Multirange Ammeter and magnitudes of the resistances

[ovoleg]

Can someone please help me out with this problem? I am not sure if what I am doing is right

The question:

The resistance of the moving coil of the galvanometer G in Fig. 26.63 is 43.0 ohms, and the galvanometer deflects full scale with a current of 0.0295 A. When the meter is connected to the circuit being measured, one connection is made to the post marked + and the other to the post marked with the desired current range. Find the magnitudes of the resistances R1, R2, and R3 that are required to convert the galvanometer to a multirange ammeter deflecting full scale with currents of 10.0 A, 1.00 A, and 0.100 A.

The picture:

My work:
I assume we are looking for the shunt resistor in each case?
Rsh=(Ifs)*(Rc)/(Ia-Ifs)

For the 10A reading, Rsh=(.0295A)(43ohms)/(10A-.0295A)=.127225ohms

For the 1A reading, Rsh+.127225=(.0295A)(43ohms)/(1A-.0295A)=1.17983ohms

For the .1A reading, Rsh+.1.17983=(.0295A)(43ohms)/(.1A-.0295A)=16.8131ohmsCan someone please help me out :)

 

[andrevdh]

You are going to have three equations each with R1,R2 and R3 in. The way it can be approached is by adding the potential differences over the branches. For the 10A circuit the potential over R1 will be equal to the sum of the potentials over the other branch
$$9.97R_1=0.03(43+R_2+R_3)$$

 

[ovoleg]

You are going to have three equations each with R1,R2 and R3 in. The way it can be approached is by adding the potential differences over the branches. For the 10A circuit the potential over R1 will be equal to the sum of the potentials over the other branch
$$9.97R_1=0.03(43+R_2+R_3)$$

Thanks I figured it out now :)

You will get 3 equations 337.983R1=R2+R3+43
32.89R1+32.89R2=43+R3
2.389R1+2.389R2+2.389R3=43

And solving for all 3 I got R1=.179948
R2=1.61971
R3=16.1907

Yay, I feel all fussy inside now. Thanks for the help!:!)