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Question

A galvanometer having internal resistance $$10\ \Omega $$ requires $$0.01 \ A$$ for a full scale deflection. To convert this galvanometer to a voltmeter of full-scale deflection at $$120\ V$$, we need to connect a resistance of:


A
11990 Ω in series
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B
11990 Ω in parallel
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C
12010 Ω in series
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D
12010 Ω in parallel
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Solution

The correct option is B $$11990\ \Omega $$ in series
Given :           $$G = 10 \Omega$$               $$I_G = 0.01  A$$              $$V = 120$$ volt
Let the resistance to be connected in series be $$R$$
$$\therefore$$            $$V = I_G  (G+R)$$
$$120  = 0.01 (10 + R)$$
$$\implies$$  $$R = 11990$$ $$\Omega$$

496987_461960_ans.png

Physics

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