Question

СРОЧНО!!!!..В проводнике индуктивностью 5 мГн сила тока в течение 0,2 с равномерно возрастает с 2 А до какого-то конечного значения. При этом в проводнике возникает ЭДС самоиндукции 0,2 В. Определите конечное значение силы тока в проводнике.

Answer 1

Пусть конечное значение I, начальное - I0;

Тогда

$E_{is}=|L\frac{\Delta I}{\Delta t}|\\ \Delta I=E_{is}*\Delta t/L=0,2*0,2/0,005=8$

 

Тогда $I-I_0=\Delta I;\\ I=I_0+\Delta I=10$ (А)

 

Всё в СИ

Answer 2

Calculation of the Final Value of Current in the Inductor

To determine the final value of the current in the inductor, we can use the formula for the self-induced electromotive force (EMF) in an inductor:

EMF = -L * (di/dt)

Where: - EMF is the self-induced electromotive force (in volts) - L is the inductance of the inductor (in henries) - di/dt is the rate of change of current with respect to time (in amperes per second)

In this case, the inductance (L) is given as 5 mH (millihenries) and the self-induced EMF is given as 0.2 V. We need to find the final value of the current (I) in the inductor.

Since the current is increasing uniformly from 2 A over a time period of 0.2 s, we can calculate the rate of change of current (di/dt) as follows:

di/dt = (final current - initial current) / time

Substituting the given values:

di/dt = (I - 2 A) / 0.2 s

Now, we can substitute the values of L, EMF, and di/dt into the formula for EMF:

0.2 V = -5 mH * [(I - 2 A) / 0.2 s]

Simplifying the equation:

0.2 V = -0.025 H * (I - 2 A)

0.2 V = -0.025 H * I + 0.05 H

0.025 H * I = 0.05 H - 0.2 V

0.025 H * I = -0.15 V

I = (-0.15 V) / (0.025 H)

I = -6 A

Therefore, the final value of the current in the inductor is -6 A.

Please note that the negative sign indicates that the current is flowing in the opposite direction to the initial current.