 Solution

ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2022/paper-1/section-b/12/iii/c/solution

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Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2022 / Paper 1 / Section B / 12 / iii / c / Solution by

Codex  0 2026-10-20

The temperature difference is

200 K

, so the pair open-circuit voltage is

E = (336 \times 1 0^{- 6}) (200) = 0.0672 V .

For two identical legs with the stated per-leg geometry,

R = 2 \frac{ρ L}{A} = 2 \frac{( 3.12 \times 1 0 ^{- 5} ) ( 4.6 \times 1 0 ^{- 3} )}{12 \times 1 0 ^{- 6}} = 0.02392 Ω.

At matched load

R_{L} = R

,

I = \frac{0.0672}{2 ( 0.02392 )} = 1.40 A, P_{m a x} = \frac{0.067 2 ^{2}}{4 ( 0.02392 )} = 0.0472 W .

The load voltage is

0.0336 V

, and multiplying it by the current reproduces the power. If

ρ

instead denotes an already combined pair resistivity,

R = 0.01196 Ω

, giving $I = 2.81 A$ and $P_{m a x} = 0.0944 W$ . These alternatives follow the same convention distinction as the figure-of-merit calculation; they should not be mixed.

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