Discussion
What percentage of maximum power is delivered to a load if a load resistance is 10 times greater than the Thevenin resistance of the source to which it is connected ?
Answer: A
The load receives 33.06% of the maximum power.
A source of Thevenin voltage V and resistance \(R_{th}\) delivers to a load \(R_{L}\) the power below, and the maximum, reached when \(R_{L} = R_{th}\), is \(V^{2}/4R_{th}\).
\[ \begin{aligned} P_{L} &= \frac{V^{2}R_{L}}{(R_{th}+R_{L})^{2}} \\ \frac{P_{L}}{P_{max}} &= \frac{4R_{th}R_{L}}{(R_{th}+R_{L})^{2}} \end{aligned} \]
With \(R_{L} = 10R_{th}\):
\[ \begin{aligned} \frac{P_{L}}{P_{max}} &= \frac{4 \times 10}{(1+10)^{2}} = \frac{40}{121} \\ &= 0.3306 = 33.06\% \end{aligned} \]
40% is what you get by leaving \(R_{th}\) out of the denominator (40/100). A tenfold mismatch costs about two thirds of the available power, which is why 25% and 35% do not fit either.
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