Maximum Power Transfer and Source Efficiency
For a linear source represented by Thevenin voltage Vth and resistance Rth, load power is Vth squared times R divided by (R+Rth) squared and is maximal at R = Rth. Efficiency there is 50 percent.
Why this shows up in the exam
Battery-load matching · Reducing a network for maximum power · Distinguishing power maximum from efficiency maximum
Learn the idea
A resistive load receives maximum power when it matches the source's Thevenin resistance. A very small load draws high current but gets little voltage; a very large load gets high voltage but little current. Their product peaks at the resistance match.
🧠 Memory hook: Match the load to Thevenin resistance for maximum power, not maximum efficiency.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- P_R = V_th² R/(R+R_th)² — load power for a resistive Thevenin source
- R = R_th — condition for maximum load power
- eta = R/(R+R_th) — fraction of source power delivered to load
How to approach it
- 1Find the network seen by the load
- 2Suppress independent sources to get Rth
- 3Set Rload = Rth and then compute power
Common slip-ups that cost marks
- •Setting load equal to emf
- •Claiming efficiency is maximum at the match
- •Using only one visible internal resistor instead of Rth
🌟 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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