Electronics - RL Circuits - Discussion

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).

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.

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.

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.

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

Ir=24/45(voltage equal in parallel).
Then,
Real power =Ir^2 * 45 = 12.8.

Why is inductive reactance not considered here?

Anyone, please explain.

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

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