• # question_answer A milliammeter of range 10 mA has a coil of resistance$1\Omega$. To use it as an ammeter of range 1 A, the required shunt must have a resistance of A)  $\frac{1}{101}\Omega$                            B)  $\frac{1}{100}\Omega$ C)  $\frac{1}{99}\Omega$                              D)  $\frac{1}{9}\Omega$

Solution :

Idea An ammeter have low resistance in parallel and its value is given by $S=\frac{{{i}_{g}}r}{i-{{i}_{g}}}$ Given, ${{i}_{g}}=10mA=0.01A,r=1\Omega ,i=1A$ As, ${{V}_{A}}-{{V}_{B}}={{i}_{g}}r=(l-{{i}_{g}})S$ $\Rightarrow$$S=\frac{{{i}_{g}}r}{(l-{{i}_{g}})}=\frac{0.01\times 1}{1-0.01}=\frac{1}{99}\Omega$ (shunt is small resistance in parallel) TEST Edge Galvanometer can be converted into voltmeter as well, for this a high resistance is connected in series with the galvanometer. $R=\frac{V}{{{I}_{g}}}-{{R}_{g}}$

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