Ammeter and Voltmeter Conversion
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Direct answer
A galvanometer becomes an ammeter when a small shunt resistance is wired in parallel, diverting the bulk of the current around the delicate coil: S = I_g G/(I - I_g) for a meter of resistance G and full-scale deflection current I_g. The same movement becomes a voltmeter when a large multiplier resistance is wired in series: R = V/I_g - G. The design logic follows from placement — an ammeter is a low-resistance series element (ideally zero), a voltmeter a high-resistance parallel element (ideally infinite). Conversion numericals are plug-ins; the deeper NEET questions concern what non-ideal meters do to the circuit they measure.
What you must remember
- Shunt formula: S = I_g G/(I - I_g); for a galvanometer of G = 100 Ω and I_g = 1 mA converted to a 1 A ammeter, S ≈ 0.1 Ω — always far smaller than G.
- Multiplier formula: R = V/I_g - G; the same galvanometer as a 10 V voltmeter needs a series resistance of 9900 Ω, total 10 kΩ.
- Resulting meter resistance: the converted ammeter's resistance is GS/(G + S), essentially the shunt; the converted voltmeter's is essentially the multiplier — the design difference in one line.
- Ideal instruments: ammeter resistance zero, voltmeter resistance infinite; every real reading carries the loading error of falling short of these ideals.
- Placement discipline: ammeter in series with the element, voltmeter across it; an ammeter connected across a battery is effectively a short circuit and can burn out.
- Shunt protection: most of the heat dissipates in the shunt, which also protects the coil from overcurrent — a parallel resistance, never a series one.
Converting one galvanometer two ways
Start with a galvanometer of G = 100 Ω that deflects fully at I_g = 1 mA. As an ammeter of range 1 A: the shunt must carry 999 mA when the coil carries 1 mA at the same 0.1 V (the coil's full-scale drop), so S = 0.1/0.999 ≈ 0.1 Ω, matching the formula S = I_g G/(I - I_g). The converted meter's resistance is now about 0.1 Ω — inserting it in series with a 10 Ω resistor changes the circuit current by barely 1 per cent, which is the whole point of the low-resistance design. As a voltmeter of range 10 V: at full scale the total resistance must pass 1 mA at 10 V, so the total is 10,000 Ω and the series multiplier is 9900 Ω. Placing this across a 10 kΩ resistor in a circuit effectively halves that branch's resistance — a severe loading error that a 100 kΩ meter would reduce to under 10 per cent. One galvanometer, two personalities: the conversion is Ohm's law applied to what the meter is allowed to do.
Loading and the traps it sets
The exam's sharpest questions are about the non-ideal meter's effect. A voltmeter of resistance R_v connected across R reads the voltage on the parallel pair R and R_v — if R_v is comparable to R the reading is systematically low, and questions give the two resistances expecting you to compute the pair before anything else. The ammeter's error runs the other way: its small series resistance raises the total circuit resistance slightly, lowering the current it is trying to measure. The second trap is the miswired meter: an ammeter in parallel with a supply is a near-short (0.1 Ω across 12 V would pass 120 A if the source could give it), while a voltmeter in series simply adds its huge resistance and kills the current. Third, the conversion formulas are direction-sensitive: increasing the ammeter's range means decreasing the shunt (more diversion), while increasing the voltmeter's range means increasing the multiplier — "to double the range" sends the two instruments in opposite directions.
Frequently asked questions
Why is a shunt connected in parallel with a galvanometer?
To divert most of the line current around the coil while both share the same voltage, letting the milliammeter-scale movement indicate currents of amperes.
How is a galvanometer converted into a voltmeter?
By connecting a high resistance in series, R = V/I_g - G, so that the full voltage range corresponds to the coil's full-scale current.
What are the ideal resistances of an ammeter and a voltmeter?
Zero for the ammeter (so it disturbs no current) and infinite for the voltmeter (so it draws none) — real meters approach but never reach these ideals.
Why should an ammeter never be connected across a battery?
Its resistance is deliberately tiny, so it would pass a destructively large current — effectively short-circuiting the source through the meter.
What happens if a voltmeter of resistance comparable to the circuit's is used?
It draws significant current and lowers the very voltage it is meant to measure — the loading error; a voltmeter's resistance should always be much larger than the element it tests.