Calculate Inductor Maximum Energy: 240V, 50Hz, 4A & Negligible Resistance

[jendrix]

Homework Statement

Assuming steady-state sinusoidal ac.A voltage of 240V at 50Hz applied to an inductor yields a current of 4A.Assume negligible resistance.Find inductance, max stored energy.

Homework Equations

E=0.5Li^2

X=2*pi*f*L

The Attempt at a Solution

V/I=X Which I found to be 60

L=60/2*pi*f
L=0.191H

E=0.5*0.191*4^2
E=1.528J

However the answer given is 3.06J , which has left me confused.

Thanks

 

[gneill]

The given voltage and current values are no doubt RMS values...

 

[PhysicoRaj]

assuming steady-state sinusoidal ac.a voltage of 240v at 50hz applied to an inductor yields a current of 4A. Assume negligible resistance. Find inductance, max stored energy.

Maximum energy is stored when maximum current flows through it.

 

[jendrix]

Of course, I need to use I pk,

so E=0.5*0.191*(4*2^1/2)^2=3.06J

Thanks :)

 

[rezeew]

for your question! Your calculations for the inductance and stored energy are correct. However, the answer given may be taking into account the peak voltage and current values, rather than the RMS values. In this case, the peak voltage would be 339.4V (240V*√2) and the peak current would be 5.66A (4A*√2). Using these values, the maximum stored energy would indeed be 3.06J, as given in the answer. It is important to clarify whether the values given are RMS or peak values in order to accurately calculate the maximum stored energy.