A galvanometer, having a resistance of 50 Ω Ω gives a full scale deflection for a current of 0.05 A. The length in meter of a resistance wire of area of cross-section 2.97× 10 –2 cm 2 that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is (Specific resistance of the wire = 5 × $10^{-7}$ Ω Ω m)
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$S = \frac{I_g G}{I - I_g}$ $= \frac{0.05 \times 50}{5 - 0.05}$ $= \frac{2.5}{4.95}$ $= \frac{250}{495}$ $= \frac{50}{99} \Omega$ $\therefore S = \frac{\rho l}{A} \text{ or } l = \frac{SA}{\rho}$ $\therefore l = \frac{50}{99} \times \frac{3 \times 10^{-6}}{5 \times 10^{-7}}$ $= 3.0 \text{ m}$
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