Question

В катушке индуктивностью 5 мгн сила тока в течении 0,5с равномерно возрастает от 0 до некоторого конечного значения. При этом в ней возникает ЭДС самоиндукции 0,2 В. Определите конечное значение силы тока в катушке.

Answer 1

эдс самоиндукции вычисляется по формуле

Eis= - L дельта I/дельта t
E is = L*I/t           I = Eis*t//L=0,2*0,5/0,005=2 A

Answer 2

Calculation of the Final 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 of the inductor is given as 5 mH (millihenries) or 0.005 H, and the self-induced EMF is given as 0.2 V. We need to find the final value of the current.

Let's assume that the initial current in the inductor is 0 A and the final current is I A. The time taken for the current to increase from 0 A to the final value is given as 0.5 s.

Using the formula for the self-induced EMF, we can rearrange it to solve for the final current:

I = -EMF * (dt/di) / L

Substituting the given values:

I = -0.2 V * (0.5 s / I) / 0.005 H

Simplifying the equation:

I = -0.2 V * (0.5 s) / (0.005 H)

I = -0.2 V * 0.5 s / 0.005 H

I = -0.1 V s / 0.005 H

I = -20 A

The negative sign indicates that the current is flowing in the opposite direction of the assumed direction. Therefore, the final value of the current in the inductor is 20 A.

Please note that the negative sign indicates the direction of the current, and the magnitude of the current is positive.