Electronics - RL Circuits - Discussion

I=V/R=24/45=0.533
P=VI=24*0.533=12.8

Another way is total I * V.
There is a power dissipated in the resistor and also low lost in Inductor.
So both need to be considered !

In parallel circuit Total current It= IL +Ir but since they are not in the same phase so cannot plus directly.

It^2 = IL^2 + Ir^2.

Ir = V/R =24/45 = 0.533.
IL = V/Xl =24/1100 = 0.0218.

It = sqr(IL^2 +Ir^2) = sqr(0.5333*0.5333 +0.02182*0.02182) = 0.53345.

It = 0.53345 ampere.

Power = VIt = 24 * 0.5334 = 12.80 (more accurate).

No power is dissipated in L...
So P=(Vpow2)/R

There is no dissipation of power at inductor.so
p=v^2/R
=(24*24)/45=12.8 is the answer

First calculate

I=V/R =24/45 = O.533

THEN P=VI
P=24*0.533
P=12.8

Good answer.

As they said only true power so inductance value will not be there and only purely resistive load is used:-.

V^2/R = (24) ^2/45 = 12.8W.

Please say why there is no power dissipation in the inductor?

Impedance z = 44.9 ohm.

I = V/Z = 0.534 A.

TRUE POWER = V*I*COS(PHI)=12.78 VOLT.

PHI = TAN inv(R/Xl) = 2.34.

If there is no losses in the inductor it is simple. This means no equivalent series or parallel are. Use 2pi x f x L to get the reactance, and then the current by ohms law.

There is no power dissipated. Current is drawn but it recirculates without any heating. The inductor charges and discharges.