In one line
A source delivers maximum power to a load when the load resistance equals the Thevenin resistance of the source network, \( R_L = R_{th} \).
Explained simply
Too small a load (nearly a short) draws large current but has almost no voltage across it, so little power. Too large a load has plenty of voltage but almost no current, so again little power. Somewhere in between is the sweet spot, like choosing the right gear on a bicycle to get the most power out of your legs.
Technically, for a network represented by \( V_{th} \) and \( R_{th} \), the load power \( P = I^2R_L \) is maximum when \( R_L = R_{th} \). At that point the efficiency is only 50%, because equal power is lost in \( R_{th} \).
How it works
- Load current: \( I = \dfrac{V_{th}}{R_{th} + R_L} \).
- Load power: \( P = \dfrac{V_{th}^2 R_L}{(R_{th}+R_L)^2} \).
- For maximum, set \( dP/dR_L = 0 \): \( (R_{th}+R_L)^2 - 2R_L(R_{th}+R_L) = 0 \).
- This simplifies to \( R_{th} + R_L = 2R_L \), so \( R_L = R_{th} \).
- Substituting back: \( P_{max} = \dfrac{V_{th}^2}{4R_{th}} \).
Key formula
\( R_L = R_{th}, \qquad P_{max} = \dfrac{V_{th}^2}{4R_{th}} \)
\( R_L \) = load resistance (\( \Omega \)), \( R_{th} \) = Thevenin resistance (\( \Omega \)), \( V_{th} \) = Thevenin voltage (V), \( P_{max} \) = maximum power (W).
Quick example
A network has \( V_{th} = 4 \) V, \( R_{th} = 2\,\Omega \). Maximum power is drawn with \( R_L = 2\,\Omega \): \( P_{max} = 16/(4\times2) = 2 \) W. Check: \( I = 4/4 = 1 \) A, \( P = 1^2 \times 2 = 2 \) W. Real use: matching an audio amplifier to a speaker, or an antenna to a receiver, where getting maximum signal power matters more than efficiency.
Common mistakes
- Writing \( P_{max} = V_{th}^2/R_{th} \); the 4 in the denominator is essential.
- Thinking maximum power means maximum efficiency; efficiency is only 50% at that point.
- Using the source's own resistance instead of the Thevenin resistance of the whole network seen by the load.
How MU asks it
Asked as "State and prove the maximum power transfer theorem" (5 marks) or "Find \( R_L \) for maximum power and the value of maximum power" (7–10 marks). For full marks: state the theorem, derive \( R_L = R_{th} \) using \( dP/dR_L = 0 \), find \( V_{th} \) and \( R_{th} \) with diagrams, then give \( R_L \) and \( P_{max} \) with units.