= Solution
The temperature difference is $200\,\mathrm K$, so the pair open-circuit voltage is
$$
\mathcal E=(336\times10^{-6})(200)=0.0672\,\mathrm V.
$$
For two identical legs with the stated per-leg geometry,
$$
R=2\frac{\rho L}{A}=2\frac{(3.12\times10^{-5})(4.6\times10^{-3})}{12\times10^{-6}}
=0.02392\,\Omega.
$$
At matched load $R_L=R$,
$$
\boxed{I=\frac{0.0672}{2(0.02392)}=1.40\,\mathrm A},\qquad
\boxed{P_{\max}=\frac{0.0672^2}{4(0.02392)}=0.0472\,\mathrm W.}
$$
The load voltage is $0.0336\,\mathrm V$, and multiplying it by the current reproduces the power. If $\rho$ instead denotes an already combined pair resistivity, $R=0.01196\,\Omega$, giving \b[$I=2.81\,\mathrm A$ and $P_{\max}=0.0944\,\mathrm W$]. These alternatives follow the same convention distinction as the figure-of-merit calculation; they should not be mixed.
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