Potentiometer — Comparing EMFs, internal resistance
Current Electricity
19
JEE Qs
8%
Hard
75
min
Master the null point principle and carefully analyze circuit connections and potential drops in both primary and secondary circuits to avoid errors.
🧮 Key Formulas
✅ Key Points for JEE
- 1A potentiometer operates on the principle of null deflection, meaning no current is drawn from the unknown source at the balancing point, ensuring a true EMF measurement.
- 2For a null point to be obtained, the potential difference across the potentiometer wire must be greater than the EMF of the cell in the secondary circuit, and the positive terminals of both primary and secondary cells must be connected to the same end of the potentiometer wire.
- 3The potential gradient (k) is the potential drop per unit length of the potentiometer wire, established by the primary circuit (driving cell and series resistance). It must be constant throughout the wire.
- 4When comparing EMFs, the ratio of the EMFs of two cells is equal to the ratio of their respective balancing lengths (E1/E2 = L1/L2) under ideal conditions.
- 5Internal resistance (r) of a cell can be determined using r = R * (L1/L2 - 1), where L1 is the balancing length for the cell's EMF (open circuit) and L2 is for its terminal voltage (when current R is drawn through an external resistance R).
⚠️ Common Mistakes
- ✕Incorrectly connecting the polarities of the primary and secondary cells; positive terminals must connect to the same end of the potentiometer wire.
- ✕Failing to ensure that the potential difference across the potentiometer wire (or its portion) is greater than the EMF of the cell being measured, which prevents a null point.
- ✕Confusing the roles of L1 and L2 in the internal resistance formula (r = R * (L1/L2 - 1)), often leading to an inverted ratio.
📝 Practice Questions
See allQ47.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.
Q27.Given below are two statements : Statement-I : The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs. Statement-II : The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries. In the light of the above statements, choose the correct answer from the options given below. (1) Both Statement-I and Statement-II are false (2) Statement-I is false but Statement-II is true (3) Both Statement-I and Statement-II are true (4) Statement-I is true but Statement-II is false
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
Q45.Which of the following resistivity ( ρ ) v/s temperature (T) curves is most suitable to be used in wire bound standard resistors? (1) (2) (3) (4)
Q47.The net current flowing in the given circuit is_______ A.
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
NCERT Chapters
- Class 12 Physics Ch 3: Current Electricity