Question

В катушке с индуктивностью 0,5 Гн при росте силы тока возникла ЭДС самоиндукции 12 В. На сколько ежесекундно увеличивалось сила тока?

Answer 1

Дано:
L=0,5Гн
ε=12B
Δt=1c
ΔI-?
ε=L*ΔI/Δt
ΔI=ε*Δt/L
ΔI=12B*1c/0,5Гн=24А

Answer 2

Calculation of the Rate of Increase of Current

To calculate the rate at which the current increases, we can use the formula for self-induced electromotive force (EMF) in an inductor:

EMF = -L * (dI/dt)

Where: - EMF is the electromotive force (voltage) induced in the inductor, - L is the inductance of the inductor, and - (dI/dt) is the rate of change of current with respect to time.

In this case, the inductance (L) is given as 0.5 H (henries), and the induced EMF is 12 V (volts).

Plugging in these values into the formula, we can solve for (dI/dt):

12 V = -0.5 H * (dI/dt)

To find the rate at which the current increases, we need to solve for (dI/dt).

Calculation:

Let's solve the equation for (dI/dt):

12 V = -0.5 H * (dI/dt)

Dividing both sides of the equation by -0.5 H:

(dI/dt) = -12 V / -0.5 H

Simplifying:

(dI/dt) = 24 A/s

Therefore, the rate at which the current increases is 24 A/s.

Please note that the calculation assumes a linear increase in current and neglects any resistance in the circuit.