Question

*18-L waves.
1) Explain how a moving - coil galvanometer is converted into an ammeter.
Derive the necessary formula.
The short wavelength limit of the Lyman series is 911.3Aº. Compute
the short wavelength limit of the Balmar Series.​

Answer 1

Answer:

Explanation:

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=) A galvanometer as such cannot be used as an ammeter because it has a appreciable resistance and it can measure only a limited current corresponding to the maximum deflection on its scale . In practice , an ammeter is made by connecting a low resistance S in the parallel with the pivoted-type moving-coil galvanometer G ( In figure ) . S is known as 'shunt' . Its value depends upon range of the required ammeter and can be calculated as follows :-

Let G be the resistance of the coil of the Galvanometer and Ig the current , which on passing through the Galvanometer , produces full scale deflection . If I is the maximum current on the measured , then apart Ig of the current I should pass through the galvanometer G and the rest (I - Ig) through the shunt S . Since G and S are in parallel , the potential difference across them will be the same :

Ig × G = ( I - Ig ) × S -----------------(1)

Ig / I = S / S + G ,

that is , only S / S + G th part of the total current will flow in the coil of the galvanometer . Again from eq. (1) , we have

S = ( Ig / I - Ig ) G -------------------(2)

If the current Ig in the coil produces a full-scale deflection , then the current I in the circuit corresponds to the full-scale deflection . Thus , with a shunt S of the above value , the galvanometer will be an ammeter of range 0 to I ampere (A) .

Suppose a current of 1.0 A in the coil of galvanometer produces a full-scale deflection . To convert it into ammeter of range 10A , a shunt is required such that when the current in the circuit is 10A , only 1.0 A flows in the coil . Substituting Ig = 1.0 A and I = 10 A in the above equation , we have

S / G = 1.0 A / 10 A - 1.0 A

S / G = 1 / 9 .

The resistance of the shunt should be only 1/9 th the resistance of the galvanometer-coil . As the shunt resistance is small , the combined resistance of the galvanometer and the shunt is very low and hence the ammeter has a much lower resistance than the galvanometer .

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