What is Shunt Resistance? – Definition, Formula, Derivation & Diagram

In this topic, you study the definition, theory, formula, derivation & diagram of ammeter shunt resistance.

The shunt resistance is essentially a low value resistance connected in parallel to the ammeter. It is used to extend the range of an ammeter.

Fig.1: Shunt Resistance.

Let there be an instrument (Galvanometer) as shown in Fig. 2 connected in series with a load and supply.

Fig. 2: Shunt Resistance circuit diagram

R_g = Resistance of the galvanometer in Ohm,

R_S = Resistance of the shunt in Ohm,

I_S = Current in shunt in Amp,

I_g = Current in the galvanometer in Amp.

Let the total current is I Amp, which is

\[\text{I }=\text{ }{{\text{I}}_{\text{g}}}+\text{ }{{\text{I}}_{\text{S}}}\text{  Amp}\]

\[{{\text{I}}_{\text{S}}}\text{ }=\text{ (I}-\text{ }{{\text{I}}_{\text{S}}}\text{)  Amp}\]

The voltage drop in the shunt and galvanometer is same so

\[{{\text{I}}_{\text{g}}}\text{ }{{\text{R}}_{\text{g}}}=\text{ }{{\text{I}}_{\text{S}}}\text{ }{{\text{R}}_{\text{S}}}\]

Now substituting for I_S

\[{{\text{I}}_{\text{g}}}\text{ }{{\text{R}}_{\text{g}}}=\text{ (I}-\text{ }{{\text{I}}_{\text{g}}}\text{) }{{\text{R}}_{\text{S}}}\]

and

\[{{\text{R}}_{\text{S}}}=\text{ }\frac{{{\text{I}}_{\text{g}}}\text{ }{{\text{R}}_{\text{g}}}}{\text{(I}-\text{ }{{\text{I}}_{\text{g}}}\text{)}}\text{ Ohm}\]

Also,

\[{{\text{R}}_{\text{S}}}=\text{ }\frac{{{\text{I}}_{\text{g}}}\text{ }{{\text{R}}_{\text{g}}}}{\text{(I / }{{\text{I}}_{\text{g}}}-\text{ 1)}}\text{ Ohm}\]

$\text{I / }{{\text{I}}_{\text{g}}}\text{ }$ is known as the multiplying factor.

Similarly

\[{{\text{I}}_{\text{g}}}\text{ }{{\text{R}}_{\text{g}}}=\text{ I }{{\text{R}}_{\text{S}}}-\text{ }{{\text{I}}_{\text{g}}}\text{ }{{\text{R}}_{\text{S}}}\]

or

\[\text{I }{{\text{R}}_{\text{S}}}=\text{ (}{{\text{R}}_{\text{S}}}+\text{ }{{\text{R}}_{\text{g}}}\text{) }{{\text{I}}_{\text{g}}}\]

or

\[\frac{\text{I}}{{{\text{I}}_{\text{g}}}}=\text{ }\frac{{{\text{R}}_{\text{S}}}+\text{ }{{\text{R}}_{\text{g}}}}{{{\text{R}}_{\text{S}}}}=\text{ }\frac{{{\text{R}}_{\text{g}}}}{{{\text{R}}_{\text{S}}}}+\text{ 1}\]

This is also known as the multiplying factor or the instrument constant and is equal to $\frac{{{\text{R}}_{\text{g}}}}{{{\text{R}}_{\text{S}}}}+\text{ 1}$

(Hence the circuit current)  = (Full scale deflection of the instrument × Instrument constant)

Example. A M. C. type instrument gives full scale deflection 15 mA and has a resistance of 3 Ω. Calculate the resistance to be connected to use this meter up to 500 m4 ammeter.

Solution. The galvanometer has the resistance of 3 Ω and the full scale deflection is 15 mA. When it is used as an ammeter 0 – 500 mA. The total current is 500 mA so the current to bypass through shunt is

500 – 15 = 485 mA

The voltage drop across both will be same, so

\[{{\text{R}}_{\text{S}}}=\text{ }\frac{{{\text{I}}_{\text{g}}}\text{ }{{\text{R}}_{\text{g}}}}{{{\text{I}}_{\text{S}}}}\]

\[{{\text{R}}_{\text{S}}}=\text{ }\frac{\text{15 }\times \text{ 1}{{\text{0}}^{-3}}\times \text{ 3}}{\text{485 }\times \text{ 1}{{\text{0}}^{-3}}}=\text{ }0.093\text{ }\Omega \]