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, $L=40\ mH=40\times 10^{-3}\ H$  

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 :

$I_t=\dfrac{V_o}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{\sqrt{2} V_{rms}}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi f L}\ sin(2\pi f t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi\times 60\times 40\times 10^{-3}}\ sin(2\pi \times 60\times \dfrac{1}{185})-\dfrac{\pi}{2})$    

$I_t=\dfrac{120\sqrt2}{15.07}\ sin(2\pi \times 60\times \dfrac{1}{185}-\dfrac{\pi}{2})$

I = 0.091 A

Energy stored in the inductor is, $U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times 40\times 10^{-3}\times (0.091)^2$

U = 0.000165 Joules

Hence, this is the required solution.