Connect moving-coil physics to practical meter loading, bridge ratios and null measurements.
JEE Main · JEE Advanced · NEET UG85 min study guide9 indexed sections
BY THE END, YOU CAN
Derive moving-coil deflection and distinguish current from voltage sensitivity.
Convert galvanometers to ammeters and voltmeters with correct resistor topology.
Quantify finite-meter loading and compare it with null measurement.
Solve Wheatstone, meter-bridge and potentiometer balances including cell internal resistance.
BEFORE YOU START
Ohm’s law, series/parallel networks and Kirchhoff rules.
Magnetic torque on a coil and linear restoring torque.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. Connecting a meter changes the circuit unless its loading is negligible
An ammeter measures current through a branch and is connected in series with that branch. A voltmeter measures potential difference between two nodes and is connected in parallel with the element or node pair of interest. These connections follow the quantities being measured: series elements share current, while parallel elements share voltage.
An ideal ammeter has zero resistance, so insertion produces no additional voltage drop. An ideal voltmeter has infinite resistance, so it draws no current. Real meters only approximate these limits. The relevant question is how their resistance compares with the circuit, not whether it is merely called “small” or “large” without a reference scale.
A finite-resistance voltmeter can alter a voltage divider by adding a parallel path. A finite-resistance ammeter adds series resistance and reduces the branch current for a fixed source. A displayed reading may be accurate for the altered circuit while differing from the value that existed before connection. Instrument calibration and circuit loading are distinct errors.
An ohmmeter supplies its own test signal and infers resistance from current and voltage. For a valid isolated-component reading, external sources must be removed and stored charge safely discharged under the device’s procedures. Other connected paths can make the meter read an equivalent network rather than the intended resistor. This lesson’s calculations are ideal low-voltage models, not instructions for probing live mains systems.
ORIGINAL WORKED EXAMPLE
Voltmeter loading of a divider
A 12 V ideal source feeds two 6 kΩ resistors in series. A voltmeter of resistance 6 kΩ is connected across the lower resistor. Find the reading and compare with the unloaded voltage.
Before connection, equal divider resistors give 6 V across the lower one.
The meter parallels the lower 6 kΩ resistor, giving equivalent 3 kΩ.
The new divider is 6 kΩ in series with 3 kΩ, so the measured lower voltage is 12(3)/(6 + 3) = 4 V.
Answer The meter reads 4 V; the unloaded node was at 6 V.
02 / UNDERSTAND & APPLY
2. A moving coil converts current into a restoring-torque balance
A moving-coil galvanometer places a coil with N turns and area A in a magnetic field. For a general loop, magnetic torque magnitude is NIAB sinθ, where θ is between the coil’s magnetic moment and the field. A radial-field design keeps the relevant torque factor at its maximum over the working deflection range, giving torque NIAB.
The suspension or spring provides a restoring torque Cθ, with C a torsional constant and θ the mechanical deflection. At steady deflection, Cθ = NIAB, so θ = (NAB/C)I. The linear current–deflection relation is the basis of a uniform analog scale in this ideal design. It is not obtained by assuming every uniform magnetic field is radial.
Current sensitivity is θ/I = NAB/C. Increasing turns, field or coil area, or reducing torsional stiffness, increases this sensitivity in the simple model. Voltage sensitivity is θ/V = NAB/(CG), where G is the coil’s electrical resistance. More turns can also increase G, so improving current sensitivity does not automatically improve voltage sensitivity by the same factor.
Damping controls how quickly the pointer settles. It should reduce oscillation without changing the final static torque balance under the ideal model. The instrument’s full-scale current I_g is the current giving its largest calibrated deflection, not a promise that larger currents are harmless. In converting it to a meter, the added resistors keep coil current within that range.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Local force-direction schematic, not a full radial pole-face drawing. A real radial-field galvanometer shapes the field to keep it parallel to the coil plane as the coil turns, giving torque NBIA balanced by the spring torque.
ORIGINAL WORKED EXAMPLE
Current and voltage sensitivity differ
A galvanometer has N = 100 turns, coil area 2.0 × 10⁻⁴ m², radial field B = 0.20 T, torsional constant C = 4.0 × 10⁻⁶ N m rad⁻¹ and resistance G = 100 Ω. Find current sensitivity, voltage sensitivity and deflection at 0.20 mA.
NAB = 100(2.0 × 10⁻⁴)(0.20) = 0.0040 N m A⁻¹.
Current sensitivity is NAB/C = 0.0040/(4.0 × 10⁻⁶) = 1000 rad A⁻¹.
Voltage sensitivity is current sensitivity/G = 10 rad V⁻¹.
At I = 0.20 mA = 2.0 × 10⁻⁴ A, deflection is 1000(2.0 × 10⁻⁴) = 0.20 rad.
Answer 1000 rad A⁻¹; 10 rad V⁻¹; deflection 0.20 rad.
03 / UNDERSTAND & APPLY
3. A shunt carries current around the sensitive coil
To make a galvanometer into an ammeter of full-scale range I > I_g, connect a small shunt resistance S in parallel with the coil resistance G. At full scale, coil current is I_g and shunt current is I − I_g. Equal branch voltage gives I_gG = (I − I_g)S, hence S = I_gG/(I − I_g).
The total ammeter resistance is G in parallel with S, not simply S unless that is a justified approximation. At full scale its voltage drop is I_gG, so R_A = I_gG/I. This small but nonzero burden voltage can alter the measured circuit. The shunt must also carry and dissipate the appropriate current and power in a real design.
If the current range is n times the galvanometer full-scale current, S = G/(n − 1). The current divides inversely with resistance because both branches share voltage. Placing the shunt in series would not protect the coil from the total branch current; it would perform a different circuit function.
An ammeter must not be connected directly across an ideal voltage source as if it were a voltmeter. Its intentionally low resistance can produce a very large current and damage equipment. In exam schematics, an ideal zero-resistance meter can also short two nodes, fundamentally changing topology rather than merely reporting a pre-existing current.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Ideal conversion schematics. G in the equations denotes coil resistance; the circled G is the galvanometer instrument. Lead resistance is neglected.
ORIGINAL WORKED EXAMPLE
Design a 1 A range
A galvanometer has G = 100 Ω and full-scale current I_g = 1.0 mA. Find the shunt for a 1.0 A ammeter range and the resulting ammeter resistance.
At full scale, coil voltage is I_gG = 0.0010(100) = 0.10 V.
Shunt current is 1.0 − 0.0010 = 0.999 A, so S = 0.10/0.999 ≈ 0.1001 Ω.
Equivalent ammeter resistance is full-scale voltage divided by total current: R_A = 0.10/1.0 = 0.10 Ω.
Answer Shunt approximately 0.100 Ω; ammeter resistance 0.10 Ω.
04 / UNDERSTAND & APPLY
4. A series multiplier limits the coil current at the desired voltage
To make a galvanometer into a voltmeter of full-scale voltage V, connect a resistance R_s in series with the coil. The same current I_g flows through both at full scale, giving V = I_g(G + R_s). Thus R_s = V/I_g − G and total voltmeter resistance R_V = V/I_g.
The desired range must be at least the coil’s own full-scale voltage I_gG if a nonnegative added series resistor is to suffice. The series resistor absorbs most of the applied voltage in a high-range meter. It is not placed in parallel, because a parallel resistor would increase current drawn rather than raise the coil’s safe voltage range in this manner.
For one galvanometer movement, the sensitivity expressed as ohms per volt is R_V/V = 1/I_g. A smaller full-scale current allows larger input resistance for the same voltage range, reducing loading. Different selected ranges can have different input resistances even when the same pointer movement is used.
When a voltmeter with finite R_V measures a source with emf ε and internal resistance r, it reads terminal voltage V = εR_V/(r + R_V) in the simple model. It approaches ε as R_V becomes very large but is below ε for positive r and finite R_V. A null method can determine emf without drawing steady current from the test source at balance.
ORIGINAL WORKED EXAMPLE
Design a 10 V range
A galvanometer has G = 100 Ω and I_g = 1.0 mA. Find the series resistance for a 10 V full-scale voltmeter and its input resistance.
Total resistance must be V/I_g = 10/0.0010 = 10000 Ω.
The completed input resistance is 10 kΩ, and its sensitivity is 1000 Ω/V.
Answer Series resistor 9.9 kΩ; total input resistance 10 kΩ.
05 / UNDERSTAND & APPLY
5. A balanced bridge compares ratios through a zero-current detector
A Wheatstone bridge has two series arms across the same supply. Name one arm A–P–B–Q–C and the other A–R–D–S–C, where P,Q,R,S are resistances and B,D are their midpoints. A galvanometer connects B to D. At null, no detector current flows and the midpoint potentials are equal.
With zero detector current, each arm is an ordinary divider. Equality of midpoint potential gives P/Q = R/S, or PS = QR for this explicitly labeled arrangement. The rule must match the actual diagram; memorizing a ratio without tracking which resistors share an arm can invert the result.
At balance, the detector resistance does not affect the ideal ratio because its current is zero. This is the advantage of a null comparison. Away from balance, the detector loads the network and its resistance matters for actual deflection. A zero reading can also result from a broken or disconnected detector in a real setup, so an operational check is separate from the mathematical null condition.
A meter bridge uses a uniform resistance wire as one divider arm. With left and right test resistances X and R across the same endpoints, and a balance contact l from the left end of a wire of total length L, the matched ordering gives X/R = l/(L − l). Uniform cross section and resistivity make wire resistance proportional to length.
End corrections, contact resistance, heating and wire nonuniformity cause systematic error. Balance near the middle generally improves sensitivity and reduces the relative impact of length-reading error. Interchanging X and R and comparing readings can help reveal some systematic effects, but does not magically remove every imperfection.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Four labelled arms represent resistors; straight sloping segments indicate their connectivity rather than their resistance magnitude. Positive source terminal is connected to A.
ORIGINAL WORKED EXAMPLE
A labeled meter-bridge balance
A uniform 100 cm bridge wire balances at 40 cm from its left end. The left gap contains unknown X and the right gap contains 6.0 Ω. Find X using the stated matched-arm geometry.
At null, X/6.0 = 40/(100 − 40).
The right wire segment is 60 cm, giving X = 6.0(40/60) = 4.0 Ω.
The smaller left resistance corresponds to the shorter left wire segment, a useful consistency check.
Answer 4.0 Ω.
06 / UNDERSTAND & APPLY
6. A potentiometer compares an emf with a known wire potential drop
A laboratory potentiometer maintains a steady current through a uniform long wire using a separate driver source and control resistance. The wire has potential gradient kₚ = voltage drop per unit length. A test cell and galvanometer are connected so that the test emf opposes the selected wire drop; a sliding contact is adjusted until detector current is zero.
At balance, ε_test = kₚ × l. Because no current is drawn from the test cell in the ideal null branch, its internal resistance causes no terminal drop and the balance measures emf. This differs from a finite-resistance voltmeter, which draws current and reads a loaded terminal voltage. The driver circuit still carries current at potentiometer balance.
For two cells compared without changing the wire’s potential gradient, ε₁/ε₂ = l₁/l₂. If the driver current changes between measurements, the ratio of lengths alone is no longer valid. The available wire drop must exceed the emf to be balanced, and the polarity must oppose correctly; otherwise no interior balance point appears.
To determine a cell’s internal resistance r, first obtain an open-circuit balance length l₁ for ε. Then connect a known load R across the cell and measure its terminal-voltage balance length l₂ while the same potentiometer gradient is maintained. Since ε/V = l₁/l₂ and V = εR/(R + r), r = R(l₁/l₂ − 1).
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Topology schematic with driver voltage sufficient for an interior balancing point. Wire gradient k is held fixed and contact resistance is irrelevant only at an exact null. The separated dashed AJ length guide is an annotation, not a conducting wire.
ORIGINAL WORKED EXAMPLE
Internal resistance from two null lengths
A cell balances at 80 cm with no external load. With a 6.0 Ω load across it, its terminal voltage balances at 60 cm on the same unchanged potentiometer gradient. Find cell internal resistance.
The ratio ε/V is l₁/l₂ = 80/60 = 4/3.
Use r = R(ε/V − 1) = 6.0(4/3 − 1).
The result is 2.0 Ω. The shorter loaded balance length is consistent with terminal voltage below emf.
Answer 2.0 Ω.
07 / UNDERSTAND & APPLY
7. Sensitivity, resolution and loading are separate qualities
Instrument sensitivity describes change in indication per change in the measured quantity. Resolution describes the smallest discernible increment, while accuracy concerns agreement with the true value under the measurement conditions. A very sensitive galvanometer can still give a biased result if its connection changes the circuit or its zero is offset.
For potentiometry, a smaller potential gradient gives a longer balance length for the same voltage and can improve fractional length-reading precision, provided the wire remains long enough for a balance. Reducing the gradient too far loses the ability to balance the cell within the available length. This is a range–sensitivity tradeoff, not an unlimited improvement.
For the meter-bridge relation X = Rl/(L − l), a small balance-length uncertainty δl produces approximate relative uncertainty δX/X ≈ [1/l + 1/(L − l)]δl when R and L are treated as exact. This expression is smallest near l = L/2 and grows sharply near the ends, explaining the central-balance preference quantitatively.
Contact heating changes wire resistivity and can spoil a uniform potential gradient. Detector zero checks, stable driver current and brief appropriate contact help preserve the model in a controlled teaching experiment. Real procedures should follow the instrument’s rating and instructions; the algebra does not authorize arbitrary connections or overrange currents.
08 / UNDERSTAND & APPLY
8. Original practice with topology and null conditions
Name the meter’s terminals or the bridge ordering before applying a formula. Draw which elements share a current and which share a voltage, then check whether the measuring branch carries current at the stated balance.
ORIGINAL PRACTICE · 1
How is an ammeter connected to measure one branch current?
Reveal answer & reasoning
In series with that branch.
It must carry the same current as the branch, while having low resistance to reduce disturbance.
ORIGINAL PRACTICE · 2
How is a voltmeter connected across a resistor?
Reveal answer & reasoning
In parallel between the resistor’s two terminals.
It measures the same potential difference and should draw little current.
ORIGINAL PRACTICE · 3
A galvanometer has G = 50 Ω and I_g = 2 mA. Find its full-scale coil voltage.
Reveal answer & reasoning
0.10 V.
I_gG = 0.002(50).
ORIGINAL PRACTICE · 4
To increase a galvanometer current range by a factor ten, what shunt is required?
Reveal answer & reasoning
G/9.
S = G/(n − 1), with n = 10.
ORIGINAL PRACTICE · 5
A 1 mA movement is used for a 5 V voltmeter. Find total input resistance.
Reveal answer & reasoning
5 kΩ.
R_V = V/I_g = 5/0.001.
ORIGINAL PRACTICE · 6
Does doubling the number of coil turns necessarily double voltage sensitivity?
Reveal answer & reasoning
No.
Current sensitivity increases if other magnetic/mechanical factors stay fixed, but coil resistance may also increase.
ORIGINAL PRACTICE · 7
In the explicitly labeled bridge, P/Q = 2 and R = 6 Ω. Find S at balance.
Reveal answer & reasoning
3 Ω.
P/Q = R/S gives 2 = 6/S.
ORIGINAL PRACTICE · 8
A 100 cm meter bridge balances at 50 cm with a 7 Ω standard. Find the unknown in the other matched gap.
Reveal answer & reasoning
7 Ω.
Equal wire lengths imply equal gap resistances.
ORIGINAL PRACTICE · 9
A potentiometer gradient is 0.020 V cm⁻¹. A cell balances at 60 cm. Find emf.
Reveal answer & reasoning
1.2 V.
ε = kₚ × l = 0.020(60).
ORIGINAL PRACTICE · 10
Two cells balance at 45 and 60 cm with unchanged gradient. Find their emf ratio.
Reveal answer & reasoning
3/4.
ε₁/ε₂ = l₁/l₂ = 45/60.
ORIGINAL PRACTICE · 11
At potentiometer null, is the driver-wire current zero?
Reveal answer & reasoning
No.
Only the test-cell detector branch has zero current; the driver maintains the wire potential gradient.
ORIGINAL PRACTICE · 12
A test cell’s emf exceeds the total driver-wire drop. Can it have a normal balance point on that wire?
Reveal answer & reasoning
No.
No available segment supplies the required opposing voltage.
ORIGINAL PRACTICE · 13
A cell has open and loaded balance lengths 90 and 60 cm with load 4 Ω. Find internal resistance.
Reveal answer & reasoning
2 Ω.
r = 4(90/60 − 1) = 4(0.5).
ORIGINAL PRACTICE · 14
Why should an ohmmeter not measure resistance in a powered external circuit?
Reveal answer & reasoning
Its own test source would interact with the external source and the reading may be invalid or damaging.
Resistance mode assumes the appropriate unpowered, isolated measurement conditions specified by the instrument.
09 / UNDERSTAND & APPLY
9. Read the circuit as part of the measurement
First identify whether the question concerns coil deflection, an ammeter conversion, a voltmeter conversion or a null comparison. For conversions, preserve the movement’s full-scale current and use the correct series or parallel equality. For null methods, write the equal-potential condition using the actual labels rather than an unlabeled memorized ratio.
Finally compare the meter resistance with the surrounding circuit and verify the operating conditions: stable driver gradient, correct polarity, a reachable balance and no double-counted resistor. State the approximations behind negligible loading. A measurement equation is reliable only when the circuit actually implements its assumptions.
Ammeter: series connection and low resistance.
Voltmeter: parallel connection and high resistance.
A radial-field moving coil has a linear torque–current relation in its working range.
Shunts are parallel; voltage multipliers are series.
Bridge and potentiometer nulls compare equal potentials with zero detector current.
Potentiometer emf measurement does not require zero current in the driver wire.
KEEP THE BIG IDEAS
Quick revision
The measuring instrument becomes part of the circuit and can change the quantity measured.
A galvanometer conversion protects the movement by controlling its current.
Null methods rely on equal potentials and stable comparison conditions.
Clear node labels prevent incorrect bridge and potentiometer ratios.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.