How will you convert a 25 μA ammeter having an internal resistanc

SOLUTION

Concept:

We can extend the range of ammeter by keeping a shunt resistance as shown:

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 will be 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:

Meter resistance (Rm) = 500 Ω

Full scale deflection current (Im) = 25 μA

Required full scale reading (I) = 10 A

∴ The required shunt resistance that should be connected in parallel/shunt will be:

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

\({R_{sh}} = \frac{{500}}{{\left( {\frac{{10}}{{25\mu}} - 1} \right)}} = 1.25\;m\Omega{\rm{ }}\)

• To increase the ranges of ammeter, we need to connect a small shunt resistance in parallel with ammeters.
• To increase the ranges of a voltmeter, we need to connect a high series of multiplier resistance in series with voltmeters.