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MHT-CET Physics · Current Electricity

Converting a Galvanometer into an Ammeter or a Voltmeter

A galvanometer of resistance G gives full-scale deflection at a small current I_g; a small shunt S in parallel lets it measure a larger current I as an ammeter, with I_g G = (I − I_g)S, and a large resistance in series lets it read a voltage V as a voltmeter, with V = I_g(G + R).

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Why this matters

24 PYQs, 4 of them HARD — the largest page in the chapter. Twelve are the ammeter: the shunt that passes a given fraction of the current, the resistance of the finished ammeter, a second shunt compared with the first, and the series resistance that keeps the main current unchanged. Twelve are the voltmeter: the series resistance for a new range, extending an existing voltmeter, and the resistance of the galvanometer from two ranges. Two cards.

Concept 1 of 2: The Shunt: Galvanometer to Ammeter

The shunt and the galvanometer are in parallel, so they share one voltage: I_g G = (I − I_g)S. If a fraction f of the current passes through the galvanometer, S = fG/(1 − f); 1% gives G/99, 5% gives G/19, 10% gives G/9. The ammeter's own resistance is G and S in parallel, which works out to fG — small, as an ammeter's must be. The current-multiplying factor is n = I/I_g = (G + S)/S, so G = S(n − 1), and a second shunt S′ gives n′ = (G + S′)/S′.

Definition

  • IgG=(I−Ig)SI_g G = (I - I_g)S, so S=IgGI−IgS = \dfrac{I_g G}{I - I_g}.
  • Fraction f through the galvanometer: S=fG1−fS = \dfrac{fG}{1 - f} (4% ⇒ G/24; S = 5 Ω ⇒ G = 120 Ω).
  • Ammeter resistance: GSG+S=fG\dfrac{GS}{G + S} = fG (0.25% ⇒ G/400).
  • Multiplying factor: n=G+SSn = \dfrac{G + S}{S}, so n′=S(n−1)+S′S′n' = \dfrac{S(n - 1) + S'}{S'}.
  • To keep the main current unchanged after shunting, add in series G−GSG+S=G2G+SG - \dfrac{GS}{G + S} = \dfrac{G^2}{G + S}.
  • Shunt S = G/10 ⇒ 111\dfrac{1}{11} of the current through G.

Shunt

S=Ig GI−IgS = \frac{I_g\,G}{I - I_g}

Worked example

A galvanometer of 50 Ω deflects fully at 2 mA. Shunt to read 1 A?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q4Easy

Example 1 · Current Electricity · Galvanometer, Ammeter, and Voltmeter Conversion

If only 5%5\% of the total current is to be passed through galvanometer of resistance GG, then the resistance of the shunt will be

Using the fraction through the shunt instead of the galvanometer

S = fG/(1 − f) with f the share through the GALVANOMETER. 5% through G is G/19; reading 5% as the shunt's share gives 19G, a resistance no ammeter could have.

Quoting the shunt as the ammeter's resistance

The ammeter is G and S in parallel, fG — slightly less than S. The question asks for the ammeter, not the shunt, when it says 'resistance of the ammeter'.
Drill 11 more on the shunt: galvanometer to ammeter

Concept 2 of 2: The Series Resistance: Galvanometer to Voltmeter

At full scale the galvanometer carries I_g, so the whole voltmeter, G plus the series resistance R, reads V = I_g(G + R), giving R = V/I_g − G = G(V/V_g − 1), with V_g = I_gG the galvanometer's own range. A series resistance n times G makes the range n + 1 times. Extending an existing voltmeter of resistance R_V from V to nV needs (n − 1)R_V more in series. Two ranges with two series resistances fix G: if R₁ gives V and R₂ gives kV, then k(G + R₁) = G + R₂. A higher range needs a higher resistance, so a voltmeter has a larger resistance than a millivoltmeter made from the same galvanometer.

Definition

  • R=VIg−G=G(VVg−1)R = \dfrac{V}{I_g} - G = G\left(\dfrac{V}{V_g} - 1\right).
  • Full scale from a scale: IgI_g = (current per division) × (divisions).
  • Extend a voltmeter R_V from V to nV: add (n−1)RV(n - 1)R_V (10 V on 50 Ω to 15 V ⇒ 25 Ω).
  • Two ranges: k(G+R1)=G+R2k(G + R_1) = G + R_2 (100 Ω for V, 1000 Ω for 2V ⇒ G = 800 Ω).
  • Series resistance nG ⇒ range × (n + 1).
  • Largest device resistance: the voltmeter of the highest range.

Series resistance

R=VIg−GR = \frac{V}{I_g} - G

Worked example

A 40 Ω galvanometer deflects fully at 5 mA. Series resistance to read 10 V?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q21Moderate

Example 2 · Current Electricity · Galvanometer, Ammeter, and Voltmeter Conversion

A galvanometer of resistance 20 Ω20\,\Omega gives a deflection of 5 divisions when 1 mA current flows through it. The galvanometer scale has 50 divisions. To convert the galvanometer into a voltmeter of range 25 volt, we should connect a resistance of

Forgetting to subtract G

V/I_g is the WHOLE voltmeter's resistance. The resistance to add is V/I_g − G.

Reading 'replaced by' as 'added to'

'A resistance of 1000 Ω is connected in series' to double the range usually REPLACES the 100 Ω: 2(G + 100) = G + 1000 gives G = 800 Ω. Where the question says 'in series with X', add it: G + X + 1500 = 2(G + X).
Drill 11 more on the series resistance: galvanometer to voltmeter

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