A galvanometer- having a resistance of 50 - 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 cm2 that can be used to convert the galvanometer into an ammeter which can read a maximum of 5A current is - Specific resistance of the wire -5 - 10-7- m

Given G=50 Ω, i_g = 0.05 Amp., i=5A, A=2.97 × 10^ 2 cm^2 and ρ = 5 × 10^ 7Ω m By using i/i_g=1 + G/S⇒ S = G.i_g/(i i_g)⇒ l = G i_g A/(i i_g) ρ on putting va ...

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Standard XIII

Physics

Galvanometer

A galvanomete...

Question

A galvanometer, having a resistance of $50 Ω$ 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 $2.97 \times 10^{- 2} c m^{2}$ that can be used to convert the galvanometer into an ammeter which can read a maximum of 5A current is : ( Specific resistance of the wire $= 5 \times 10^{- 7} Ω m$ )

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Solution

The correct option is C 3
Given $G = 50 Ω, i_{g} = 0.05 A m p ., i = 5 A, A = 2.97 \times 10^{- 2} c m^{2}$ and $ρ = 5 \times 10^{- 7} Ω - m$
By using $\frac{i}{i_{g}} = 1 + \frac{G}{S} \Rightarrow S = \frac{G . i_{g}}{(i - i_{g})} \Rightarrow l = \frac{G i_{g} A}{(i - i_{g}) ρ}$ on putting values l = 3 m.

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The correct option is C 3Given G=50Ω,ig=0.05 Amp., i=5A, A=2.97×10−2 cm2 and ρ=5×10−7Ω−m By using iig=1+GS⇒S=G.ig(i−ig)⇒l=GigA(i−ig)ρ on putting values l = 3 m.