A moving coil instrument gives a full scale deflection of 10 mA when the potential difference across its terminals is 100 mV. Calculate the shunt resistance for full scale deflection which corresponds to 200 A?

Concept:

We can extend the range of the ammeter by keeping a shunt resistance.

Here Rm = internal resistance of the coil

Rsh = Shunt resistance

I = Required full-scale range

Im = Full scale deflection of current

As the two resistances, Rm and Rsh are in parallel, the voltage drop across the resistance is equal.

\({I_m}{R_m} = \left( {I - {I_m}} \right){R_{sh}}\)

\({R_m} = \left( {\frac{I}{{{I_m}}} - 1} \right){R_{sh}}\)

\(⇒ {R_{sh}} = \frac{{{R_m}}}{{\left( {\frac{I}{{{I_m}}} - 1} \right)}}\)

\(⇒ {R_{sh}} = \frac{{{R_m}}}{{\left( {m - 1} \right)}}\)

Where \(m = \frac{I}{{{I_m}}}\)

‘m’ is called multiplying power

Calculation:

Given that,

Full-scale deflection voltage (Vm) = 100 mV

Full scale reading (I) = 10 mA =  0.01 A.

Let's consider

Meter resistance = Rm

Full scale deflection current = Im

m = 200/(0.01) = 20000
R_m = 100 mV / 10mA = 10 Ω

\(⇒ {R_{sh}} = \frac{{{R_m}}}{{\left( {\frac{I}{{{I_m}}} - 1} \right)}}\)

= \(\frac{10}{20000 - 1}\)

= 500.02 µΩ