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In galvanometer $5\%$ of the total current in the circuit passes through it. If the resistance of the galvanometer is G the shunt resistance 'S' connected to the galvanometer is

\[\begin {array} {1 1} (1)\;19\;G & \quad (2)\;\frac{G}{19} \\ (3)\;20 \;G & \quad (4)\;\frac{G}{20} \end {array}\]
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$(2) \frac{G}{19}$
answered Nov 7, 2013 by pady_1

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