IOM 2016, Теория — Задача 2. «Instantaneous switching-on»

• 1) $\Phi = \frac{\mu_0\mu S}{l}[N_1 I_1 + N_2 I_2]$

• 2) $\begin{cases} U(t) - N_1\frac{\mu_0\mu S}{l}\left[N_1\frac{dI_1}{dt} + N_2\frac{dI_2}{dt}\right] = I_1 r, \\ -N_2\frac{\mu_0\mu S}{l}\left[N_1\frac{dI_1}{dt} + N_2\frac{dI_2}{dt}\right] = I_2 R. \end{cases}$

• 3) $I_1(0) = I_2(0) = 0$

• 4) 2.72

  I2(2τ)/I2(4τ) ≈ e ≈ 2,72.

• 5) $I_2^{\max} \approx \frac{k\mathcal{E}}{R + k^2 r}, \quad k \equiv \frac{N_2}{N_1} = 5$ ≈ $0.6$ A

• 6) $I_2(t) \approx -\frac{k\mathcal{E}}{R + k^2 r}\cdot\exp\left[-\frac{R}{(R + k^2 r)\tau}t\right]$

• 7) A jump from 0 to about 0,6 A at t ≈ 0, then an exponential decay to almost zero by t = 8τ (see the plot in the official solution).

• 8) $I_1^{\max} = \frac{\mathcal{E}}{r}$ ≈ $6$ A

• 9) $I_1(t) \approx \frac{\mathcal{E}}{r}\left[1 - \frac{R}{R + k^2 r}\exp\left(-\frac{R}{(R + k^2 r)\tau}t\right)\right]$

• 10) A rapid rise to about 3 A at t ≈ 0, then an exponential approach to 6 A (see the plot in the official solution).

TABLE OF ANSWERS:

Proposed Solution:

1. Since the core length $l < 1$ m the system can be regarded as quasistationary, so a magnetic flux confined in the core and currents in the windings are related as $\Phi = \dfrac{\mu_0\mu S}{l}[N_1 I_1 + N_2 I_2]$ (positive direction in the windings is the same).

2. If the voltage across the primary winding circuit varies during the switch closure as $U_1 = U(t)$, the Kirchhoff’s loop rule yields (a winding resistance is negligible):

3. Since the current in the windings cannot change instantaneously, the initial condition at $t = 0$ (the moment the switch starts being closed) must read $I_1(0) = I_2(0) = 0$.

4. By multiplying the first equation by $N_2$ and the second one by $N_1$ and then subtracting the equations one obtains $I_1 = \dfrac{U(t)}{r} + \dfrac{R}{kr}I_2$ at any time (here $k \equiv \dfrac{N_2}{N_1}$). Substituting this expression into the second equation one obtains the following equation for the current in secondary winding: $\dfrac{dI_2}{dt} + \dfrac{R}{R + k^2 r}\cdot\dfrac{I_2}{\tau} = -\dfrac{k}{R + k^2 r}\dfrac{dU}{dt}$. Therefore, as soon as the voltage across the primary winding circuit reaches the battery emf and remains approximately constant thereafter ($\dfrac{dU}{dt} \approx 0$), the dependence $I_2(t)$ will be approximately given by $I_2(t) \approx C\cdot\exp\left[-\dfrac{R}{(R + k^2 r)\tau}t\right] = C\cdot\exp\left[-\dfrac{t}{2\tau}\right]$. The last equation is evaluated by using numerical values of $R$, $r$, and $k$. One can see that $\dfrac{I_2(2\tau)}{I_2(4\tau)} \approx e \approx 2{,}72$.

5, 6) However, the above expression does not allow one to determine the final answer for $I_2(t)$ right away since the constant $C \ne 0$ must be specified as the current at the end of the voltage «instantaneous switching-on». To do this one has to specify an appropriate function $U(t)$ such that $U(0) = 0$ and $U(\tau_0) = \mathcal{E}$, where $\tau_0 \ll \tau$ is a switching-on time of the battery (the time it takes the switch resistance to drop almost to zero). For example consider $U(t) \approx \mathcal{E}\dfrac{t}{\tau_0}$ during the time interval from zero to $\tau_0$. Then for $t \le \tau_0$ one obtains:

(the term with $I_2$ can be neglected since $I_2 \le k\dfrac{\mathcal{E}}{R}$ and $\dfrac{\tau_0}{\tau} \ll 1$), where $I_2(0) = 0$. Thus in this time interval $I_2(t) \approx -\dfrac{k}{R + k^2 r}\cdot\dfrac{\mathcal{E}}{\tau_0}t$. This implies that at the end of «instantaneous switching-on» the current in secondary winding is opposite to the current in primary winding and it reaches its maximum equal to $I_2^{\max} \approx \dfrac{k\mathcal{E}}{R + k^2 r} \approx 0{,}6$ A. Subsequently the current decays as

7. Since $\dfrac{RT}{(R + k^2 r)\tau} = 5$ the current in secondary winding at $T = 0{,}5$ sec is less than 0,7% of the maximum value. An approximate plot of $I_2(t)$ is shown in the figure.

8, 9, 10) Using the expression obtained above one finds for the current in primary winding:

Thus after the switching-on starts the current in primary winding rapidly (for $\tau_0 \ll \tau$) increases to $\dfrac{k^2\mathcal{E}}{R + k^2 r} \approx 3$ A and then exponentially approaches the «stationary» value $\dfrac{\mathcal{E}}{r} = 6$ A by almost reaching it at the time $T$. An approximate plot of $I_1(t)$ is shown in the figure.

Notice that the maximum current in primary winding after the switch has been closed (the answer to 8)) can be found simply by analysing the «stationary» mode when the currents are constant and emf in both windings is zero.

Remark: Of course, the linear dependence used to model $U(t)$ of the «instantaneous switching-on» is not a must: any reasonable model gives the same result providing $\tau_0 \ll \tau$. The linear model is just the simplest. For example one can define $U(t) \approx \mathcal{E}\dfrac{t}{\tau_0}\left(2 - \dfrac{t}{\tau_0}\right)$ for $t \le \tau_0$ (a smooth crossover to $U(t) \approx \mathcal{E}$ for $t > \tau_0$). Quite effective solution follows if one chooses $U(t) = \mathcal{E}\cdot[1 - \exp(-t/\tau_0)]$ (this expression can be used for any $t$, so the system evolution has not to be split in two stages). In this case a bit more complicated differential equation yields the solution:

One can see that qualitatively time dependence of the currents at $\tau_0 \ll \tau$ is the same (e.g. the value of $I_2^{\max}$) as in the «linear» model.