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in Moving Charges and Magnetism by (26.7k points)
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A boy connects a galvanometer directly to a cell of emf 1.5 v to measure a current through a load 1Ω. 

1. Which instrument can be used to measure the current in such a circuit? 

2. What changes should be made in the galvanometer to measure such a high current? Explain using a circuit diagram.

3. The boy connected the galvanometer into a high current measuring device and he connected the device parallel to the load. What will be the observation. Justify.

1 Answer

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Best answer

1. Ammeter

2. 

A galvanometer can be converted into an ammeter by a low resistance (shunt) connected parallel to it. 

Theory:

Let G be the resistance of the galvanometer, giving full deflection fora current Ig . To convert it into an ammeter, a suitable shunt resistance ‘S’ is connected in parallel. In this arrangement, Ig current flows through Galvanometer and remaining (I-Ig) current flows through shunt resistance.

Since G and S are parallel

P.d Across G = p.d across S

Ig x G =(I-Ig)S

S = \(\frac{I_g\,G}{(I-I_g)}\)

Connecting this shunt resistance across galvanometer we can convert a galvanometer into ammeter.

3. Ammeter is a low resistance device. Hence it draws high current. This high current will damage it.

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