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Galvanometer conversion questions look like formula substitution, and they mostly are. The marks are lost in three places: putting the resistor in the wrong place, forgetting that the galvanometer's own resistance is part of the circuit, and not seeing why a meter changes the thing it measures. This post explains how the instrument works, converts one galvanometer both ways with numbers, and shows the loading effect that exam setters like.
Is this in the JEE and board syllabus?
Yes. The JEE (Main) 2026 syllabus lists the moving coil galvanometer, its sensitivity and conversion to ammeter and voltmeter, and its experimental skills list includes the resistance and figure of merit of a galvanometer by the half deflection method. The JEE (Advanced) 2026 brochure lists the moving coil galvanometer, voltmeter, ammeter and their conversions. NCERT covers it in Class 12 Chapter 4, section 4.10. Check your year's bulletin.
How does a moving coil galvanometer work?
A coil of N turns and area A sits in a radial magnetic field B. A current I produces a torque NIAB, and a spring supplies a restoring torque kφ. At balance, φ = (NAB/k) I, so the deflection is proportional to the current. NCERT defines current sensitivity as φ/I = NAB/k.
The same section explains why a bare galvanometer is not an ammeter: it gives full-scale deflection for currents of the order of microamperes, and its resistance would change the current if placed in series. For voltmeters, NCERT says we usually want the disturbance from the meter to stay below one per cent, which is why the series resistance must be large.
Worked example: one galvanometer, two meters
A galvanometer has resistance G = 50 Ω and gives full-scale deflection at I_g = 2.0 mA.
As an ammeter reading up to 2.0 A. Put a shunt S in parallel. At full scale, 2.0 mA goes through the coil and the rest, 1.998 A, through the shunt. The voltage across both is the same:
I_g G = (I − I_g) S, so S = (0.002 × 50)/1.998 ≈ 0.050 Ω.
The ammeter's total resistance is G and S in parallel, which works out to (I_g G)/I = 0.10/2.0 = 0.050 Ω. Small, as an ammeter should be.
As a voltmeter reading up to 10 V. Put a resistor R in series. At full scale the current is still 2.0 mA:
I_g (G + R) = V, so R = V/I_g − G = 10/0.002 − 50 = 5000 − 50 = 4950 Ω.
The voltmeter's total resistance is 5000 Ω.
Why does the meter change the reading?
Connect two 10 kΩ resistors in series across an ideal 10 V battery. The true voltage across each is 5.0 V.
With our 5000 Ω voltmeter across one resistor, that resistor and the meter in parallel make 10 kΩ × 5 kΩ / 15 kΩ ≈ 3.33 kΩ. The reading is 10 × 3.33/13.33 = 2.5 V. Half the true value.
With a more sensitive galvanometer (full scale at 50 μA), the 10 V voltmeter has 10/0.00005 = 200 kΩ. Now the parallel pair is about 9.52 kΩ and the reading is 10 × 9.52/19.52 ≈ 4.88 V, within about 2.5% of the truth.
This is why a good voltmeter has a high resistance and why its quality depends on the galvanometer's sensitivity, not just the series resistor.
Does a more sensitive galvanometer always help?
Not in every way. NCERT points out that if you double the number of turns, current sensitivity doubles, but the coil's resistance is also likely to double because the wire is longer. Voltage sensitivity, φ/V = NAB/(kR), stays the same. So the change that improves an ammeter is not automatically the one that improves a voltmeter.
What mistakes do students make?
Forgetting G. The voltmeter series resistor is V/I_g minus G, not V/I_g.
Using I instead of I − I_g for the shunt current. With 2 A and 2 mA it barely matters; with 10 mA and 2 mA it matters a lot.
Thinking current is fixed by the battery. Adding a meter changes the circuit's resistance and so the current. Engelhardt and Beichner (2004), who built a 29-question circuits diagnostic, found that the idea of the battery as a constant source of current was the one students used most in interviews, and that students often gave the properties of current to voltage or resistance.
For the circuit rules that underpin these questions, see our posts on Kirchhoff's law mistakes and on the meter bridge and potentiometer.
A practice routine
Convert three galvanometers, each to two ammeter ranges and two voltmeter ranges. Write the total meter resistance every time.
Do two loading problems like the 10 kΩ example, one for a voltmeter and one for an ammeter.
If your school lab has one, find the figure of merit of a galvanometer by the half deflection method and compare it with the label.
When the numbers come out right but you cannot say why the shunt goes in parallel, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can check your circuit logic, not just the arithmetic.
Before your next circuits test, redo the 50 Ω galvanometer conversion from memory and check the total resistance of each meter.
