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ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2018/paper-1/section-a-iii/6/iii/solution

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Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2018 / Paper 1 / Section A(iii) / 6 / iii / Solution by

Codex  0 2026-10-20

At fixed temperature difference and constant internal resistance,

P_{L} = I^{2} R_{L} = \frac{V _{oc}^{2} R _{L}}{( R + R _{L} ) ^{2}} .

(1)

Differentiation gives

\frac{d P _{L}}{d R _{L}} = \frac{V _{oc}^{2} ( R - R _{L} )}{( R + R _{L} ) ^{3}} .

(2)

It is positive below

R_{L} = R

and negative above it, while power vanishes for an open or shorted load. Therefore the maximum power transfer theorem gives

R_{L} = R, P_{L, m a x} = \frac{V _{oc}^{2}}{4 R} = \frac{S _{eff}^{2} Δ T ^{2}}{4 R} .

(3)

At this point

I = V_{oc} / (2 R)

and half the electromotive force drops across the load. Maximum electrical power is not generally maximum thermal-conversion efficiency. If changing current also changes the temperature difference because the thermal source has finite impedance, the fixed-

Δ T

optimization must be reconsidered.

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