Question

A 40.0-mH inductor is connected to a North American electrical outlet (ΔVrms = 120 V, f = 60.0 Hz). Assuming the energy stored in the inductor is zero at t = 0, determine the energy stored at t = 1 185 s.

Answer 1

Explanation:

It is given that,

Inductance,
`L=40\ mH=40* 10^(-3)\ H`

RMS value of voltage,
`v_(rms)=120\ V`

Frequency, f = 60 Hz

We need to find the energy stored at t = (1 /185) s. It is assumed that energy stored in the inductor is zero at t = 0. So,

The current flowing through the inductor is given by :

`I_t=(V_o)/(X_L)\ (sin\ \omega t-(\pi)/(2))`

`I_t=(√(2) V_(rms))/(X_L)\ (sin\ \omega t-(\pi)/(2))`

`I_t=(120√(2))/(2\pi f L)\ sin(2\pi f t-(\pi)/(2))`

`I_t=(120√(2))/(2\pi* 60* 40* 10^(-3))\ sin(2\pi * 60* (1)/(185))-(\pi)/(2))`

`I_t=(120\sqrt2)/(15.07)\ sin(2\pi * 60* (1)/(185)-(\pi)/(2))`

I = 0.091 A

Energy stored in the inductor is,
`U=(1)/(2)LI^2`

`U=(1)/(2)* 40* 10^(-3)* (0.091)^2`

U = 0.000165 Joules

Hence, this is the required solution.