Electromagnetic induction > AC electrical circuits > Problem 11.4.25

Statement

$11.4.25.$ [Insert the problem statement]

Solution

$I_{L_{n-1}} + I_{C_n} - I_{L_n} = 0$ - charge conservation at point A.

$V_{C_{n-1}} + V_{L_{n-1}} - V_{C_n} = 0$ - potential difference between A and B is same going both ways along the circuit.

As the problem states we have sinusoidal wave running along the circuit such that there is $\phi$ phase difference between adjacent units of the circuit in voltage. Thus $V_n = V_0 sin(\omega t +n\phi)$.
$I_n = \frac{dQ_n}{dt} = \omega C V_0 cos(\omega t +n\phi)$. Let's define ${I_n}^$ as the current flowing through capacitor n with phase lag of $\pi/2$. Then we get the following expression: $V_{C_{n-1}} = \frac{1}{\omega C} {I_n}^$. In inductive element current lags behind by $\pi/2$ so we will define ${I_{L_n}}^$ as current flowing through inductive element n with phase ahead of $I_{L_n}$ by $\pi/2$. This will give us expression $V_{L_{n-1}} = \omega L {I_{L_{n-1}}}^$.
Plugging expressions for voltages into potential difference equation between points A and B and after simplifications we end up with $I_{L_{n}} = \frac{V_0}{\omega L} 2sin(\omega t + (n+\frac{1}{2})\phi)sin\frac{\phi}{2}$. Plugging it into charge conservation equation at point A and simplifying it we get $\omega C V_0 cos(\omega t + n\phi) + \frac{V_0}{\omega L} \cdot 2sin\frac{\phi}{2}\cdot2cos(\omega t + n \phi)sin(-\frac{\phi}{2}) = 0$. This equation must hold for any moment in time. So we get $\omega C + \frac{4}{\omega L}sin^2(\frac{\phi}{2}) = 0$. Solving this gives us $\phi = \pm 2arcsin(\frac{\omega \sqrt{LC}}{2})$. Choosing direction of wave propagation such that sign of $\phi$ is + we get $\phi = 2arcsin(\frac{\omega \sqrt{LC}}{2})$.
Wave travels along one unit with length $l$ in time $\tau \implies v\tau = l$. For phase difference between circuit units we can write $\phi = \omega \tau$. From these equations we get $v = \frac{l\omega}{2arcsin(\frac{\omega \sqrt{LC}}{2})}$. For limit $\omega \to 0$ we get $v_{\omega \to 0} = \frac{l}{\sqrt{LC}}$.

Answer

[Insert a concise answer or boxed result]