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PhysicsEasyClass 12

Potentiometer + Wheatstone Bridge

Current Electricity

42

JEE Qs

5%

Hard

60

min

Master the null deflection principle for both instruments and practice identifying effective circuit components to correctly apply formulas and analyze complex configurations.

🧮 Key Formulas

P/Q = R/S (Balanced Wheatstone Bridge condition)
V_wire = k * L_wire (Potentiometer Principle, V_wire is potential drop across length L_wire, k is potential gradient)
k = I_primary * (R_wire / L_total_wire) (Potential gradient, where I_primary is current in primary circuit)
I_primary = E_driver / (R_driver_internal + R_series_external + R_total_wire)
E1 / E2 = L1 / L2 (Comparison of EMFs using potentiometer)
r = R_shunt * ((L1 / L2) - 1) (Internal resistance of a cell, where L1 is balance length without shunt R_shunt, L2 is balance length with R_shunt)

✅ Key Points for JEE

  • 1A Wheatstone Bridge is balanced when the ratio of resistances in adjacent arms is equal, leading to zero current through the galvanometer and thus zero potential difference across its terminals.
  • 2Potentiometer works on the principle that for a uniform wire carrying a constant current, the potential drop across any segment is directly proportional to its length (uniform potential gradient).
  • 3The potentiometer is a null deflection instrument, meaning it draws no current from the unknown EMF source at the balance point, providing highly accurate measurements.
  • 4The potential gradient of the potentiometer wire must be less than the EMF of the cell being measured, otherwise, a null point cannot be obtained.
  • 5Polarity is crucial in potentiometer circuits; the positive terminals of all cells (driver and unknown) must be connected to the same end of the potentiometer wire.

⚠️ Common Mistakes

  • Incorrectly identifying the arms (P, Q, R, S) in a Wheatstone bridge, especially when the circuit is drawn in a non-standard configuration.
  • Calculating potential gradient incorrectly by using the full EMF of the driver cell instead of the actual potential drop across the potentiometer wire.
  • Forgetting to consider the internal resistance of the driver cell or any series resistance in the primary circuit when calculating the primary current and potential gradient.
  • Not ensuring that the EMF of the primary driver cell is greater than the EMF of the secondary cell being measured (essential for obtaining a balance point).
  • Incorrectly connecting the polarities of the cells in the potentiometer circuit, leading to no null point or incorrect readings.

📝 Practice Questions

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Q47.The value of current I in the electrical circuit as given below, when potential at A is equal to the potential at B, will be ____ A.

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2025·Assertion ReasoningMedium

Q34. Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance Rp = 1Ω as shown in the figure. An external resistance of Re = 2Ω is connected via the sliding contact. The electric current in the circuit is : (1) 0.9 A (2) 1.35 A (3) 0.3 A (4) 1.0 A 2025 (22 Jan Shift 1) JEE Main Previous Year Paper

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Q47.The net current flowing in the given circuit is_______ A.

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Q30.Refer to the circuit diagram given in the figure. which of the following observations are correct? A. Total resistance of circuit is 6Ω B. Current in Ammeter is 1 A C. Potential across AB is 4 Volts. D. Potential across 2025 (23 Jan Shift 1) JEE Main Previous Year Paper CD is 4 Volts E. Total resistance of the circuit is 8Ω. Choose the correct answer from the options given below: (1) A, B and D Only (2) A, B and C Only (3) A, C and D Only (4) B, C and E Only

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NCERT Chapters

  • Class 12 Physics Ch 3: Current Electricity