A galvanometer,having a resistance of 50 Omeg | vedclass

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(C) Resistance of galvanometer,$G = 50 \Omega$. Full scale current,$i_g = 0.05 	ext{ A}$. Maximum current to be measured,$i = 5 	ext{ A}$. Area of c

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$3$

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(C) Resistance of galvanometer,$G = 50 \Omega$. Full scale current,$i_g = 0.05 	ext{ A}$. Maximum current to be measured,$i = 5 	ext{ A}$. Area of cross-section,$A = 2.97 	imes 10^{-2} 	ext{ cm}^2 = 2.97 	imes 10^{-6} 	ext{ m}^2$. Specific resistance,$ho = 5 	imes 10^{-7} \Omega	ext{-m}$. To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel. $S = \frac{i_g G}{i - i_g} = \frac{0.05 	imes 50}{5 - 0.05} = \frac{2.5}{4.95} = \frac{250}{495} = \frac{50}{99} \Omega$. Using the formula for resistance,$S = ho \frac{l}{A}$,we get $l = \frac{S \cdot A}{ho}$. $l = \frac{50}{99} 	imes \frac{2.97 	imes 10^{-6}}{5 	imes 10^{-7}} = \frac{50}{99} 	imes \frac{29.7}{5} = 10 	imes 0.3 = 3 	ext{ m}$.

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