Subfield: Electromagnetism


Question: a) Determine how $\Phi$ depends on $\omega, \mathrm{L}$ and $\mathrm{C}$ ( $\omega$ is the angular frequency of the sine wave).

Solution:
a)

![](https://cdn.mathpix.com/cropped/2023_12_21_4abc4f73a4283ca2dd46g-1.jpg?height=454&width=1100&top_left_y=1937&top_left_x=615)

Current law: 
$$
\quad \mathrm{I}_{\mathrm{L}_{\mathrm{n}-1}}+\mathrm{I}_{\mathrm{C}_{\mathrm{n}}}-\mathrm{I}_{\mathrm{L}_{\mathrm{n}}}=0
\tag{1}
$$

Voltage law: 
$$
\quad \mathrm{V}_{\mathrm{C}_{\mathrm{n}-1}}+\mathrm{V}_{\mathrm{L}_{\mathrm{n}-1}}-\mathrm{V}_{\mathrm{C}_{\mathrm{n}}}=0
\tag{2}
$$



Capacitive voltage drop: 
$$
\mathrm{V}_{\mathrm{C}_{\mathrm{n}-1}}=\frac{1}{\omega \cdot \mathrm{C}} \cdot \tilde{\mathrm{I}}_{\mathrm{C}_{\mathrm{n}-1}}
\tag{3}
$$

Note: In eq. (3) $\tilde{\mathrm{I}}_{\mathrm{C}_{\mathrm{n}-1}}$ is used instead of $\mathrm{I}_{\mathrm{C}_{\mathrm{n}-1}}$ because the current leads the voltage by $90^{\circ}$.

Inductive voltage drop: 
$$
\quad \mathrm{V}_{\mathrm{L}_{\mathrm{n}-1}}=\omega \cdot \mathrm{L} \cdot \tilde{\mathrm{I}}_{\mathrm{L}_{\mathrm{n}-1}}
\tag{4}
$$

Note: In eq. (4) $\tilde{\mathrm{I}}_{\mathrm{L}_{\mathrm{n}-1}}$ is used instead of $\mathrm{I}_{\mathrm{L}_{\mathrm{n}-1}}$ because the current lags behind the voltage by $90^{\circ}$.

The voltage $\mathrm{V}_{\mathrm{C}_{\mathrm{n}}}$ is given by: 
$$
\mathrm{V}_{\mathrm{C}_{\mathrm{n}}}=\mathrm{V}_{0} \cdot \sin (\omega \cdot \mathrm{t}+\mathrm{n} \cdot \varphi)
\tag{5}
$$

Formula (5) follows from the problem.

From eq. (3) and eq. (5): 
$$
\mathrm{I}_{\mathrm{C}_{\mathrm{n}}}=\omega \cdot \mathrm{C} \cdot \mathrm{V}_{0} \cdot \cos (\omega \cdot \mathrm{t}+\mathrm{n} \cdot \varphi)
\tag{6}
$$

From eq. (4) and eq. (2) and with eq. (5)

$$
I_{L_{n-1}}=\frac{V_{0}}{\omega \cdot L} \cdot\left[2 \cdot \sin \left(\omega \cdot t+\left(n-\frac{1}{2}\right) \cdot \varphi\right) \cdot \sin \frac{\varphi}{2}\right]
\tag{7}
$$
$$
I_{L_{n}}=\frac{V_{0}}{\omega \cdot L} \cdot\left[2 \cdot \sin \left(\omega \cdot t+\left(n+\frac{1}{2}\right) \cdot \varphi\right) \cdot \sin \frac{\varphi}{2}\right]
\tag{8}
$$

Eqs. (6), (7) and (8) must satisfy the current law. This gives the dependence of $\varphi$ on $\omega, \mathrm{L}$ and $\mathrm{C}$.

$0=\mathrm{V}_{0} \cdot \omega \cdot \mathrm{C} \cdot \cos (\omega \cdot \mathrm{t}+\mathrm{n} \cdot \varphi)+2 \cdot \frac{\mathrm{V}_{0}}{\omega \cdot \mathrm{L}} \cdot \sin \frac{\varphi}{2} \cdot\left[2 \cdot \cos (\omega \cdot \mathrm{t}+\mathrm{n} \cdot \varphi) \cdot \sin \left(-\frac{\varphi}{2}\right)\right]$

This condition must be true for any instant of time. Therefore it is possible to divide by $\mathrm{V}_{0} \cdot \cos (\omega \cdot \mathrm{t}+\mathrm{n} \cdot \varphi)$.

Hence $\omega^{2} \cdot \mathrm{L} \cdot \mathrm{C}=4 \cdot \sin ^{2}\left(\frac{\varphi}{2}\right)$. The result is

$$
\varphi=2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right) with 0 \leq \omega \leq \frac{2}{\sqrt{\mathrm{L} \cdot \mathrm{C}}}
\tag{9}
$$

Final answer: $\varphi=2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right)$ with $0 \leq \omega \leq \frac{2}{\sqrt{\mathrm{L} \cdot \mathrm{C}}}$