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PhysicsMediumClass 12

Galvanometer — Conversion to ammeter and voltmeter

Magnetic Effects of Current

13

JEE Qs

8%

Hard

75

min

Always draw the circuit diagram for the converted instrument to correctly apply Ohm's Law and series/parallel resistor rules for current and voltage distribution.

🧮 Key Formulas

R_s = (I_g * R_g) / (I - I_g)
R_v = (V / I_g) - R_g
R_eq_ammeter = (R_g * R_s) / (R_g + R_s)
R_eq_voltmeter = R_g + R_v

✅ Key Points for JEE

  • 1A galvanometer is converted to an ammeter by connecting a small resistance (shunt resistance, R_s) in parallel with it.
  • 2A galvanometer is converted to a voltmeter by connecting a large resistance (series resistance, R_v) in series with it.
  • 3An ideal ammeter has zero resistance and is connected in series in a circuit to measure current; a real ammeter has a small but finite resistance.
  • 4An ideal voltmeter has infinite resistance and is connected in parallel across components to measure potential difference; a real voltmeter has a very large but finite resistance.
  • 5The full-scale deflection current (I_g) and the galvanometer's internal resistance (R_g) are critical parameters for conversion calculations.

⚠️ Common Mistakes

  • Confusing the series and parallel connections for ammeter and voltmeter conversions.
  • Incorrectly using Kirchhoff's laws or Ohm's law to calculate shunt/series resistance, e.g., using total current through galvanometer for ammeter conversion.
  • Mixing up I_g (full-scale deflection current of galvanometer) with I (total current range of ammeter) or V (total voltage range of voltmeter).
  • Ignoring the internal resistance of the galvanometer (R_g) in calculations.

📝 Practice Questions

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Q43.Match List - I with List - II. List - I List - II (A) Magnetic induction (I) Ampere meter 2 (B) Magnetic intensity (II) Weber (C) Magnetic flux (III) Gauss (D) Magnetic moment (IV) Ampere/meter (1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (3) (A)-(III), (B)-(II), (C)-(I), (D)-(IV) (4) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

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Q46.The magnetic field inside a 200 turns solenoid of radius 10 cm is 2.9 × 10−4Tesla. If the solenoid carries a current of 0.29 A , then the length of the solenoid is ________ πcm .

2025·NumericalMedium

Q47.A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A . If the rate of change of potential difference between the plates is 7 × 108 V/s then the integer value of the distance between the parallel plates is ( Take, ϵ0 = 9 × 10−12 mF , π = 227 ) ________ μm . 2025 (29 Jan Shift 2) JEE Main Previous Year Paper

2025·NumericalMedium

Q37.If B is magnetic field and μ0 is permeability of free space, then the dimensions of (B/μ0) is (1) ML2 T−2 A−1 (2) MT−2 A−1 (3) L−1 A (4) LT−2 A−1

2025·ConceptualEasy

Q46.A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2 × 105 ms−1 . When the electric field is switched off, the proton moves along a circular path of radius 2 cm . The magnitude of electric field is x × 104 N/C. The value of x is _______ Take the mass of the proton = 1.6 × 10−27 kg.

2025·NumericalMedium

Q50.Two long parallel wires X and Y , separated by a distance of 6 cm , carry currents of 5A and 4A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance 2025 (22 Jan Shift 2) JEE Main Previous Year Paper of 4 cm from wire Y is x × 10−5 T. The value of x is__________ . Take permeability of free space as μ0 = 4π × 10−7SI units.

2025·NumericalMedium

NCERT Chapters

  • Class 12 Physics Ch 3: Current Electricity
  • Class 12 Physics Ch 4: Moving Charges and Magnetism