JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Current Electricity

A galvanometer having a resistance of

20Ω

and 30 divisions on both sides has figure of merit 0.005 ampere /division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt is:

(A) $100Ω$
(B) $120Ω$
(C) $80Ω$ ✓ Correct
(D) $125Ω$

Answer: (C)

Solution

To turn a galvanometer into a voltmeter a large resistance

R

is placed in series with it. The full-scale current is the figure of merit times the number of divisions,

I_{g} = 0.005 \times 30 = 0.15

A. At full-scale deflection this current must carry the whole

15

V across the galvanometer and the series resistance together,

V = I_{g} (R_{g} + R)

, so

15 = 0.15 (20 + R)

. Hence

20 + R = 100

and

R = 80 Ω

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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