Single Choice

Maximum current that can pass through galvanometer is $$0.002A$$ and resistance of galvonameter is $$R_g = 50\Omega$$. find out shunt resistance to convert in into ammeter of range $$0.5 A$$.

A$$0.5 \Omega$$
B$$0.2 \Omega$$
Correct Answer
C$$0.7 \Omega$$
D$$0.9 \Omega$$

Solution

$$Rs = \dfrac{I_g R_g}{I - I_g} = \dfrac{2 \times 10^{-3} \times 50}{0.5 - 2 \times 10^{-3}} = \dfrac{0.1}{0.498} = 0.2 \Omega$$

SIMILAR QUESTIONS

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