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
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 idea behind this question
A source with internal resistance $R_{th}$ delivers the most power to a load equal to $R_{th}$. At that point the source and load share the voltage equally, so the load receives $V_{th}^2/(4R_{th})$ - and exactly the same power is lost inside the source, which is why matching maximises power but not efficiency.
Try a variation
A source has $V_{th} = 12$ V and $R_{th} = 4\ \Omega$. What is the maximum power it can deliver, in watts?
Show the answer
Answer: $9$ W
$12^2 / (4 \times 4) = 9$ W.
Other mistakes to avoid
- Thinking maximum power transfer means maximum efficiency. At the matched load the efficiency is 50%.
- Applying the result to a source whose resistance can change. It assumes $R_{th}$ is fixed and only $R_L$ varies.