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Galvanometer conversion and measurement devices

Understand how to convert a galvanometer into an ammeter or voltmeter using shunt and series resistances, and the principles behind these measuring instruments.

Why this shows up in the exam

You may be asked to design or analyze circuits involving measurement of current and voltage using modified galvanometers.

How NEET tests this

Numerical · 2 Qs

Learn the idea

A galvanometer can be turned into a voltmeter by adding a series resistance so that the full‑scale current produces the required voltage range; the key insight is that the total resistance seen by the source must equal V_range ÷ I_g.

🧠 Memory hook: Volt‑meter = V over I_g minus coil; think “V = I_g × (coil + extra)”.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Full‑scale deflection current of galvanometer = I_g
  • Coil resistance of galvanometer = R_g
  • Series resistance for voltmeter: R_s = (V_range ÷ I_g) – R_g
  • Shunt resistance for ammeter: R_sh = (I_g ÷ (I_required – I_g)) × R_g
  • Voltmeter must have high resistance, ammeter must have low resistance

How to approach it

  1. 1Read the required voltage (or current) range and the galvanometer’s I_g and R_g
  2. 2For a voltmeter compute total resistance needed: V_range ÷ I_g
  3. 3Subtract the coil resistance to obtain the series resistance R_s
  4. 4Connect R_s in series with the galvanometer; for an ammeter compute shunt resistance and connect in parallel

Worked example — watch it click

A galvanometer has a coil of resistance 100 ohm and gives a full scale deflection for 30 mA current. If it is to work as a voltmeter of 30 volt range, the resistance required to be added will be:

  • ✅900 Ω
  • B)1800 Ω
  • C)500 Ω
  • D)1000 Ω

The concept behind this problem

The example asks for the series resistance that makes the galvanometer read full scale at 30 V, directly applying the voltmeter formula R_s = (V ÷ I_g) – R_g.

Step by step

  1. 1For voltmeter, V = I_g(G + R_s). 30 = 0.03(100 + R_s), so 1000 = 100 + R_s, giving R_s = 900 Ω.

Watch out

Students often forget to subtract the 100 Ω coil, giving 1000 Ω instead of the correct 900 Ω.

Common slip-ups that cost marks

  • •Using the coil resistance instead of subtracting it when finding R_s
  • •Mixing up series and parallel connections for voltmeter and ammeter
  • •Using I_g directly for shunt formula without subtracting I_g from the desired current

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 1NEET 2010

A galvanometer has a coil of resistance 100 ohm and gives a full scale deflection for 30 mA current. If it is to work as a voltmeter of 30 volt range, the resistance required to be added will be:

Push further

More challenging

2 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 2

A galvanometer has a resistance of 60 Ω and gives a full-scale deflection for a current of 10 mA. To convert it into a voltmeter of range 0-15 V, a series resistance should be connected. What is the value of this resistance?

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