PhysicsCBSE Class 12 · Practical, Section A, experiment 4

Resistance of a galvanometer by half-deflection method and its figure of merit

Updated 16 October 2026 · Fact-checked

With a high resistance R in series, note the galvanometer deflection θ. Then add a shunt S across the galvanometer and adjust it until the deflection is θ/2. The galvanometer resistance is G = RS / (R − S), and the figure of merit, the current per division, is k = E / ((R + G)θ).

Aim

To determine resistance of a galvanometer by half-deflection method and to find its figure of merit.

Apparatus and material required

Moving coil galvanometer
Battery or battery eliminator (0 to 6 V)
High resistance box RBOX 1 (0 to 10 kΩ)
Low resistance box RBOX 2 (0 to 200 Ω), used as the shunt
Two one-way keys
Voltmeter (to measure the emf E)
Connecting wires and sandpaper

Principle

A moving coil galvanometer works on the principle that a current-carrying coil in a magnetic field experiences a torque. In a radial field the deflection is proportional to the current: $I = k θ$ , where $k$ is the figure of merit, the current needed for a deflection of one division.

With only the high resistance R in series (key K2 open), the current is $I_{g} = \frac{E}{R + G} = k θ$ .

When the shunt S is connected across the galvanometer (K2 closed) and adjusted so that the deflection falls to $θ /2$ , the galvanometer current is $\frac{E S}{R ( G + S ) + GS} = \frac{k θ}{2}$ . Dividing the two equations and simplifying gives

$G = \frac{R S}{R - S}$ .

Because R (thousands of ohms) is much larger than S (about a hundred ohms), $G \approx S$ . Then $k = \frac{E}{( R + G ) θ}$ .

Galvanometer: half-deflection method and figure of merit: formulas and key facts

Galvanometer resistance: $G = \frac{R S}{R - S}$; R: high resistance in series; S: shunt for half deflection.
Approximation when R ≫ S: $G \approx S$
Figure of merit: $k = \frac{I}{θ} = \frac{E}{( R + G ) θ}$; Unit: ampere per division (A/div).
Current sensitivity: $\frac{θ}{I} = \frac{1}{k}$; Divisions per ampere; for a moving coil galvanometer it equals NAB/k_s, where k_s is the torsion constant.
Key facts: Take out the high resistance before closing K1
Choose θ an even number of divisions
Keep R fixed while adjusting S
Emf E must stay constant
Larger R makes G ≈ S a better approximation

Procedure

1Clean the ends of the connecting wires and make neat, tight connections: battery, key K1, high resistance box R and galvanometer in series; the low resistance box S with key K2 across the galvanometer.
2Measure the emf E of the battery with the voltmeter.
3Take out a high resistance (about 5 kΩ) from RBOX 1 before closing K1, so a large current does not damage the galvanometer.
4Close K1 (K2 open) and adjust R until the deflection θ is large and an even number of divisions. Note R and θ.
5Keeping R the same, close K2 and adjust the shunt S until the deflection is exactly θ/2. Note S, then open K2.
6Repeat for four more values of R (five sets in all), each with an even θ.
7Calculate G and k for each set and take the means.

Observation table

Draw the table like this in your record. The numbers are sample readings to show how it is filled in; use your own.

Resistance and figure of merit of the galvanometer (E = 2.0 V; 30 divisions on each side of zero)
S. No.	R (Ω)	θ (div)	S for θ/2 (Ω)	θ/2 (div)	$G = R S / (R - S)$ (Ω)	$k = E / ((R + G) θ)$ (× 10⁻⁵ A/div)
1	7000	28	98	14	99.4	1.006
2	8300	24	99	12	100.2	0.992
3	10000	20	99	10	100.0	0.990
4	11000	18	99	9	99.9	1.001
5	12500	16	99	8	99.8	0.992

Sample readings to show how the table is filled in; they are not real measurements. S is read from a box in 1 Ω steps, so G scatters by a few tenths of an ohm, and k by about 1%.

Calculation

Work out G and then k for each row; take the means.

1.Row 3: $G = \frac{R S}{R - S} = \frac{10000 \times 99}{10000 - 99} = \frac{990000}{9901} = 99.99 \approx 100.0 Ω$
2.Row 3: $k = \frac{E}{( R + G ) θ} = \frac{2.0}{( 10000 + 100.0 ) \times 20} = \frac{2.0}{202000} = 0.990 \times 1 0^{- 5} A/div$
3.Mean $G = \frac{99.4 + 100.2 + 100.0 + 99.9 + 99.8}{5} = 99.86 \approx 99.9 Ω$
4.Mean $k = \frac{1.006 + 0.992 + 0.990 + 1.001 + 0.992}{5} \times 1 0^{- 5} = 0.996 \times 1 0^{- 5} \approx 1.0 \times 1 0^{- 5}$ A/div (E is known only to 2 significant figures)
5.Check: S = 98 Ω to 99 Ω is close to G = 99.9 Ω, as expected because R is about 100 times S.

Result

1. Resistance of the galvanometer by half-deflection method, G = ____ Ω (sample: 99.9 Ω). 2. Figure of merit of the galvanometer, k = ____ A/div (sample: 1.0 × 10⁻⁵ A/div).

Precautions

Take out a high resistance from RBOX 1 before closing K1, or the galvanometer coil may be damaged.
Adjust R so that θ is an even number of divisions, so θ/2 can be read exactly.
Keep R unchanged while adjusting the shunt S.
Use a battery or eliminator whose emf stays constant.
Use R as large as practicable compared with S.
Keep all connections and resistance box plugs tight and clean.

Sources of error

Plugs of the resistance boxes may be loose or dirty, adding unknown resistance.
The emf of the battery may not stay constant during the readings.
The internal resistance of the battery is neglected in the formula for k.
Parallax in reading the pointer position, especially for θ/2.

Viva questions with answers

Say your answer out loud first, then open the card. Examiners want one or two clear sentences, not a paragraph.

1.What is a galvanometer?Show answer
A sensitive instrument that detects small currents. A moving coil galvanometer works on the torque on a current-carrying coil in a magnetic field.
2.Define the figure of merit of a galvanometer. What is its unit?Show answer
The current needed to deflect the pointer by one division, k = I/θ. Unit: ampere per division.
3.What is current sensitivity? How is it related to the figure of merit?Show answer
The deflection per unit current, θ/I. It is the reciprocal of the figure of merit.
4.Why is a high resistance taken out from the box before closing K1?Show answer
To keep the current small. Without it, a large current would flow through the delicate coil and could burn it.
5.Why should R be much larger than S?Show answer
Then closing the shunt hardly changes the total current, so half the current goes through S and half through G, and G ≈ S. The larger R is, the better this holds.
6.Why is the deflection θ taken as an even number?Show answer
So that half of it, θ/2, is a whole number of divisions and can be read exactly.
7.What is a shunt?Show answer
A low resistance connected in parallel with a galvanometer, which takes part of the current.
8.Why is the magnetic field in a moving coil galvanometer made radial?Show answer
So the plane of the coil is always parallel to the field, the torque is NIAB at every position, and the deflection is proportional to the current, giving a linear scale.
9.What is the function of the soft iron core in the coil?Show answer
It concentrates the field lines and makes the field radial and stronger, so the galvanometer is more sensitive.
10.How can the current sensitivity of a galvanometer be increased?Show answer
By increasing the number of turns, the area of the coil or the magnetic field, or by using a suspension with a smaller torsion constant.
11.Which has the highest resistance: a galvanometer, an ammeter or a voltmeter? Which the lowest?Show answer
A voltmeter has the highest resistance and an ammeter the lowest; the galvanometer lies between them.
12.Can you find G using one-third deflection instead of half?Show answer
Yes. If the deflection is reduced to θ/n with R ≫ S, the galvanometer current is 1/n of the total, so G ≈ (n − 1)S. The exact formula changes accordingly.

Viva practice: these questions one at a time, with the answer hidden until you ask, and a tally of what to revise.

Common mistakes in the practical exam

The slips examiners see most often, and what to write instead.

Mistake: Closing K1 with zero resistance in RBOX 1.
A large current can burn the galvanometer coil; examiners note it.
Do this: Take out the 5 kΩ plug first, then close K1.
Mistake: Writing G = S as the final answer without using the formula.
Do this: Calculate G = RS/(R − S) for each set and then remark that G ≈ S because R ≫ S.
Mistake: Changing R while adjusting the shunt.
Do this: Keep R fixed for each set; only S changes when K2 is closed.
Mistake: Giving the figure of merit without a unit or in amperes only.
Do this: Write k in ampere per division, for example 1.0 × 10⁻⁵ A/div.
Mistake: Using the deflection on both sides of zero as the full scale.
Do this: Read θ on one side only, and note the number of divisions on that side.

Galvanometer: half-deflection method and figure of merit: frequently asked questions

Why does the next experiment need this one?

Converting a galvanometer into a voltmeter or an ammeter (Section A, experiment 5) needs the galvanometer's resistance G and figure of merit k, which this experiment gives.

Can G be found from a graph?

Yes. Since RS = G(R − S), a graph of RS (y-axis) against R − S (x-axis) is a straight line whose slope is G.

Based on: CBSE Physics (042) curriculum 2026-27, Class XII practicals, Section A experiment 4; NCERT Laboratory Manual Physics Class XII, Experiment 6 (half-deflection method); NCERT Physics Part I, Textbook for Class XII, Chapter 4, section on the moving coil galvanometer