Finding Cell EMF and Internal Resistance from Loads
For a discharging cell connected to load R, E = I(R+r) and terminal voltage is IR. Two distinct loads produce simultaneous linear equations for E and r.
Why this shows up in the exam
Estimating battery health · Two-load numerical problems · Interpreting terminal-voltage measurements
Learn the idea
Two load measurements can reveal a cell's emf and internal resistance. Changing the external resistor changes current and therefore the hidden internal drop. Each operating point supplies one equation for the same E and r.
🧠 Memory hook: Same hidden E and r must fit both loads.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- E = I₁(R₁+r) = I₂(R₂+r) — same cell measured with two loads
- r = (V₁-V₂)/(I₂-I₁) — from V = E-Ir with signed consistent data
How to approach it
- 1Write one full equation per trial
- 2Eliminate E or r
- 3Substitute back and check both measurements
Common slip-ups that cost marks
- •Using different emf values for the two trials
- •Confusing load resistance with internal resistance
- •Mixing terminal voltage and 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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