JEE MAIN - Physics (2017 (Offline) - No. 5)
When a current of 5 mA is passed through a galvanometer having a coil of resistance 15$$\Omega $$, it shows full
scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a
voltmeter of range 0 – 10V is:
4.005 × 103 $$\Omega $$
1.985 × 103 $$\Omega $$
2.535 × 103 $$\Omega $$
2.045 × 103 $$\Omega $$
Explanation
Given : Current through the galvanometer,
ig = 5 × 10–3 A
Galvanometer resistance, G = 15 $$\Omega $$
Let resistance R to be put in series with the galvanometer to convert it into a voltmeter.
V = ig (R + G)
10 = 5 × 10–3 (R + 15)
$$ \therefore $$ R = 2000 – 15 = 1985
= 1.985 × 103 $$\Omega $$
ig = 5 × 10–3 A
Galvanometer resistance, G = 15 $$\Omega $$
Let resistance R to be put in series with the galvanometer to convert it into a voltmeter.
V = ig (R + G)
10 = 5 × 10–3 (R + 15)
$$ \therefore $$ R = 2000 – 15 = 1985
= 1.985 × 103 $$\Omega $$
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