A galvanometer of resistance $100 \Omega$ has 50 divisions on its scale and has sensitivity of $20 \mu \mathrm{~A} /$ division. It is to be converted to a voltmeter with three ranges, of $0-2 \mathrm{~V}, 0-10 \mathrm{~V}$ and $0-20 \mathrm{~V}$. The appropriate circuit to do so is :

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Current Electricity

A galvanometer of resistance

100Ω

has 50 divisions on its scale and has sensitivity of

20 μ A /

division. It is to be converted to a voltmeter with three ranges, of

0 - 2 V, 0 - 10 V

and

0 - 20 V

. The appropriate circuit to do so is :

Answer: (A)

The full-scale galvanometer current is

I_{g} = 50 \times 20 \times 1 0^{- 6} = 1 mA

. A voltmeter is made by placing a multiplier resistance in series, chosen from

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

. For the

0 - 2 V

range,

2 = 1 0^{- 3} (100 + R_{1})

gives

R_{1} = 1900 Ω

. Tapping further along for

0 - 10 V

10 = 1 0^{- 3} (2000 + R_{2})

gives an added

R_{2} = 8000 Ω

. For

0 - 20 V

20 = 1 0^{- 3} (10000 + R_{3})

gives a further

R_{3} = 10000 Ω

. The correct arrangement is these three multipliers in series with the galvanometer, with the three ranges taken off successive taps at

1900 Ω

8000 Ω

and

10000 Ω

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option