Priority 5 for the next ILTS. The last section of the chapter — no vectors, no geometry, just two formulas and Ohm's law. The most crammable marks in Chapter 4.
This section is where the whole chapter turns into a real instrument you can hold. Everything here is built on one thing you already know: a current loop in a magnetic field feels a torque.
1. What problem is being solved?
Electric current is invisible. You cannot see it, hear it or feel it. So how do you measure it? The trick is to turn something invisible into something visible — a needle moving across a scale.
A galvanometer does exactly that. It is a device that detects and measures small currents by turning them into a visible deflection.
The instrument turns something invisible into something you can read off a scale. Everything else in this section is about making that reading trustworthy.
2. How it is built
A coil of many turns of thin insulated wire, wound on a light frame.
A permanent magnet with specially curved (concave) pole pieces.
A soft iron core sitting inside the coil.
A spring (or a suspension wire) attached to the coil.
A pointer fixed to the coil, moving over a scale.
The curved poles and the iron core are not decoration — together they are what makes the field radial, and that is what makes the scale usable.
3. How it works — the tug of war
Send a current through the coil. From Priority 1 we know a current loop in a magnetic field feels a torque, so the coil starts to turn:
Magnetic torque = N I A B
But the spring resists. The more the coil turns, the harder the spring pulls back:
Restoring torque = k θ
where k is the torsional constant — how stiff the spring is. The coil stops turning when the two exactly balance:
N I A B = k θ ⇒ θ = (N A B / k) I
Exactly like pulling a door against its closer. The needle stops where the two pulls match — and because both bars grow in step, the scale can be marked in equal divisions.
Picture it: pulling a door open against a strong door-closer spring. The harder you pull, the further it opens, and it settles where your pull and the spring's pull match. Pull twice as hard, it opens twice as far.
The key result: θ ∝ I. The deflection is directly proportional to the current. That means the scale can be marked in equal steps — a linear scale — which is exactly what makes the instrument usable.
4. Why the field must be radial
Look back at the torque formula from Priority 1: τ = NIAB sin θ. That sin θ is a problem. As the coil rotates, θ changes, so the torque would change too — and the scale would be squashed at one end and stretched at the other.
The fix is clever. The magnet's poles are curved and a soft iron core is placed inside, so the field lines always point towards the centre — like the spokes of a bicycle wheel. This is called a radial field. However far the coil turns, its plane always stays along the field, so sin θ = 1 at all times.
In a radial field the coil's plane always stays along B, so the torque is NIAB at every position. That is the only reason the divisions can be equally spaced.
Two jobs of the soft iron core:
It makes the field radial, so sin θ stays 1 and the scale is linear.
It makes the field stronger, so the instrument is more sensitive.
5. Sensitivity — two different meanings
Current sensitivity is how much deflection you get per unit current:
Current sensitivity = θ / I = N A B / k
Voltage sensitivity is how much deflection you get per unit voltage. Since V = I G, where G is the galvanometer's own resistance:
Voltage sensitivity = θ / V = N A B / k G
The favourite conceptual question in this section. To make the instrument more sensitive, why not just add more turns?
Adding turns does increase current sensitivity, because N appears on top. But more turns also means more wire, which means more resistance G — and G sits on the bottom of the voltage sensitivity. The two effects cancel.
So increasing the number of turns increases current sensitivity but does NOT necessarily increase voltage sensitivity. This exact NCERT sentence has been examined repeatedly.
Three bars grow, one refuses to move. This is the exact NCERT sentence NEET keeps examining: more turns raises current sensitivity but not necessarily voltage sensitivity.
6. Why a galvanometer cannot be used as it is
A galvanometer is too good at its job. It is extremely sensitive, so even a small current makes the needle slam to the end of the scale. And it has a moderate resistance of its own, which disturbs any circuit you put it into.
So we modify it in two different ways to make two different instruments.
The galvanometer is too good at its job. So we modify it in two different ways, giving two different instruments.
7. Making an ammeter — add a shunt in PARALLEL
An ammeter is connected in series in a circuit, so all the current must pass through it. But the galvanometer can only take a tiny current Ig without breaking.
The solution: give most of the current a bypass road — a small resistance S connected in parallel, called a shunt. The galvanometer takes its small share and the shunt carries the rest.
Picture it: a narrow footbridge next to a wide motorway. Almost all the traffic takes the motorway; only a trickle crosses the footbridge. Count the trickle and you know the total, because you know the ratio.
Since the two are in parallel, the voltage across them is equal:
Ig G = (I − Ig) S ⇒ S = Ig G / (I − Ig)
A narrow footbridge beside a wide motorway. Almost all the traffic takes the motorway — but because you know the ratio, counting the trickle tells you the whole flow.
Two useful consequences:
The fraction of the current that goes through the galvanometer is Ig/I = S / (S + G).
The combined resistance of the ammeter is GS/(G+S), which is very small — exactly what an ammeter needs, so it barely disturbs the circuit.
8. Making a voltmeter — add a resistance in SERIES
A voltmeter is connected in parallel across whatever you are measuring. It must draw almost no current, or it changes the very voltage it is trying to measure.
The solution: put a large resistance R in series with the galvanometer, so only a trickle of current can get through.
V = Ig (G + R) ⇒ R = V / Ig − G
The total resistance G + R is now very large — exactly what a voltmeter needs.
Compare the two dots crawling through here with the crowd streaming through the shunt above. Shunt = parallel = small. Series resistance = large. They do opposite jobs.
The memory hook that prevents every mix-up in this section:
An ammeter must be invisible to the circuit, so it needs almost no resistance → small shunt in parallel.
A voltmeter must be invisible to the current, so it needs almost infinite resistance → large resistance in series.
9. The range-multiplier shortcut
Many questions say "extend the range n times". If n = I / Ig for an ammeter, or n = V / (IgG) for a voltmeter, the formulas simplify beautifully:
Instrument
Shortcut
Ammeter (range × n)
S = G / (n − 1)
Voltmeter (range × n)
R = G (n − 1)
Notice they are mirror images — one divides, the other multiplies.
The two bars move in opposite directions as the range grows. One formula divides by (n − 1), the other multiplies by it — perfect mirror images, so remembering one gives you the other.
10. Deflection-change problems
A whole family of questions says: "connecting a resistance in parallel reduced the deflection from 50 divisions to 10 — find G." The key idea is one line:
Deflection is proportional to the current through the galvanometer. So the ratio of the two deflections equals the fraction of current the galvanometer keeps:
θ₂ / θ₁ = S / (S + G)
Adding the shunt sends most of the current the other way, so the needle falls back in exactly the same proportion. Set the ratio of deflections equal to the fraction of current kept and G drops out in one step.
11. The ideal instruments
Instrument
Connected
Ideal resistance
Why
Ammeter
In series
Zero
So it does not reduce the current it is measuring
Voltmeter
In parallel
Infinite
So it does not draw current from the circuit
Galvanometer
—
Moderate
Detects small currents; too sensitive to use directly
Watch where the dots go. The ammeter sits in the road, so it must not block it. The voltmeter sits off to one side, so nothing should turn up into it. Opposite jobs, opposite ideal resistances.
Part 2 — Formula sheet
The whole of section 4.10 on one page. Every question in Part 3 uses one of these.
Quantity
Formula
Watch out for
Working principle
N I A B = k θ
Magnetic torque balanced by the spring.
Deflection
θ = (N A B / k) I
θ ∝ I ⇒ linear scale.
Current sensitivity
θ / I = N A B / k
Units: divisions per ampere (or per mA).
Voltage sensitivity
θ / V = N A B / k G
Note the extra G in the denominator.
Relation between the two
Vs = Is / G ⇒ G = Is / Vs
Handy for finding G when both are given.
Shunt for an ammeter
S = Ig G / (I − Ig)
S is small and goes in parallel.
Shunt, range-multiplier form
S = G / (n − 1), n = I / Ig
Faster when the question says "range × n".
Current split
Ig / I = S / (S + G)
Also equals the ratio of deflections.
Current through the shunt
Is / I = G / (S + G)
The two fractions add up to 1.
Resistance of the ammeter
RA = G S / (G + S)
Always less than S — very small.
Series resistance for a voltmeter
R = V / Ig − G
R is large and goes in series.
Series R, range-multiplier form
R = G (n − 1), n = V / (IgG)
Mirror image of the shunt formula.
Resistance of the voltmeter
RV = G + R
Very large.
Deflection ratio
θ₂ / θ₁ = S / (S + G)
Used to find an unknown G.
Full-scale voltage
Vg = Ig G
Often given instead of Ig.
Ideal ammeter
R = 0
Connected in series.
Ideal voltmeter
R = ∞
Connected in parallel.
The one distinction that decides most marks here.
Ammeter
Voltmeter
What you add
Shunt S
Resistance R
How it is connected
Parallel
Series
Formula
S = IgG / (I − Ig)
R = V/Ig − G
Size of the added resistance
Very small
Very large
Effect on total resistance
Decreases it
Increases it
Instrument goes into the circuit
In series
In parallel
Notice the neat reversal: the shunt is in parallel but the ammeter goes into the circuit in series; the extra resistance is in series but the voltmeter goes into the circuit in parallel.
Part 3 — 50 questions with step-by-step solutions
Attempt each one on paper first, then open the solution. Every solution follows the same five steps: Given → Asked → Concept → Formula → Solution.
Q1Working principle
In a moving coil galvanometer, the deflection θ of the coil carrying current I is related to I by:
(a) θ = k I / NAB
(b) θ = NAB I / k
(c) θ = NAB / kI
(d) θ = k NAB I
Show step-by-step solution
GivenCoil of N turns, area A, radial field B, torsional constant k
AskedRelation between θ and I
ConceptAt rest the magnetic torque is exactly balanced by the spring's restoring torque.
FormulaN I A B = k θ
SolutionMagnetic torque on the coil = N I A B (the radial field keeps sin θ = 1). Restoring torque of the spring = k θ. At equilibrium these are equal: N I A B = k θ θ = N A B I / k, so θ ∝ I.
Answer: θ = NAB I / k
Q2Working principle
The principle on which a moving coil galvanometer works is:
(a) a current-carrying coil in a magnetic field experiences a torque
(b) a current-carrying coil in a magnetic field experiences a net force
(c) a magnetic field is produced by a current
(d) a moving magnet induces a current
Show step-by-step solution
GivenMoving coil galvanometer
AskedUnderlying principle
ConceptThe coil rotates, which means a turning effect — not a net force.
Formulaτ = N I A B
SolutionA current loop in a magnetic field feels a torque but no net force. That torque turns the coil against the spring. So the principle is the torque on a current-carrying coil.
Answer: a current-carrying coil in a magnetic field experiences a torque
Q3Radial field
The magnetic field in a moving coil galvanometer is made radial so that:
(a) the coil rotates faster
(b) the scale becomes linear
(c) the resistance decreases
(d) the coil does not overheat
Show step-by-step solution
GivenCurved pole pieces and a soft iron core
AskedPurpose of the radial field
ConceptIn a radial field the plane of the coil always stays along B, so sin θ = 1 no matter how far it turns.
Formulaτ = N I A B sin θ, with sin θ = 1 always
SolutionIn an ordinary field, τ would depend on sin θ and change as the coil turned. A radial field keeps the coil plane always along B, so sin θ = 1 at every position. Then τ = NIAB throughout, giving θ ∝ I — a uniform, linear scale.
Answer: the scale becomes linear
Q4Soft iron core
The soft iron core in a moving coil galvanometer serves to:
(a) reduce the resistance of the coil
(b) make the field radial and stronger
(c) support the spring
(d) insulate the coil
Show step-by-step solution
GivenSoft iron cylinder placed inside the coil
AskedFunction of the core
ConceptThe core shapes the field lines and concentrates them.
FormulaRadial field, increased B
SolutionIt makes the field radial, keeping sin θ = 1 and the scale linear. It also concentrates the field lines, making B stronger and the instrument more sensitive. Both effects together are its purpose.
Answer: make the field radial and stronger
Q5Working principle
In a moving coil galvanometer, the magnetic torque on the coil is balanced by:
(a) the weight of the coil
(b) the restoring torque of the spring
(c) the force of the magnet
(d) friction at the pivot
Show step-by-step solution
GivenCoil at rest at a steady deflection
AskedWhat balances the magnetic torque
ConceptThe coil settles where the two competing turning effects match exactly.
FormulaN I A B = k θ
SolutionThe current produces a magnetic torque NIAB that turns the coil. The spring produces a restoring torque kθ that opposes the turning. The coil stops where these two are equal.
Answer: the restoring torque of the spring
Q6Working principle
The scale of a moving coil galvanometer is linear because:
(a) the coil has many turns
(b) the deflection is proportional to the current
(c) the spring is very stiff
(d) the magnet is permanent
Show step-by-step solution
Givenθ = NABI/k
AskedReason for the linear scale
ConceptLook at the relation between θ and I — it is a simple direct proportion.
Formulaθ = (N A B / k) I
SolutionWith N, A, B and k all constant, θ = constant × I. So doubling the current doubles the deflection. Equal current steps give equal deflection steps — a linear scale.
Answer: the deflection is proportional to the current
Q7Sensitivity
The current sensitivity of a moving coil galvanometer is given by:
(a) NAB / k
(b) NAB / kG
(c) k / NAB
(d) kG / NAB
Show step-by-step solution
GivenCoil of N turns, area A, field B, torsional constant k
AskedExpression for current sensitivity
ConceptCurrent sensitivity means deflection per unit current — rearrange θ = NABI/k.
FormulaCurrent sensitivity = θ / I
SolutionFrom θ = NABI/k, divide both sides by I. θ / I = N A B / k
Answer: NAB / k
Q8Sensitivity
The voltage sensitivity of a moving coil galvanometer of resistance G is:
(a) NAB / k
(b) NAB / kG
(c) NABG / k
(d) kG / NAB
Show step-by-step solution
GivenGalvanometer of resistance G
AskedExpression for voltage sensitivity
ConceptVoltage sensitivity is deflection per unit voltage. Use V = IG to convert.
FormulaVoltage sensitivity = θ / V = θ / (I G)
Solutionθ / V = (θ / I) × (1/G) = (N A B / k) × (1/G) = N A B / k G
Answer: NAB / kG
Q9Sensitivity · concept
If the number of turns in a galvanometer coil is doubled, then:
(a) both current and voltage sensitivity double
(b) current sensitivity doubles but voltage sensitivity may not change
(c) voltage sensitivity doubles but current sensitivity does not change
(d) neither changes
Show step-by-step solution
GivenNumber of turns N doubled
AskedEffect on the two sensitivities
ConceptMore turns means more wire, which means more resistance G. G appears in the voltage sensitivity but not in the current sensitivity.
FormulaI_s = NAB/k; V_s = NAB/kG
SolutionCurrent sensitivity ∝ N, so doubling N doubles it. But doubling the turns also roughly doubles the coil resistance G. Voltage sensitivity = NAB/kG, and both N and G double — so the two effects cancel. Hence voltage sensitivity may not increase at all.
Answer: current sensitivity doubles but voltage sensitivity may not change
Q10Sensitivity · relation
A galvanometer has current sensitivity 10 divisions per mA and voltage sensitivity 2 divisions per mV. Its resistance is:
(a) 2 Ω
(b) 5 Ω
(c) 10 Ω
(d) 20 Ω
Show step-by-step solution
GivenCurrent sensitivity = 10 div/mA, voltage sensitivity = 2 div/mV
AskedGalvanometer resistance G
ConceptVoltage sensitivity is current sensitivity divided by G, so dividing one by the other gives G.
A moving coil galvanometer has 150 equal divisions, current sensitivity 10 divisions per mA and voltage sensitivity 2 divisions per mV. To make each division read 1 volt, the resistance to be connected in series is:
(a) 9995 Ω
(b) 99995 Ω
(c) 10005 Ω
(d) 995 Ω
Show step-by-step solution
Given150 divisions, I_s = 10 div/mA, V_s = 2 div/mV, each division to read 1 V
AskedSeries resistance R
ConceptFind G first from the two sensitivities, then the full-scale current, then use the voltmeter formula.
FormulaG = I_s/V_s; R = V/I_g − G
SolutionG = 10/2 = 5 Ω Full-scale current: 150 divisions ÷ 10 div/mA = 15 mA = 0.015 A Each division reads 1 V, so full scale = 150 V R = 150/0.015 − 5 = 10000 − 5 = 9995 Ω
Answer: 9995 Ω
Q12Sensitivity · concept
To increase the current sensitivity of a galvanometer, one should:
(a) increase the torsional constant k
(b) decrease the area of the coil
(c) increase the number of turns N
(d) decrease the magnetic field B
Show step-by-step solution
GivenCurrent sensitivity = NAB/k
AskedHow to increase it
ConceptN, A and B are on top, and k is on the bottom — so increase the first three or decrease k.
FormulaI_s = N A B / k
SolutionIncreasing N, A or B increases the sensitivity. Increasing k (a stiffer spring) decreases it. So increasing the number of turns is correct.
Answer: increase the number of turns N
Q13Ammeter
A galvanometer is converted into an ammeter by connecting:
(a) a high resistance in series
(b) a low resistance in parallel
(c) a low resistance in series
(d) a high resistance in parallel
Show step-by-step solution
GivenConversion of a galvanometer into an ammeter
AskedWhat is connected and how
ConceptAn ammeter goes in series in a circuit and must have very low resistance, so the extra path must be in parallel.
FormulaS = I_g G / (I − I_g)
SolutionMost of the current must bypass the delicate galvanometer. A small resistance in parallel (a shunt) provides that bypass. It also makes the combined resistance very small, as an ammeter requires.
Answer: a low resistance in parallel
Q14Ammeter · numeric
A galvanometer of resistance 100 Ω gives full-scale deflection for 1 mA. To convert it into an ammeter reading up to 10 A, the shunt required is about:
(a) 0.01 Ω
(b) 0.1 Ω
(c) 1 Ω
(d) 10 Ω
Show step-by-step solution
GivenG = 100 Ω, I_g = 1 mA = 10⁻³ A, I = 10 A
AskedShunt resistance S
ConceptThe voltage across the galvanometer equals that across the shunt, since they are in parallel.
A millivoltmeter of range 25 mV is to be converted into an ammeter of range 25 A. The shunt required is:
(a) 0.001 Ω
(b) 0.01 Ω
(c) 1 Ω
(d) 0.05 Ω
Show step-by-step solution
GivenFull-scale voltage across the instrument = 25 mV, new range I = 25 A
AskedShunt resistance S
ConceptThe full-scale voltage across the galvanometer is already given, so the shunt simply has to carry 25 A at that same voltage.
FormulaS = V_g / I (approximately, since I ≫ I_g)
SolutionThe shunt is in parallel, so it has the same 25 mV across it. It must carry essentially the whole 25 A. S = 25 × 10⁻³ / 25 = 1 × 10⁻³ = 0.001 Ω
Answer: 0.001 Ω
Q17Ammeter · numeric
A galvanometer of resistance 240 Ω allows only 4% of the main current to pass through it after a shunt is connected. The shunt resistance is:
(a) 5 Ω
(b) 8 Ω
(c) 10 Ω
(d) 20 Ω
Show step-by-step solution
GivenG = 240 Ω, I_g / I = 4% = 0.04
AskedShunt resistance S
ConceptUse the current-division formula rather than the shunt formula — it is faster when a percentage is given.
FormulaI_g / I = S / (S + G)
Solution0.04 = S / (S + 240) 0.04S + 9.6 = S 9.6 = 0.96S S = 10 Ω
Answer: 10 Ω
Q18Ammeter · numeric
A galvanometer of resistance 36 Ω is changed into an ammeter using a shunt of 4 Ω. The fraction of the total current passing through the galvanometer is:
(a) 1/40
(b) 1/10
(c) 1/4
(d) 1/140
Show step-by-step solution
GivenG = 36 Ω, S = 4 Ω
AskedFraction I_g / I
ConceptCurrent divides between two parallel paths in inverse proportion to their resistances.
FormulaI_g / I = S / (S + G)
SolutionI_g / I = 4 / (4 + 36) = 4 / 40 = 1/10
Answer: 1/10
Q19Ammeter · numeric
The resistance of an ammeter is 13 Ω and it reads up to 100 A. To extend its range to 750 A, the shunt required is:
(a) 1 Ω
(b) 2 Ω
(c) 4 Ω
(d) 6.5 Ω
Show step-by-step solution
GivenG = 13 Ω, I_g = 100 A, I = 750 A
AskedShunt resistance S
ConceptThe existing ammeter now plays the role of the galvanometer, taking 100 A while the shunt takes the rest.
A moving coil galvanometer of resistance 100 Ω is used as an ammeter with a shunt of 0.1 Ω. Its maximum deflection current is 100 mA. The minimum circuit current that produces full-scale deflection is:
(a) 100 A
(b) 100.1 A
(c) 1001 A
(d) 10.01 A
Show step-by-step solution
GivenG = 100 Ω, S = 0.1 Ω, I_g = 100 mA = 0.1 A
AskedTotal current I for full-scale deflection
ConceptRearrange the current-division relation to find the total current when the galvanometer carries its maximum.
A galvanometer of resistance 100 Ω is shunted by 0.1 Ω. The effective resistance of the resulting ammeter is about:
(a) 0.0999 Ω
(b) 0.1 Ω
(c) 100.1 Ω
(d) 50 Ω
Show step-by-step solution
GivenG = 100 Ω, S = 0.1 Ω in parallel
AskedEffective resistance of the ammeter
ConceptThe two are in parallel, so use the parallel combination formula. The result is always slightly less than the smaller resistance.
FormulaR_A = G S / (G + S)
SolutionR_A = (100 × 0.1) / (100 + 0.1) = 10 / 100.1 ≈ 0.0999 Ω — very small, as an ammeter should be.
Answer: 0.0999 Ω
Q24Ammeter · concept
An ammeter should have very low resistance because:
(a) it is connected in parallel and must not draw current
(b) it is connected in series and must not reduce the circuit current
(c) it protects the galvanometer
(d) it increases the sensitivity
Show step-by-step solution
GivenAmmeter in a circuit
AskedReason for low resistance
ConceptAn ammeter is inserted into the circuit itself, so any resistance it adds changes the very current it is measuring.
FormulaR_A = GS / (G + S), very small
SolutionAn ammeter is connected IN SERIES, so the whole current passes through it. If it had significant resistance, it would reduce that current. So it must have as little resistance as possible — ideally zero.
Answer: it is connected in series and must not reduce the circuit current
Q25Ammeter · concept
When a shunt is connected across a galvanometer, the combined resistance:
(a) increases
(b) decreases
(c) stays the same
(d) becomes infinite
Show step-by-step solution
GivenShunt connected in parallel with a galvanometer
AskedEffect on the total resistance
ConceptAdding any resistance in parallel always lowers the combined resistance below the smaller of the two.
FormulaR = GS / (G + S)
SolutionParallel combination is always smaller than either individual resistance. Since S is very small, the combination is smaller still. So the resistance decreases.
Answer: decreases
Q26Ammeter · concept
An ideal ammeter has a resistance of:
(a) zero
(b) infinity
(c) equal to G
(d) equal to S
Show step-by-step solution
GivenIdeal ammeter
AskedIts resistance
ConceptIt must not disturb the circuit at all when inserted in series.
FormulaR_A → 0
SolutionAn ammeter goes in series with the circuit. Any resistance would reduce the current being measured. So the ideal value is zero.
Answer: zero
Q27Voltmeter
A galvanometer is converted into a voltmeter by connecting:
(a) a low resistance in parallel
(b) a high resistance in series
(c) a high resistance in parallel
(d) a low resistance in series
Show step-by-step solution
GivenConversion of a galvanometer into a voltmeter
AskedWhat is connected and how
ConceptA voltmeter goes in parallel across a component and must draw almost no current, so it needs a very large total resistance.
FormulaR = V / I_g − G
SolutionA large resistance in series restricts the current to a trickle. It also makes the total resistance very large, as a voltmeter requires. So: a high resistance in series.
Answer: a high resistance in series
Q28Voltmeter · numeric
A galvanometer has a coil resistance of 100 Ω and gives full-scale deflection for 30 mA. To work as a voltmeter of range 30 V, the resistance to be added in series is:
(a) 500 Ω
(b) 900 Ω
(c) 1000 Ω
(d) 1800 Ω
Show step-by-step solution
GivenG = 100 Ω, I_g = 30 mA = 0.03 A, V = 30 V
AskedSeries resistance R
ConceptThe full voltage must be shared between the galvanometer and the added resistance, with only I_g flowing.
FormulaR = V / I_g − G
SolutionV / I_g = 30 / 0.03 = 1000 Ω (this is the total needed) R = 1000 − 100 R = 900 Ω
Answer: 900 Ω
Q29Voltmeter · numeric
A galvanometer of 50 Ω resistance has 25 divisions. A current of 4 × 10⁻⁴ A gives a deflection of one division. To convert it into a voltmeter of range 25 V, the resistance required is:
(a) 2450 Ω
(b) 2500 Ω
(c) 2550 Ω
(d) 245 Ω
Show step-by-step solution
GivenG = 50 Ω, 25 divisions, 4 × 10⁻⁴ A per division, V = 25 V
AskedSeries resistance R
ConceptFirst find the full-scale current from the per-division figure, then apply the voltmeter formula.
FormulaI_g = (current per division) × (number of divisions); R = V/I_g − G
SolutionI_g = 25 × 4 × 10⁻⁴ = 10⁻² A = 0.01 A V / I_g = 25 / 0.01 = 2500 Ω R = 2500 − 50 = 2450 Ω
Answer: 2450 Ω
Q30Voltmeter · numeric
A galvanometer of resistance 80 Ω gives full-scale deflection for a potential difference of 20 mV across it. The resistance required to convert it into a voltmeter of range 5 V is:
(a) 19920 Ω
(b) 20000 Ω
(c) 19980 Ω
(d) 24920 Ω
Show step-by-step solution
GivenG = 80 Ω, full-scale voltage across G = 20 mV, V = 5 V
AskedSeries resistance R
ConceptThe full-scale voltage is given instead of the current, so find I_g from Ohm's law first.
A galvanometer of resistance G is converted into a voltmeter whose range is n times the voltage that would give full-scale deflection on its own. The series resistance required is:
(a) G / (n − 1)
(b) G (n − 1)
(c) nG
(d) G / n
Show step-by-step solution
GivenRange multiplied by n
AskedSeries resistance R in terms of G and n
ConceptSubstitute V = n I_g G into the voltmeter formula.
FormulaR = V / I_g − G
SolutionR = (n I_g G) / I_g − G = nG − G = G (n − 1) Note this is the mirror image of the shunt formula S = G/(n − 1).
Answer: G (n − 1)
Q32Voltmeter · concept
A voltmeter should have very high resistance because:
(a) it is connected in series and must not reduce the current
(b) it is connected in parallel and must not draw current from the circuit
(c) it protects the shunt
(d) it makes the scale linear
Show step-by-step solution
GivenVoltmeter in a circuit
AskedReason for high resistance
ConceptA voltmeter is placed across a component; if it drew significant current it would change the voltage it is measuring.
FormulaR_V = G + R, very large
SolutionA voltmeter is connected IN PARALLEL across the component. If it drew appreciable current, it would alter the circuit and the reading. So it must have a very high resistance — ideally infinite.
Answer: it is connected in parallel and must not draw current from the circuit
Q33Voltmeter · concept
An ideal voltmeter has a resistance of:
(a) zero
(b) infinity
(c) equal to G
(d) equal to R
Show step-by-step solution
GivenIdeal voltmeter
AskedIts resistance
ConceptIt must draw no current at all from the circuit it measures.
FormulaR_V → ∞
SolutionA voltmeter is connected in parallel. Infinite resistance means it draws no current and disturbs nothing. So the ideal value is infinity.
Answer: infinity
Q34Voltmeter · numeric
A voltmeter of resistance 1000 Ω gives full-scale deflection when 100 mA flows through it. The shunt required to convert it into an ammeter reading 1 A at full scale is:
(a) 100 Ω
(b) 111.1 Ω
(c) 11.1 Ω
(d) 1.11 Ω
Show step-by-step solution
GivenG = 1000 Ω, I_g = 100 mA = 0.1 A, new range I = 1 A
AskedShunt resistance S
ConceptAlthough the instrument is called a voltmeter, for this conversion it plays the role of the galvanometer.
When a 12 Ω resistor is connected in parallel with a moving coil galvanometer, its deflection falls from 50 divisions to 10 divisions. The resistance of the galvanometer is:
(a) 12 Ω
(b) 24 Ω
(c) 48 Ω
(d) 60 Ω
Show step-by-step solution
GivenS = 12 Ω, deflection falls from 50 to 10 divisions
AskedGalvanometer resistance G
ConceptDeflection is proportional to the current through the galvanometer, so the ratio of deflections equals the fraction of current it keeps.
On connecting a shunt of 10 Ω, the deflection in a moving coil galvanometer falls from 40 divisions to 6 divisions. The galvanometer resistance is about:
(a) 36.7 Ω
(b) 46.7 Ω
(c) 56.7 Ω
(d) 66.7 Ω
Show step-by-step solution
GivenS = 10 Ω, deflection falls from 40 to 6 divisions
AskedGalvanometer resistance G
ConceptSame method — set the deflection ratio equal to the current-division fraction.
A galvanometer of resistance 50 Ω is connected to a 3 V battery through a series resistance of 2950 Ω, giving a full-scale deflection of 30 divisions. To reduce the deflection to 20 divisions, the extra resistance to be added in series is:
(a) 1050 Ω
(b) 1500 Ω
(c) 1550 Ω
(d) 2050 Ω
Show step-by-step solution
GivenG = 50 Ω, series R = 2950 Ω, V = 3 V, deflection 30 → 20 divisions
AskedExtra series resistance
ConceptCurrent is proportional to deflection. Find the new current, then the new total resistance, then subtract what is already there.
FormulaI ∝ θ; R_total = V / I
SolutionOriginal total resistance = 50 + 2950 = 3000 Ω, so I₁ = 3/3000 = 1 mA for 30 divisions. For 20 divisions: I₂ = 1 × (20/30) = 2/3 mA New total resistance = 3 / (2/3 × 10⁻³) = 4500 Ω Extra resistance = 4500 − 3000 = 1500 Ω
Answer: 1500 Ω
Q38Shunt · series combination
A galvanometer of resistance G is shunted by a resistance S. The resistance that must be added in series with this combination to keep the main circuit current unchanged is:
(a) G² / (G + S)
(b) GS / (G + S)
(c) G + S
(d) S² / (G + S)
Show step-by-step solution
GivenGalvanometer G shunted by S, main current to stay unchanged
AskedSeries resistance to be added
ConceptThe combination has a lower resistance than G alone, so add just enough to bring the total back up to G.
FormulaR_parallel = GS/(G + S)
SolutionOriginal resistance in the circuit = G After shunting = GS/(G + S) Extra needed = G − GS/(G + S) = [G(G + S) − GS] / (G + S) = G² / (G + S)
Answer: G² / (G + S)
Q39Comparison
Which of the following correctly describes how the instruments are connected in a circuit?
(a) Ammeter in parallel, voltmeter in series
(b) Ammeter in series, voltmeter in parallel
(c) Both in series
(d) Both in parallel
Show step-by-step solution
GivenUse of an ammeter and a voltmeter
AskedCorrect connections
ConceptAn ammeter must carry the current being measured; a voltmeter must sit across the potential difference being measured.
FormulaAmmeter: series; Voltmeter: parallel
SolutionTo measure current, the ammeter must be in the current's path ⇒ series. To measure potential difference, the voltmeter must be across the two points ⇒ parallel. (Note the reversal: the shunt is in parallel with the galvanometer, but the ammeter is in series with the circuit.)
Answer: Ammeter in series, voltmeter in parallel
Q40Comparison
Which quantity is the SAME for the galvanometer and its shunt?
(a) the current through them
(b) the potential difference across them
(c) their resistances
(d) their power dissipation
Show step-by-step solution
GivenGalvanometer and shunt connected in parallel
AskedThe common quantity
ConceptComponents in parallel always share the same potential difference; that is exactly the relation used to derive the shunt formula.
FormulaI_g G = (I − I_g) S
SolutionThey are connected in parallel. Parallel components have the same potential difference across them. That is why I_g G = (I − I_g) S.
Answer: the potential difference across them
Q41Concept
A galvanometer cannot be used directly as an ammeter because:
(a) it has zero resistance
(b) it is too sensitive and would be damaged by large currents
(c) it gives a non-linear scale
(d) it works only with alternating current
Show step-by-step solution
GivenGalvanometer used directly in a circuit
AskedReason it cannot be used as an ammeter
ConceptThe galvanometer is designed to detect tiny currents, so a large current would drive it far past full scale.
FormulaS = I_g G / (I − I_g)
SolutionA galvanometer gives full-scale deflection for a very small current I_g. A large circuit current would damage the coil. It also has a significant resistance, which would disturb the circuit. So a shunt is added to fix both problems.
Answer: it is too sensitive and would be damaged by large currents
Q42Concept · numeric
In an ammeter, 0.2% of the main current passes through the galvanometer of resistance G. The resistance of the ammeter is approximately:
(a) G / 500
(b) G / 499
(c) 500 G
(d) 499 G
Show step-by-step solution
GivenI_g / I = 0.2% = 0.002, galvanometer resistance G
AskedResistance of the ammeter
ConceptFind S from the current-division relation, then compute the parallel combination.
FormulaI_g/I = S/(S+G); R_A = GS/(G+S)
Solution0.002 = S/(S + G) ⇒ 0.002S + 0.002G = S ⇒ S = 0.002G/0.998 ≈ G/499 R_A = GS/(G + S). Since S ≪ G, R_A ≈ S More precisely R_A = I_g G / I = 0.002 G = G/500
Answer: G / 500
Q43Assertion–Reason
Assertion (A): To convert a galvanometer into an ammeter, a small resistance is connected in parallel with it. Reason (R): The small resistance increases the combined resistance of the combination.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenConversion of a galvanometer into an ammeter
AskedTruth of A and R
ConceptCheck the second statement carefully — a parallel resistance always lowers the combined value.
FormulaR_A = GS / (G + S)
SolutionA is TRUE: a shunt (small resistance in parallel) is indeed used. R is FALSE: a parallel resistance DECREASES the combined resistance, it does not increase it. So A is true but R is false.
Answer: A is true but R is false
Q44Assertion–Reason
Assertion (A): The magnetic field in a moving coil galvanometer is made radial. Reason (R): A radial field keeps the plane of the coil always parallel to the field, so the torque is NIAB at all positions.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenRadial field in a galvanometer
AskedTruth of A and R, and whether R explains A
ConceptBoth statements are correct, and the second gives exactly the reason for the first.
Formulaτ = NIAB sin θ, with sin θ = 1 in a radial field
SolutionA is TRUE: the field is made radial using curved poles and a soft iron core. R is TRUE: in a radial field the coil plane always stays along B, so sin θ = 1 and τ = NIAB. That constant torque is precisely WHY the radial field is used — it gives a linear scale. So R correctly explains A.
Answer: Both A and R are true, and R is the correct explanation of A
Q45Assertion–Reason
Assertion (A): Increasing the number of turns of a galvanometer coil always increases its voltage sensitivity. Reason (R): Voltage sensitivity is given by NAB/kG.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is false but R is true
(d) Both A and R are false
Show step-by-step solution
GivenEffect of increasing N on voltage sensitivity
AskedTruth of A and R
ConceptThe formula in R is right, but it is exactly what shows the assertion to be wrong, because G also grows with N.
FormulaV_s = N A B / k G
SolutionR is TRUE: voltage sensitivity is indeed NAB/kG. A is FALSE: increasing N also increases the coil resistance G, and the two effects cancel. So A is false but R is true.
Answer: A is false but R is true
Q46Match the following
Match each item with its correct description:
(a) Shunt (b) Series resistance (c) Radial field (d) Soft iron core
(i) Converts a galvanometer into a voltmeter (ii) Converts a galvanometer into an ammeter (iii) Strengthens the field (iv) Makes the scale linear
(a) a–ii, b–i, c–iv, d–iii
(b) a–i, b–ii, c–iv, d–iii
(c) a–ii, b–i, c–iii, d–iv
(d) a–iv, b–iii, c–ii, d–i
Show step-by-step solution
GivenFour components of the instrument
AskedCorrect matching
ConceptEach part has one clear job in the design.
FormulaS = I_gG/(I−I_g); R = V/I_g − G
Solution(a) Shunt ⇒ ammeter conversion ⇒ (ii) (b) Series resistance ⇒ voltmeter conversion ⇒ (i) (c) Radial field ⇒ keeps sin θ = 1 ⇒ linear scale ⇒ (iv) (d) Soft iron core ⇒ concentrates and strengthens the field ⇒ (iii)
Answer: a–ii, b–i, c–iv, d–iii
Q47Match the following
Match each quantity with its expression:
(a) Current sensitivity (b) Voltage sensitivity (c) Shunt for an ammeter (d) Series resistance for a voltmeter
(i) V/I_g − G (ii) NAB/k (iii) I_gG/(I − I_g) (iv) NAB/kG
(a) a–ii, b–iv, c–iii, d–i
(b) a–iv, b–ii, c–iii, d–i
(c) a–ii, b–iv, c–i, d–iii
(d) a–iii, b–i, c–ii, d–iv
Show step-by-step solution
GivenFour standard results from section 4.10
AskedCorrect matching
ConceptThese four expressions cover essentially every numerical in this section.
FormulaStandard galvanometer formulas
Solution(a) Current sensitivity = NAB/k ⇒ (ii) (b) Voltage sensitivity = NAB/kG ⇒ (iv) (c) Shunt = I_gG/(I − I_g) ⇒ (iii) (d) Series resistance = V/I_g − G ⇒ (i)
Answer: a–ii, b–iv, c–iii, d–i
Q48Match the following
Match each instrument with its ideal resistance and connection:
(i) Very large, in series with G (ii) Ideally zero, connected in series (iii) Ideally infinite, connected in parallel (iv) Very small, in parallel with G
(a) a–ii, b–iii, c–iv, d–i
(b) a–iii, b–ii, c–iv, d–i
(c) a–ii, b–iii, c–i, d–iv
(d) a–iv, b–i, c–ii, d–iii
Show step-by-step solution
GivenInstruments and the resistances added to make them
AskedCorrect matching
ConceptKeep separate how the instrument joins the circuit and how the extra resistance joins the galvanometer.
FormulaAmmeter: series, low R; Voltmeter: parallel, high R
Solution(a) Ammeter ⇒ ideally zero resistance, connected in series ⇒ (ii) (b) Voltmeter ⇒ ideally infinite resistance, connected in parallel ⇒ (iii) (c) Shunt ⇒ very small, in parallel with G ⇒ (iv) (d) Voltmeter's added resistance ⇒ very large, in series with G ⇒ (i)
Answer: a–ii, b–iii, c–iv, d–i
Q49Recognition
A question states: 'A galvanometer of resistance 50 Ω gives full-scale deflection for 2 mA. What resistance converts it to read up to 10 V?' Which formula applies?
(a) S = I_gG/(I − I_g)
(b) R = V/I_g − G
(c) θ = NABI/k
(d) R_A = GS/(G + S)
Show step-by-step solution
GivenG = 50 Ω, I_g = 2 mA, target 10 V
AskedThe correct formula
ConceptThe target is given in volts, so this is a voltmeter conversion — series resistance.
FormulaR = V / I_g − G
SolutionThe range is quoted in VOLTS ⇒ voltmeter conversion. Voltmeter ⇒ high resistance in SERIES. R = 10/0.002 − 50 = 5000 − 50 = 4950 Ω
Answer: R = V/I_g − G
Q50Concept · summary
Which of the following statements is INCORRECT?
(a) A shunt is connected in parallel with a galvanometer
(b) A voltmeter has a very high resistance
(c) The deflection of a moving coil galvanometer is proportional to the square of the current
(d) The radial field makes the galvanometer scale linear
Show step-by-step solution
GivenStatements about the moving coil galvanometer
AskedThe incorrect statement
ConceptCheck each against the working relation θ = NABI/k.
Formulaθ = N A B I / k
Solution'Shunt in parallel' — correct. 'Voltmeter has high resistance' — correct. 'Deflection proportional to the SQUARE of the current' — INCORRECT. θ ∝ I, a direct proportion. 'Radial field makes the scale linear' — correct.
Answer: The deflection of a moving coil galvanometer is proportional to the square of the current
Before the next ILTS — how to use this pack
Read Part 1 once. This section is short and mechanical — no vectors, no geometry, no directions to get wrong.
Copy the ammeter-versus-voltmeter table by hand, then cover it and rewrite it. That single table answers about a third of the questions here.
Q1–Q13 (working and sensitivity) in one sitting. Mostly conceptual.
Q14–Q28 (ammeter) next. Learn both routes — the shunt formula and the current-division fraction — because questions give the data in both forms.
Q29–Q37 (voltmeter) the following sitting. Same skeleton, one formula.
Q38–Q50 cover deflection-change problems, assertion–reason and matching. Save for final revision.
The four mistakes that cost the most marks in this topic
Swapping the two conversions. Shunt = small = parallel = ammeter. Extra resistance = large = series = voltmeter.
Forgetting to subtract G in the voltmeter formula. R = V/I_g minus G, not just V/I_g.
Not converting mA to A. A full-scale current of 30 mA is 0.03 A.
Saying more turns always increases voltage sensitivity. It does not — G rises too.
The one-minute self-test. If she can answer these four without hesitating, this section is secure:
Why is the field in a galvanometer made radial?
Ammeter — shunt in parallel or resistance in series? And which way round for a voltmeter?
Why does doubling the number of turns not necessarily double the voltage sensitivity?
A shunt is connected and the deflection drops from 40 to 10 divisions. What single relation do you write down first?