JEE Advance - Physics (2014 - Paper 1 Offline - No. 18)
A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 $$\Omega$$ resistance, it can be converted into a voltmeter of range 0-30V. If connected to a $${{2n} \over {249}}\Omega $$ resistance, it becomes an ammeter of range 0-1.5 A. The value of n is
Answer
5
Explanation
$${i_g}(G + 4990) = V$$
$$ \Rightarrow {6 \over {1000}}(G + 4990) = 30$$
$$ \Rightarrow G + 4990 = {{30,000} \over 6} = 5000$$
$$ \Rightarrow G = 10\,\Omega $$
$${V_{ab}} = {V_{cd}}$$
$$ \Rightarrow {i_g}G = (1.5 - {i_g})S$$
$$ \Rightarrow {6 \over {1000}} \times 10 = \left( {1.5 - {6 \over {1000}}} \right)S$$
$$ \Rightarrow S = {{60} \over {1494}} = {{2n} \over {249}}$$
$$ \Rightarrow n = {{249 \times 30} \over {1494}} = {{2490} \over {498}} = 5$$
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