= Two-current resistor model
{title2=$R_P=(2a+r)\parallel(2b+r),\quad R_{AP}=(a+b+r)/2$}
Consider two identical magnetic layers separated by a nonmagnetic layer, between reservoirs that impose the same terminal <electric potential difference> on both <spin> channels. Neglect <spin-flip scattering> and assume each layer obeys <Ohm's law>. In a magnetic layer, define positive channel <electrical resistances> $a,b>0$. Let the spacer contribute $r\geq0$ to \b[each] channel. These are circuit parameters, not measured <iron> or <chromium> resistivities.
For parallel layer <magnetizations>, one channel meets $a,r,a$ and the other $b,r,b$. For opposite layer <magnetizations>, each meets one $a$, one $b$, and the spacer. The two channels share the same terminal <electric potential difference> and run in parallel. Writing $x\parallel y=xy/(x+y)$ gives
$$
R_P=\frac{(2a+r)(2b+r)}{2(a+b+r)},\qquad
R_{AP}=\frac{a+b+r}{2},\qquad
R_{AP}-R_P=\frac{(a-b)^2}{2(a+b+r)}\geq0.
$$
The inequality is algebra, given the specified network. It shows how a difference between spin-channel responses can yield a difference between magnetic configurations even though all the constituent <resistors> are ohmic.
For chosen $a=1\,\Omega$, $b=4\,\Omega$, $r=1\,\Omega$, the parallel channels have total <electrical resistances> $3$ and $9\,\Omega$, while the opposite-alignment channels each have $6\,\Omega$. Thus \b[$R_P=2.25\,\Omega$ and $R_{AP}=3.00\,\Omega$]. At a common terminal <electric potential difference> $1\,\mathrm V$, the total <electric currents> are $4/9\,\mathrm A$ and $1/3\,\mathrm A$, respectively.
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