Maximum power delivered to a load from a Thevenin source
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A network reduces to a Thevenin equivalent of $V_{th} = 20\ \text{V}$ in series with $R_{th} = 5\ \Omega$. A load resistance $R_L$ is connected across the terminals and adjusted for maximum power transfer.
The maximum power delivered to $R_L$, in watts, is ________.
Show the step-by-step derivation
Answer
20 W W
Step-by-step derivation
- Write the power delivered to the load as a function of $R_L$. The series current is $$I = \frac{V_{th}}{R_{th} + R_L},$$ so $$P_L = I^2 R_L = \frac{V_{th}^2\,R_L}{(R_{th}+R_L)^2}.$$
- Maximise by differentiating with respect to $R_L$ and setting the result to zero. Using the quotient rule: $$\frac{dP_L}{dR_L} = V_{th}^2 \cdot \frac{(R_{th}+R_L)^2 - R_L \cdot 2(R_{th}+R_L)} {(R_{th}+R_L)^4}.$$
- Setting the numerator to zero and cancelling one factor of $(R_{th}+R_L)$: $$(R_{th}+R_L) - 2R_L = 0 \;\Rightarrow\; R_L = R_{th}.$$ This is the maximum power transfer theorem. Here $R_L = 5\ \Omega$.
- Substitute $R_L = R_{th}$ back into the power expression: $$P_{max} = \frac{V_{th}^2 R_{th}}{(2R_{th})^2} = \frac{V_{th}^2 R_{th}}{4R_{th}^2} = \frac{V_{th}^2}{4R_{th}}.$$
- Evaluate with the given numbers: $$P_{max} = \frac{20^2}{4 \times 5} = \frac{400}{20} = \mathbf{20\ W}.$$
- Efficiency footnote. At this operating point the source resistance dissipates an equal $20$ W, so the efficiency is only $50\%$. Maximum power transfer and maximum efficiency are different design goals - power systems deliberately avoid this condition.
The trap this question is built around
Using $P = V_{th}^2 / R_L$ and answering $80$ W. Half the source voltage is dropped across $R_{th}$ at the matched condition, so the load only ever sees $V_{th}/2 = 10$ V, giving $10^2/5 = 20$ W.