Potentiometer Comparison of EMFs
If the potential gradient k is unchanged, E = kl and E1/E2 = l1/l2. A valid null requires the test emf not exceed the available drop and correct polarity.
Why this shows up in the exam
Comparing cell emfs · Calibrating a voltmeter · Finding an unknown potential without loading
Learn the idea
At null, a potentiometer balance length is proportional to the test emf without drawing current from it. A uniform wire carrying steady primary current has a constant potential gradient. The test cell balances when its emf equals the drop along a matching wire length.
🧠 Memory hook: Same gradient: emf follows balance length.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- E = k l — null condition on a uniform potentiometer wire
- E₁/E₂ = l₁/l₂ — emf comparison at unchanged potential gradient
How to approach it
- 1Confirm the same potential gradient
- 2Locate which length belongs to each emf
- 3Use the length ratio and check range
Common slip-ups that cost marks
- •Using the ratio when primary current changed
- •Treating balance length as proportional to current directly
- •Ignoring polarity or an out-of-range emf
🌟 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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