MixedJEE Physics · Original learning card10 original chapter questions

Converting a Galvanometer into a Voltmeter

For galvanometer resistance G and full-scale current I_g converted to voltmeter range V, series resistance R satisfies V = I_g(G + R), so R = V/I_g - G.

Why this shows up in the exam

Designing voltmeter ranges · Finding multiplier resistance · Comparing loading effects

Learn the idea

A large series resistance limits current so a galvanometer can measure a larger voltage. The added resistor drops most of the applied voltage while keeping coil current at or below its safe full-scale value.

🧠 Memory hook: Voltmeter means a large resistor in series.

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

Formulas & facts to keep ready

  • R = V/I_g - G — series resistance required for voltmeter range V
  • R_V = G + R = V/I_g — total voltmeter resistance
  • V_g = I_g G — bare galvanometer full-scale voltage

How to approach it

  1. 1Set the safe coil current to I_g
  2. 2Write total series resistance V/I_g
  3. 3Subtract the coil resistance

Common slip-ups that cost marks

  • •Putting the multiplier in parallel
  • •Forgetting to subtract G
  • •Using the desired voltage as the drop only across the added resistor

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

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 10

A charge of 2 microC moves perpendicular to a 3 T magnetic field at 4 x 10^5 m/s. Find the magnetic force.

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