alleryStart learning →For MAXIMUM POWER TRANSFER from a source with internal resistance $R_S$ to a load $R_L$, we need
A$R_L = 0$ (short)
B$R_L = R_S$
C$R_L \to \infty$ (open)
D$R_L = 2R_S$
Answer & Solution
Correct answer: B. $R_L = R_S$
1. Total resistance: $R_S + R_L$. Current: $I = V/(R_S + R_L)$.
2. Power in load: $P_L = I^2 R_L = V^2 R_L/(R_S + R_L)^2$.
3. Maximise: $dP_L/dR_L = 0$ gives $R_L = R_S$.
4. At this match, $P_L = V^2/(4R_S)$. Source dissipates equal power, so EFFICIENCY = 50% only.
5. Used in audio amplifier output matching, antenna feed-line matching, etc.
6. Options A, C, D give zero power transfer or sub-optimal.
_Source: Tony Kuphaldt, "Lessons in Electric Circuits — DC", Vol I, Ch 10 (Max power transfer)._
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