The resistance of a galvanometer is 90, Omega | vedclass

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(B) Let the main current be $I$ and the galvanometer resistance be $G = 90\, \Omega$. Given that the current through the galvanometer $I_g = 10\% 	ex

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$10\, \Omega$ in parallel

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(B) Let the main current be $I$ and the galvanometer resistance be $G = 90\, \Omega$. Given that the current through the galvanometer $I_g = 10\% 	ext{ of } I = \frac{I}{10}$. To allow only a fraction of the current to pass through the galvanometer,a shunt resistor $S$ must be connected in parallel with it. The formula for the shunt resistance is $S = \frac{I_g \cdot G}{I - I_g}$. Substituting the values: $S = \frac{(\frac{I}{10}) \cdot 90}{I - \frac{I}{10}} = \frac{9I}{0.9I} = \frac{90}{9} = 10\, \Omega$. Thus,a resistor of $10\, \Omega$ should be connected in parallel.

The resistance of a galvanometer is $90\, \Omega$. If only $10\%$ of the main current is to flow through the galvanometer,how should a resistor be connected and what should be its value?

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