Explain how a galvanometer can be converted into an ammeter of a given range. Derive an expression for shunt resistance and current for full scale deflection. Find the effective resistance of ammeter.

 

 
 
 
 
 

Answers (1)
S safeer

The galvanometer is used as an ammeter by connecting the low resistance wire in parallel with the galvanometer.

Thus, the potential difference between the voltage and the shunt resistance are equal.

Such as, 

The galvanometer and the shunt resistance are connected in parallel with the circuit, hence their potential are equal,

V_{g}=V_{s}

V_{g} = potential across galvanometer

V_{s} = potential across the shunt

Such that we know that,

V_{g}=GI_{g}and 

V_{s}=S(I-I_{g})

on placing the values, we have,

GI_{g} = S(I-I_{g})

where,

I is the total current coming to the circuit

I-I_g = current across a shunt

I_{g} = galvanometer current

G = Galvanometer resistance

S = Shunt resistance

Thus, shunt resistance is given as:

S =\frac{GI_{g}}{(I-I_{g})}

The effective resistance of ammeter is given as:

R_{A}=\frac{GS}{G+S}

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