When a shunt of 4,Omega is connected to a gal | vedclass

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(D) Let the initial current through the galvanometer be $I_g = I/5$. The shunt resistance is $S = 4\,\Omega$. The current through the shunt is $I_s =

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$I/13$

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(D) Let the initial current through the galvanometer be $I_g = I/5$. The shunt resistance is $S = 4\,\Omega$. The current through the shunt is $I_s = I - I_g = I - I/5 = 4I/5$.Since the galvanometer and shunt are in parallel,the potential difference across them is equal:$I_g \cdot G = I_s \cdot S$$(I/5) \cdot G = (4I/5) \cdot 4$$G = 16\,\Omega$Now,an additional shunt of $2\,\Omega$ is connected in parallel to the existing $4\,\Omega$ shunt. The new equivalent shunt resistance $S'$ is:$1/S' = 1/4 + 1/2 = (1+2)/4 = 3/4$$S' = 4/3\,\Omega$Let the new current through the galvanometer be $I'$. The total current $I$ remains the same. The potential difference across the galvanometer and the new shunt is equal:$I' \cdot G = (I - I') \cdot S'$$I' \cdot 16 = (I - I') \cdot (4/3)$$12I' = I - I'$$13I' = I$$I' = I/13$

When a shunt of $4\,\Omega$ is connected to a galvanometer,the deflection reduces to $1/5$ of its initial value. If an additional shunt of $2\,\Omega$ is connected in parallel,what will be the new deflection?

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