Ammeter: galvanometer + low-resistance shunt in parallel. Voltmeter: galvanometer + high resistance in series.
-- NCERT Class 12 Physics, Ch. 4, p. 130Ammeter Voltmeter Conversion
8 MCQs9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Ammeter Voltmeter Conversion, explained for NEET
A galvanometer is a delicate instrument. It gives full-scale deflection for a current of the order of microamperes, and its coil has a resistance of a few ohms to a few hundred ohms. Neither property suits it for direct measurement in a circuit. Conversion fixes both — and the direction of the fix is where repeaters lose marks.
Ammeter: shunt in parallel. An ammeter must carry the full circuit current while disturbing the circuit as little as possible, so it needs low resistance. Connect a small resistance — the shunt — in parallel with the galvanometer coil. Most of the current bypasses the coil through the shunt; only the fraction the coil can survive passes through it. The combination's resistance is lower than the coil's alone, which is exactly what you want in series with a circuit.
Voltmeter: high resistance in series. A voltmeter sits across two points and must draw almost no current, so it needs high resistance. Connect a large resistance in series with the galvanometer coil. The combination's resistance is higher than the coil's alone, so it siphons off very little current from the element it is measuring.
NCERT Class 12 Physics Chapter 4, page 130, states this plainly: the shunt is small and parallel for the ammeter; the added resistance is large and in series for the voltmeter. An ideal ammeter has zero resistance; an ideal voltmeter has infinite resistance.
Bridge to NEET. Questions on this topic are typically single-fact or single-step: identify which connection converts which instrument, or reason about what happens to the effective resistance. The frequency is modest — roughly one question every two or three years — but it is a full four marks for a fact you can hold in one sentence.
Watch-out. The two conversions are mirror images, and under time pressure the mirror flips. Anchor on the purpose, not the wiring: ammeter goes in series, so it must be low, so the extra resistance is parallel. Voltmeter goes in parallel, so it must be high, so the extra resistance is in series. Series instrument → parallel resistor. Parallel instrument → series resistor.
Can you answer these Ammeter Voltmeter Conversion MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
To convert a moving coil galvanometer into an ammeter, the additional resistance is connected in which way relative to the galvanometer coil?
Show answer and why every option is right or wrong
Answer: B. B is correct. An ammeter must have low resistance, and a small shunt connected in parallel both lowers the effective resistance and diverts most of the current away from the delicate coil — NCERT Class 12 Physics Chapter 4, page 130.
Why A is wrong: A is wrong because a high resistance in series is the voltmeter conversion; it would raise the meter's resistance and choke the circuit current it is meant to measure.
Why C is wrong: C is wrong because a high resistance in parallel would divert almost nothing, leaving nearly the full current in the coil and destroying it.
Why D is wrong: D is wrong because any series addition raises the total resistance; an ammeter sits in series with the circuit and must add as little resistance as possible.
MCQ 2Easy RecallPractice
To convert a moving coil galvanometer into a voltmeter, the additional resistance is connected in which way relative to the galvanometer coil?
Show answer and why every option is right or wrong
Answer: A. A is correct. A voltmeter is connected across two points and must draw negligible current, so a large resistance in series raises the total resistance and keeps the current drawn tiny — NCERT Class 12 Physics Chapter 4, page 130.
Why B is wrong: B is wrong because a low parallel resistance is the ammeter conversion; it lowers the effective resistance, which is the opposite of what a voltmeter needs.
Why C is wrong: C is wrong because a low series resistance barely raises the total resistance, so the meter would still draw appreciable current and disturb the potential difference being measured.
Why D is wrong: D is wrong because a parallel addition always lowers the combined resistance below the coil's own value, and a voltmeter needs the combined resistance to be as high as possible.
MCQ 3Easy RecallPractice
An ideal voltmeter and an ideal ammeter have which resistances respectively?
Show answer and why every option is right or wrong
Answer: C. C is correct. An ideal voltmeter draws no current at all, requiring infinite resistance; an ideal ammeter introduces no potential drop, requiring zero resistance — NCERT Class 12 Physics Chapter 4, page 130.
Why A is wrong: A is wrong because it reverses the pair: zero resistance belongs to the ammeter and infinite resistance to the voltmeter, not the other way round.
Why B is wrong: B is wrong because a zero-resistance voltmeter placed across a circuit element would short it out entirely.
Why D is wrong: D is wrong because an infinite-resistance ammeter placed in series would block the circuit current completely, so nothing could be measured.
MCQ 4Direct ApplicationPractice
A shunt of resistance S is connected in parallel with a galvanometer of coil resistance G, where S is much smaller than G. The resistance of the resulting ammeter is:
Show answer and why every option is right or wrong
Answer: D. D is correct. Two resistances in parallel give a combination smaller than either one, so with S much smaller than G the parallel value sits just below S — NCERT Class 12 Physics Chapter 4, page 130.
Why A is wrong: A is wrong because adding any resistance in parallel can only reduce the combined resistance; a parallel connection never raises it above either branch.
Why B is wrong: B is wrong because it treats the shunt as having no effect, whereas the whole purpose of the shunt is to lower the meter's resistance well below G.
Why C is wrong: C is wrong because it applies series-style reasoning, where the total lies between the parts; in parallel the result falls below the smaller branch, not between the two.
MCQ 5Direct ApplicationPractice
A galvanometer of coil resistance G is converted into a voltmeter by adding a resistance R in series, with R much larger than G. The resistance of the resulting voltmeter is:
Show answer and why every option is right or wrong
Answer: A. A is correct. Series resistances add, so the voltmeter's resistance is R + G, which with R much larger than G is just above R — NCERT Class 12 Physics Chapter 4, page 130.
Why B is wrong: B is wrong because a series addition can never bring the total below one of its parts; the total must be at least G.
Why C is wrong: C is wrong because it describes a parallel combination, which is the ammeter conversion, not the voltmeter conversion.
Why D is wrong: D is wrong because halving would follow from two equal resistances in parallel, and this is a series connection of unequal resistances.
MCQ 6Concept TrapPractice
A student converts a galvanometer into an ammeter and then inserts it in series into a circuit. Compared with the current before insertion, the circuit current afterwards is:
Show answer and why every option is right or wrong
Answer: C. C is correct. A real ammeter has small but non-zero resistance, so it adds a little to the circuit's total resistance and lowers the current slightly; only an ideal zero-resistance ammeter would leave it exactly unchanged — NCERT Class 12 Physics Chapter 4, page 130.
Why A is wrong: A is wrong because reduction to zero would need infinite meter resistance, which is the ideal-voltmeter property, not an ammeter's.
Why B is wrong: B is wrong because inserting any resistance in series cannot increase the current; adding resistance always reduces it.
Why D is wrong: D is wrong because 'exactly unchanged' holds only for an ideal ammeter of zero resistance; a real converted galvanometer retains a small resistance.
MCQ 7Direct ApplicationPractice
In an ammeter formed by shunting a galvanometer, the shunt resistance is made smaller while the coil resistance stays the same. The fraction of the total current passing through the galvanometer coil:
Show answer and why every option is right or wrong
Answer: A. A is correct. In a parallel pair, current divides in inverse proportion to resistance, so lowering the shunt resistance sends a larger share through the shunt and a smaller share through the coil — NCERT Class 12 Physics Chapter 4, page 130.
Why B is wrong: B is wrong because it inverts the current-division rule; the branch with lower resistance takes more current, so shrinking the shunt takes current away from the coil, not towards it.
Why C is wrong: C is wrong because the division ratio depends on the two resistances, so changing one of them must change the split.
Why D is wrong: D is wrong because an equal split requires the two branches to have equal resistance, which is not stated and is in fact moved further away by shrinking the shunt.
MCQ 8CalculationPractice
A galvanometer is converted into a voltmeter with a series resistance, and separately an identical galvanometer is converted into an ammeter with a shunt. Both are then connected, correctly, across and in series with the same resistor in a circuit. Which statement about the two instruments' resistances is correct?
Show answer and why every option is right or wrong
Answer: C. C is correct. The series addition raises the voltmeter above the coil resistance while the parallel shunt drops the ammeter below it, which is precisely why each meter suits its own connection — NCERT Class 12 Physics Chapter 4, page 130.
Why A is wrong: A is wrong because only the parallel shunt lowers resistance; the voltmeter's series resistance necessarily raises it.
Why B is wrong: B is wrong because only the series addition raises resistance; the ammeter's parallel shunt necessarily lowers it.
Why D is wrong: D is wrong because it swaps the two conversions — the mirror-image error this topic invites when the wiring is memorised without the purpose behind it.
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How do you solve a Ammeter Voltmeter Conversion question? A worked example
- 1
Given
A moving coil galvanometer has coil resistance G = 50. Ω (exact as stated by the problem). It is shunted by a resistance S = 5.0 × 10⁻¹ Ω to make an ammeter.
- 2
Required
The resistance of the resulting ammeter, and the fraction of the total current that flows through the galvanometer coil.
- 3
Concept
Converting a galvanometer to an ammeter means placing a small shunt in parallel so the combination has low resistance and the coil carries only a small share of the current.
- 4
Formula
Two resistances in parallel combine as the product over the sum. Current divides between parallel branches in inverse proportion to their resistances, so the coil's share is S/(G + S).
- 5
Substitution
Combination: (50. × 0.50)/(50. + 0.50). Coil's current share: 0.50/(50. + 0.50).
- 6
Calculation
Combination = 25/50.5 = 4.95 × 10⁻¹ Ω. Coil's share = 0.50/50.5 = 9.9 × 10⁻³, that is about 0.99 percent. The numbers 2 (in the parallel form) and the percentage conversion factor 100 are exact counting values and do not limit the significant-figure count; the two measured resistances, given to two significant figures, set the precision at two significant figures.
- 7
Final answer
The ammeter's resistance is 5.0 × 10⁻¹ Ω — just below the shunt value, and a hundred times below the bare coil. Roughly 1.0 percent of the total current passes through the coil; the rest goes through the shunt.
- 8
Common trap
Adding the two resistances instead of combining them in parallel, giving 50.5 Ω. That number is larger than the bare galvanometer, which is the signature of the voltmeter conversion. Any ammeter answer that comes out above the coil resistance is wrong by inspection — stop and check the connection before checking the arithmetic.
- 9
Similar NEET-style question
A galvanometer of coil resistance 20. Ω is to be converted into a voltmeter by adding a series resistance of 1.98 × 10³ Ω. State whether the voltmeter's resistance is above or below 20. Ω, and by roughly what factor. *(Answer: above, by a factor of about 1.0 × 10² — series resistances add, giving 2.00 × 10³ Ω.)*
What to remember before solving Ammeter Voltmeter Conversion questions
More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 11 formulas · 5 question patterns from its other lessons.
Ammeter Voltmeter Conversion questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
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Sources
NCERT refs: Class 12 Physics Chapter 4, p.130
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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