# Galvanometer has a coil with a resistance

part 1: A galvanometer has a coil with a resistance of 81 W. A current of 131 mA causes full scale deflection. Calculate the value of the series resistor that is required in order to use the galvanometer as a Voltmeter with 40 V full scale.

so to make a voltmeter, the galvanometer and resistor have to be in series. so for the equivilant resistor i would have R+Rg=Reg

so by kirchoff's rules (loop rule) i conclude that the V=0 for the entire loop

Vo-ReqI=0
Vo=(R+Rg)I
40=(R+81)1.31

part 2: What shunt resistor is necessary to convert the galvanometer into an Ammeter which deflects full scale for 1.1 A?

so for an ammeter, the resistor and galvanometer are in parallel
Req=(1/R+1/Rg)^-1

and i am going to use the loop rule again.

Vo=RegI
Vo=I(1/R+1/Req)^-1
40=1.1(1/R+1/81)^-1

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berkeman
Mentor
Two things on part 1. First, when you cut and paste in the text that has character set formatting, it changed the 81 Ohms in the first sentence to 81 W ("W" is the Ohms symbol in the Symbol character set for Windows). If you want to keep it Ohms, use latex like this:

$$81 \Omega$$

You can press the Quote button on this post to see the format of the latex for the line above.

Second, 131mA = 0.131 A. You used 1.31 in your 3rd equation for some reason, instead of 0.131.

Does that get you going again?

ok...i got the first one...but i am still confused on the second part

where am i going wrong?

berkeman
Mentor
On part 2, when 131mA is going through the galvanometer's resistance, how much voltage is that across the galvanometer? And to make the full 1.1A, that's 131mA plus how much extra current through the shunt resistor? And that much current at the previously calculated voltage implies that the shunt resistor is what value of resistance?

EDIT -- fixed a typo of mine.

ok so the voltage drop across the galve. is = to the voltage drop across Rs

and Vg=IR=.131*81=10.61

so Vr=10.61=RI and that is 10.61=R(1.1-.131)=10.95 ohms

berkeman
Mentor
Looks good to me. Good job!