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: In the circuit shown below, a capacitor $C=4 \mathrm{~F}$, inductor $L=5 \mathrm{H}$, and resistors $R_{1}=3 \Omega$ and $R_{2}=2 \Omega$ are placed in a diamond shape and are then fed an alternating current with peak voltage $V_{0}=1 \mathrm{~V}$ of unknown frequency. Determine the magnitude of the maximum instantaneous output voltage shown in the diagram.
<img_4324>

Solution:
We use the method of phasors. Consider the following phasor diagram:
<img_4316>

We define angles as $\angle A P B=\angle A Q B=\gamma, \angle A O B=2 \gamma, \angle B P Q=\alpha, \angle A Q P=\beta$. Note that $\overrightarrow{P A}=I_{2} \omega L, \overrightarrow{A Q}=I_{2} R, \overrightarrow{P B}=I_{1} R_{1}, \overrightarrow{B Q}=1 / I_{1} \omega C$. We seek to maximize the length of $\mathrm{AB}$. We can write via law of cosines that

$$
\mathrm{AB}=\sqrt{r^{2}+r^{2}-2 r \cos (2 \gamma)}=2 r \sin (\gamma)
$$

where $r=V_{0}$ is the radius of the circle of which phasors are inscribed in. Since $\gamma=\alpha+\beta$, we can then rewrite the length of $\mathrm{AB}$ to be

$$
\mathrm{AB}=r(\cos \alpha \cos \beta-\sin \alpha \sin \beta)=V_{0}\left(\frac{R_{1} R_{2}}{V_{0}^{2}}-\frac{L}{C V_{0}^{2}}\right) I_{1} I_{2}=V_{0}\left(R_{1} R_{2}-\frac{L}{C}\right)\left(\frac{1}{\left|Z_{1}\right|\left|Z_{2}\right|}\right)
$$

where the product of both complex exponentials is simply

$$
\left|Z_{1}\right|\left|Z_{2}\right|=\sqrt{\omega^{2} L^{2}+R_{2}^{2}} \sqrt{\frac{1}{\omega^{2} C^{2}}+R_{1}^{2}}
$$

which implies the best frequency of the circuit is $\omega=\sqrt{\frac{R_{2}^{2}}{L^{2} R_{1} C}}$. This means with algebra, the lowest value of $\left|Z_{1}\right|\left|Z_{2}\right|=\frac{L}{C}-R_{1} R_{2}$. Hence, the maximum value of $\mathrm{AB}$ is simply

$$
\mathrm{AB}=V_{0}\left(\frac{R_{1} R_{2}-\frac{L}{C}}{R_{1} R_{2}+\frac{L}{C}}\right)
$$

Final answer: 0.65 V