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# Physics/magnetic effect of electric cuurent

Question
A galvanometer has resistance of 16ohm and gives a full scale deflection when a current of 20mA is passed through it.The only shunt resistance available is 0.04ohm which is not appropriate to convert galvanometer into an ammeter.How much resistance should be connected in series with coil of galvanometer so that range of ammeter is 10A?

Hello jay,

The problem write-up seems to have some problems. For one thing, a galvanometer is an ammeter.
http://en.wikipedia.org/wiki/Galvanometer
It's just that it is typically only for small currents unless external modifications have been done.

With the 0.04 Ohm shunt installed in  parallel with the coil, the shunt and the coil would each have the same voltage, V, across them. Let the currents in the coil and shunt be called Ic and Is. From Ohm's Law
V = 16 Ohms*Ic
V = 0.04 Ohms*Is
Setting the 2 expressions for V equal to each other
16 Ohms*Ic = 0.04 Ohms*Is
Is = (16/0.04)*Ic = 400*Ic
At full scale, the coil current must be 20 ma, so
Is = 400*20 ma = 8 Amps
So the current flowing into the network (galvanometer+shunt) is 8.02 A. The write-up says "The only shunt resistance available is 0.04ohm which is not appropriate to convert galvanometer into an ammeter." But that setup would be reasonable, if we wanted an ammeter where 8.02 A is full scale. The only thing is that it would not be an ammeter whose range is 10 A.

Seems like a different parallel shunt resistor would yield an ammeter whose full scale deflection comes with 10 A flowing into the galvanometer+shunt network. The write-up talks about adding a series resistor. Adding a series resistor to make a voltage divider is how a galvanometer is converted into a voltmeter. See the 3rd paragraph in the section titled Operation.
http://en.wikipedia.org/wiki/Galvanometer#Operation

I suggest a parallel shunt resistor, R, calculated this way:
Is + Ic = 10 A
Let Is = Ic*k
Ic*k + Ic = 10 A
Ic is 20 ma when meter is at full scale.
0.02 a*(k+1) = 10 A
k+1 = 10 A / 0.02 A = 500
k = 499
Again let V be the voltage across the parallel network.
V = Ic*16 Ohms
V = Is*R = 499*Ic*R
499*Ic*R = Ic*16 Ohms
R = 16 Ohms/499 = 0.032 Ohms
Note: check my math.

I hope this helps,
Steve

Physics

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