Ncert Exercises & Solutions

4.28 A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Resistance of the galvanometer coil, G = 15

Ω

Current for which the galvanometer shows full scale deflection,

I_{g} = 4 mA = 4 \times 10^{- 3} A

Range of the ammeter is 0, which needs to be converted to 6 A.

∴ Current, I = 6 A

A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:

S = \frac{I_{g} G}{I - I_{g}}

= \frac{4 \times 10^{- 3} \times 15}{6 - 4 \times 10^{- 3}}

S = \frac{6 \times 10^{- 2}}{6 - 0.004} = \frac{0.06}{5.996}

\approx 0.01 Ω = 10 mΩ

Hence, a shunt resistor is to be connected in parallel with the galvanometer.

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