Potentiometer Measurement of Internal Resistance
With balance length l1 for emf and l2 for terminal voltage across load R at unchanged gradient, E/V = l1/l2 and r = R(l1/l2 - 1).
Why this shows up in the exam
Measuring cell internal resistance · Comparing terminal voltage under different shunts · Battery-condition experiments
Learn the idea
Comparing open-circuit and loaded balance lengths reveals a cell's internal resistance. The open-circuit balance measures emf because no current is drawn. Adding a known shunt load lowers terminal voltage; the fractional drop reveals the hidden internal resistance.
🧠 Memory hook: Open length gives E; loaded length gives V; their drop reveals r.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- E/V = l₁/l₂ — same potentiometer gradient
- r = R(l₁-l₂)/l₂ — cell internal resistance with known load R
How to approach it
- 1Identify open and loaded balance lengths
- 2Write E/V as the length ratio
- 3Substitute the known load and solve r
Common slip-ups that cost marks
- •Swapping l1 and l2
- •Using the external shunt as if it were in series
- •Changing the primary gradient between readings
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 2 ohm resistor. Find the circuit current.
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