flow through the shunt while the same current ig will flow through the galvanometer but my textbook says it will show a deflection of I in this case so I was just confused here.I see now that the sensitivity will get changed due to change in current

So here since the current will become 11 times the previous case therefore sensitivity becomes 1/11 times the original

 

[gleem]

Your on the right track but can you show your reasoning . That is if you wanted to decrease the sensitivity to 1/5 what is the value of the shunt you wold use?

 

[palkia]

1/4 times the galvanometer resistance in my opinion

 

[gleem]

In physics opinion do not carry much weight but you are correct. Please show your work so that I know that you understand what is going on.

 

[palkia]

Okay.

The current in the moving coil ammeter is given by

S=Ig G/I-Ig

Using this ,I calculated the current through the galvanometer which came out to be 11 ig when I plugged S=G/10

 

[gleem]

Very good. I am satisfied. In solving problems it is sometime best first to think about what is happening than automatically writing down equations.which may or may not be relevant as in this case. You were led astray by the term sensitivity from which you thought the equation for the sensitivity of the galvanometer which in this case is irrelevant.

 

[palkia]

Yea.I was actually aiming for that only but later I realized It wasn't going to help me.

Thanks for the help