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QJEEAdv14Ph1a35: JEE Advanced 2014 Physics - Class 12 JEEAdv14Ph1a35 Solution: A galvanometer gives full scale deflection with 0.006 A…

A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 Ω\Omega resistance, it can be converted into a voltmeter of range 0-30V. If connected to a 2n249Ω{{2n} \over {249}}\Omega resistance, it becomes an ammeter of range 0-1.5 A. The value of n is

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JEE Advanced 2014 Paper 1 Offline Physics - Current Electricity Question 17 English Explanation 1
ig(G+4990)=V{i_g}(G + 4990) = V

⇒61000(G+4990)=30\Rightarrow {6 \over {1000}}(G + 4990) = 30

⇒G+4990=30,0006=5000\Rightarrow G + 4990 = {{30,000} \over 6} = 5000

⇒G=10 Ω\Rightarrow G = 10\,\Omega

Vab=Vcd{V_{ab}} = {V_{cd}}

⇒igG=(1.5−ig)S\Rightarrow {i_g}G = (1.5 - {i_g})S

JEE Advanced 2014 Paper 1 Offline Physics - Current Electricity Question 17 English Explanation 2
⇒61000×10=(1.5−61000)S\Rightarrow {6 \over {1000}} \times 10 = \left( {1.5 - {6 \over {1000}}} \right)S

⇒S=601494=2n249\Rightarrow S = {{60} \over {1494}} = {{2n} \over {249}}

⇒n=249×301494=2490498=5\Rightarrow n = {{249 \times 30} \over {1494}} = {{2490} \over {498}} = 5

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