A galvanometer has a resistance of 20,Omega a | vedclass

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(B) Given: Galvanometer resistance $R_g = 20\,\Omega$,full-scale voltage $V_g = 0.2\,V$,and target current $I = 10\,A$.First,calculate the full-scale

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$0.02\,\Omega$ resistor connected across it

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(B) Given: Galvanometer resistance $R_g = 20\,\Omega$,full-scale voltage $V_g = 0.2\,V$,and target current $I = 10\,A$.First,calculate the full-scale deflection current $I_g$ of the galvanometer:$I_g = \frac{V_g}{R_g} = \frac{0.2\,V}{20\,\Omega} = 0.01\,A$.To convert a galvanometer into an ammeter of range $I$,a shunt resistance $S$ must be connected in parallel with the galvanometer coil.The formula for shunt resistance is:$S = \frac{I_g R_g}{I - I_g}$.Substituting the values:$S = \frac{0.01 	imes 20}{10 - 0.01} = \frac{0.2}{9.99} \approx 0.02\,\Omega$.Therefore,a $0.02\,\Omega$ resistor should be connected in parallel (across) the galvanometer.

$A$ galvanometer has a resistance of $20\,\Omega$ and reads full-scale when $0.2\,V$ is applied across it. To convert it into a $10\,A$ ammeter,the galvanometer coil should have a

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