Subfield: Electromagnetism

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: For any circuit network made of batteries and resistors, if we know the voltages of all the batteries and the resistance values of all the resistors, we can calculate all the electrical currents. However, if we know the voltages of all the batteries and all the currents, it is still not enough to uniquely determine the resistance values of all the resistors. Consider a sail-shape circuit network, in which we connect points $\mathrm{H}$ and $\mathrm{N}$ with a $\mathcal{E}_{\mathrm{HN}}=10 \mathrm{~V}$ battery, points $\mathrm{A}$ and $\mathrm{M}$ with a $\mathcal{E}_{\mathrm{AM}}=20 \mathrm{~V}$ battery. The electrical currents in this network have directions and magnitudes (in $\mathrm{mA}$ ) as shown the figure. The possible resistance values of resistors $R_{\alpha}, R_{\beta}, R_{\gamma}$ is not a single point (corresponds to an unique solution) but a confined region in the three-dimensional $\left(R_{\alpha}, R_{\beta}, R_{\gamma}\right)$-space. Determine the volume of this region (in $\Omega^{3}$ ).

<img_4279>

Solution:
Let us call the nodes on this sail-shape circuit network as in figure A. Since the electrical current always flow toward lower potential, we have the diagram in figure $\mathrm{B}$ for the relation between the electrical potential of the nodes (the arrow indicate high-to-low). Use the notation $\mathcal{V}(\bullet)$ to indicate the electrical potential of the point

The trick for this problem is noticing that the Ohm's laws in our circuit network can always be trivially satisfied for every possible solution following the comparison relationship in figure $\mathrm{B}$, the voltage differences $\mathcal{V}(\mathrm{H})-\mathcal{V}(\mathrm{N})=10 \mathrm{~V}, \mathcal{V}(\mathrm{A})-\mathcal{V}(\mathrm{M})=20 \mathrm{~V}$ and the equality of in-out currents (in $\mathrm{H} 5+7$ equals out $\mathrm{N}-4+16$, in $\mathrm{A} 8+15$ equals out $\mathrm{M} 6+17$ ) of the batteries. That indeed responsible to why the resistance values of the resistors in this network are not an unique point but rather a region in the 14-dimensional space.

Define:

$X=\mathcal{V}(\mathrm{P})-\mathcal{V}(\mathrm{M})>0, \quad Y=\mathcal{V}(\mathrm{Q})-\mathcal{V}(\mathrm{U})>0, Z=\mathcal{V}(\mathrm{A})-\mathcal{V}(\mathrm{S})>0, a=\mathcal{V}(\mathrm{U})-\mathcal{V}(\mathrm{M})>0$

We want to find the volume in the 3 -dimensional subspace of possible $\left(R_{\alpha}, R_{\beta}, R_{\gamma}\right)$ :
$$
\int d R_{\alpha} d R_{\beta} d R_{\gamma}=\frac{\int d X d Y d Z}{I_{\alpha} I_{\beta} I_{\gamma}}
$$
From figure $B$ and the fixed voltage differences $\mathcal{V}(\mathrm{H})-\mathcal{V}(\mathrm{N})=10 \mathrm{~V}, \mathcal{V}(\mathrm{A})-\mathcal{V}(\mathrm{M})=20 \mathrm{~V}$, we can write down the inequality equations to specify the boundary of $\{(X, Y, Z)\}$ :
$$
a<20, \quad Y<30-a, \quad a<X<a+Y, \quad Z<\min [(20-a), y]
$$
which we can solve to obtain a polygonal-region as shown in figure $\mathrm{C}$ (basically, just carving out from the $30 \times 30 \times 20$ rectangle representing maximum bounds of $X, Y, Z)$. Thus:
$$
\int d X d Y d Z=30 \times 30 \times 20-30 \times\left(\frac{1}{2} \times 20 \times 20\right)-20 \times\left(\frac{1}{2} \times 10 \times 10\right)=1.1 \times 10^{4} \mathrm{~V}^{3}
$$
Plug this answer back in Eq. (21) and use $I_{\alpha}=6 \mathrm{~mA}, I_{\beta}=3 \mathrm{~mA}, I_{\gamma}=15 \mathrm{~mA}$ :
$$
\int d R_{\alpha} d R_{\beta} d R_{\gamma}=\frac{\int d X d Y d Z}{I_{\alpha} I_{\beta} I_{\gamma}}=\frac{1.1 \times 10^{4} \mathrm{~V}^{3}}{6 \mathrm{~mA} \times 3 \mathrm{~mA} \times 15 \mathrm{~mA}} \approx 4.0741 \times 10^{10} \Omega^{3} .
$$

<img_4319>

(B)

<img_4308>

(C)

<img_4313>

Final answer: $4.0741 \times 10^{10}$ $\Omega^{3}$