The electric-power industry is interested in finding a way to store electric energy during times of low demand for use during peak-demand times. One way of achieving this goal isto use large inductors.a)What inductance L would be needed to store energy E = 3.0 kWh (kilowatt-hours) in a coil carrying current I = 300A?

Respuesta :

Answer:

240 H

Explanation:

The energy stored in an inductor is given by:

[tex]E=\frac{1}{2}LI^2[/tex]

where

L is the inductance of the inductor

I is the current

In this problem, we know:

[tex]I=300 A[/tex] is the current in the inductor

[tex]E=3.0 kWh = 3000 Wh[/tex] is the energy stored. We need to convert it into Joules:

[tex]E=3000 Wh=3000 Wh \cdot (3600 s/h)=1.08\cdot 10^7 J[/tex]

So, we can now solve the equation to find L, the inductance:

[tex]L=\frac{2E}{I^2}=\frac{2(1.08\cdot 10^7 J)}{(300 A)^2}=240 H[/tex]