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show that power dissipated by a pure inductor exited by a
sinusoidal voltage source v = vm sin wt is zero
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Answer Posted / sarmadahmed
Let Vmcos(wt+X) be the voltage and Imcos(wt+Y) be the current through the inductor.
Now the power is given by
P=V*I=VmImcos(wt+X)cos(wt+Y)
By using a trigonometric identity:
P= 0.5 * VmIm * cos(X-Y) + 0.5 * VmIm * cos(2wt+X+Y)
This expression has 2 terms which both must be zero.
The first term is a constant (independent of time) but notice that in an inductor Voltage leads Current by 90 degrees so X-Y = 90 and cos(90) = 0 so first term is zero.
Now for average power integrate the above expression for a cycle and divide it by time period, since integral of every sinusoid for a complete cycle is zero the next term will essentially be zero too.
So P = 0 for an inductor.
Email me if you still have any doubt or query.
sarmadahmed@live.com
| Is This Answer Correct ? | 3 Yes | 0 No |
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