Question 356 of 380

A moving coil galvanometer allows a full scale current of ${10}^{-4}A$ . A series resistance of $2\times {10}^{4}\Omega$ is required to convert the galvanometer into a voltmeter of range $0-5V$ . Therefore, the value of shunt resistance required to convert the above galvanometer into an ammeter of range $0-10mA$ is:

A$100\Omega$B$200\Omega$C$300\Omega$D$10\Omega$

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Answer: (C) $300\Omega$

${i}_{g}={10}^{-4}A$
$i=10\times {10}^{-3}A$
${R}_{V}=20 k\Omega$

${i}_{g}G+{i}_{g} {R}_{V}=5$

${i}_{g}G=\left(i-{i}_{g}\right){R}_{A}$
${R}_{A}=\dfrac{{i}_{g}G}{i-{i}_{g}}=\dfrac{{i}_{g}\left(5-{i}_{g}{R}_{V}\right)}{\left(i-{i}_{g}\right){i}_{g}}$
${R}_{A}=\dfrac{5-{i}_{g}{R}_{V}}{i-{i}_{g}}\approx 300$ Ω