Question 321 markMCQSection D

A galvanometer with coil of resistance 20 Ω shows full scale deflection for a current of 5 mA. To convert it into an ammeter of range (0 – 10 A), a resistance of

Moving Charges and Magnetism
Converting a galvanometer into an ammeter (shunt)

Options

(A)0.05 Ω should be connected in series with it.
(B)0.05 Ω should be connected in parallel with it.
(C)0.01 Ω should be connected in parallel with it.
(D)0.01 Ω should be connected in series with it.

Official answer

Correct option: (C) 0.01 Ω should be connected in parallel with it.


An ammeter is made by adding a small shunt resistance S in parallel with the galvanometer, so most of the current bypasses the coil:


  • S=IgG/(I−Ig)S = I_g G / (I - I_g)
  • S=(5×10−3×20)/(10−5×10−3)S = (5 \times 10^{-3} \times 20) / (10 - 5 \times 10^{-3})
  • S=0.1/9.995≈S = 0.1 / 9.995 \approx 0.01 Ω

Because I_g (5 mA) is tiny compared with I (10 A), the shunt is very low resistance and connected across (parallel to) the galvanometer.

Answer keywords:shunt resistancegalvanometer to ammeterS = IgG/(I − Ig)parallel low resistancefull scale deflection currentrange 0 to 10 A0.01 ohm

Marking scheme

  1. 1.1 mark: correct option (C) 0.01 Ω in parallel.
  2. 2.Reasoning credit for using the shunt formula S=IgG/(I−Ig)S = I_g G /(I - I_g) with correct substitution.
  3. 3.No marks for 'series' options or for the 0.05 Ω value.

Hint

Ammeter = low shunt in parallel: S=Ig⋅G/(I−Ig)S = I_g \cdot G/(I - I_g); here Ig≪II_g \ll I so S≈Ig⋅G/IS \approx I_g \cdot G/I.

Quick oral answer

To read 10 A the galvanometer needs a low shunt in parallel; S=IgG/(I−Ig)=0.005×20/9.995≈0.01S = I_g G/(I - I_g) = 0.005 \times 20 / 9.995 \approx 0.01 ohm, so option C.

Common mistakes

  1. 1.Connecting the resistance in series (that makes a voltmeter, not an ammeter).
  2. 2.Forgetting to subtract IgI_g in the denominator, or using I = 5 mA instead of 10 A for the range.
  3. 3.Arithmetic slip giving 0.05 Ω instead of the correct 0.01 Ω.

Analysis and explanation

Concept:


A galvanometer can carry only a small full-scale current Ig. To read a large current I, a low-resistance shunt S is placed in parallel with it. The shunt diverts the excess current (I − Ig) while the galvanometer still carries only I_g at full scale. Equal potential drop across the parallel branches gives IgG=(I−Ig)SI_g G = (I - I_g) S.


Working:


  • S=IgG/(I−Ig)=(0.005×20)/(10−0.005)≈0.1/9.995≈0.01S = I_g G / (I - I_g) = (0.005 \times 20)/(10 - 0.005) \approx 0.1/9.995 \approx 0.01 Ω.
  • Since Ig≪II_g \ll I, the denominator ≈ I, confirming S≈0.01S \approx 0.01 Ω.

Exam trap:


Students often pick 0.05 Ω by using I=10I = 10 with G-only arithmetic errors, or wrongly choose 'series'. Remember: ammeter → low shunt in parallel; voltmeter → high resistance in series.


Real-world:


Multi-range ammeters simply switch between several shunt resistors; clamp meters achieve the same non-contact.

Interesting facts

An ideal ammeter has zero resistance; a good shunt makes the meter's overall resistance a small fraction of an ohm so it barely disturbs the circuit.

Shunts for very large currents (hundreds of amperes) are precision manganin bars, chosen because manganin's resistance hardly changes with temperature.

The same coil can be a microammeter or a 10 A ammeter just by changing the shunt — the movement never changes, only the parallel resistor.

Question details

Q32 of 39 questions

Board
CBSE
Class
Class 12
Subject
Physics
Year
2026
Section
Section D
Marks
1
Type
MCQ
Chapter
Moving Charges and Magnetism
Topic
Converting a galvanometer into an ammeter (shunt)
Bloom's level
apply
Syllabus topic
Unit III: Magnetic Effects of Current and Magnetism — Conversion of Galvanometer into Ammeter (shunt resistance)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why must the shunt be connected in parallel and be very small?

In parallel, the galvanometer and shunt have the same voltage across them, so the large excess current (I−Ig)(I - I_g) is diverted through the low-resistance shunt while the coil still carries only its safe full-scale current IgI_g. A very small shunt also keeps the ammeter's overall resistance tiny, so inserting it in a circuit hardly changes the current being measured.

How does this differ from converting a galvanometer into a voltmeter?

For a voltmeter you add a high resistance in series so the meter draws negligible current and reads potential difference. For an ammeter you add a low shunt in parallel so it can carry a large current. Series-high-resistance = voltmeter; parallel-low-shunt = ammeter.