Galvanometer Resistance and Conversion
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Direct answer
A galvanometer is a sensitive current meter — full-scale deflection at a tiny current I_g such as 1 mA — and its coil resistance G (typically tens to hundreds of ohms) decides how it joins real circuits. To measure large currents, a shunt S = I_gG/(I − I_g) is wired in parallel, bypassing all but I_g and producing an ammeter of very small resistance. To measure voltage, a series multiplier R = V/I_g − G produces a voltmeter of large resistance. The figure of merit, I_g itself, is the current for full-scale deflection — the smaller it is, the more sensitive the instrument. An ideal ammeter would have zero resistance and an ideal voltmeter infinite resistance; real ones merely approach these, and their imperfections are examinable physics.
What you must remember
- Shunt conversion (ammeter): S = I_gG/(I − I_g) in parallel; the converted meter's resistance is approximately S — tiny, so series insertion barely disturbs the circuit.
- Multiplier conversion (voltmeter): R = V/I_g − G in series; the voltmeter's total resistance is V/I_g, hence the "ohms per volt" rating (1 mA meter = 1000 Ω/V).
- Figure of merit: I_g, the current producing full-scale deflection; sensitivity rises as I_g falls.
- Standard numbers: G = 100 Ω, I_g = 1 mA converts to a 1 A ammeter with S = (10^-3 × 100)/(1 − 10^-3) ≈ 0.1 Ω; to a 10 V voltmeter with R = 10/10^-3 − 100 = 9900 Ω.
- Why the designs differ: an ammeter must share the current (parallel shunt); a voltmeter must drop the surplus voltage (series multiplier) — structure follows function.
- Insertion effects: an ammeter adds a little resistance and reads marginally low; a voltmeter draws current and loads the circuit it measures — a high-resistance voltmeter distorts less.
- Null-method link: in a balanced Wheatstone bridge no current flows through the galvanometer, so its resistance does not matter — the point of null methods.
Converting one meter twice
Start with G = 100 Ω, I_g = 1 mA. Ammeter conversion for 1 A: the coil may carry at most 1 mA, so the shunt must bypass 0.999 A; S = I_gG/(I − I_g) = (0.001 × 100)/0.999 ≈ 0.1 Ω, and the parallel combination's resistance is essentially 0.1 Ω — inserted in a 10 Ω circuit it changes the current by about 1%. Voltmeter conversion for 10 V: total resistance must be V/I_g = 10000 Ω, of which the coil provides 100 Ω, so the multiplier is 9900 Ω. Now the cautionary tale: measure the voltage across a 1000 Ω resistor with a 1000 Ω voltmeter and the parallel combination drops to 500 Ω — the measured voltage falls to half its true value. The instrument changed what it measured, which is the whole argument for sensitive (high-resistance) voltmeters.
Where NEET sets the trap
The multiplier formula punishes the forgotten subtraction: options include both V/I_g and V/I_g − G, separated by exactly G. The shunt formula is inverted in distractors (I − I_g)/(I_gG), producing plausible but wrong small numbers. Conceptual items: an ammeter connected in parallel or a voltmeter in series (wrong topology, possible damage — a metre with near-zero resistance across a supply is a short); increasing an ammeter's range requires a smaller shunt; and the ideal-instrument assertions (zero and infinite resistances) recur as statement-evaluation items. The half-deflection method for measuring G — prescribed in the board practical, and hence fair game — uses a known resistance box and the condition that equal deflection means equal current division; knowing the logic survives better than memorising the final expression.
Frequently asked questions
How is a galvanometer converted into an ammeter?
By a low-resistance shunt in parallel, S = I_gG/(I − I_g), so that only I_g passes through the coil and the rest bypasses it.
How is a galvanometer converted into a voltmeter?
By a high resistance in series, R = V/I_g − G, so full-scale coil current corresponds to the full-scale voltage V across the combination.
What series resistance turns a 100 Ω, 1 mA galvanometer into a 10 V voltmeter?
R = V/I_g − G = 10/0.001 − 100 = 9900 Ω — the coil's own 100 Ω is part of the needed 10 kΩ total.
Why must an ammeter have very low resistance?
It sits in series with the circuit; any appreciable resistance would reduce the very current being measured — the reading would drop on connection.
What is the figure of merit of a galvanometer?
The current required for full-scale deflection; a smaller figure of merit means a more sensitive meter (and a higher-resistance voltmeter per volt of range).