# Electronics - RL Circuits - Discussion

Discussion Forum : RL Circuits - General Questions (Q.No. 3)

3.

What is the true power of a 24 Vac parallel

*RL*circuit when*R*= 45 and*X*= 1100 ?_{L}Discussion:

12 comments Page 1 of 2.
Pawan said:
5 years ago

Why is inductive reactance not considered here?

Anyone, please explain.

Anyone, please explain.

Priti said:
6 years ago

Ir=24/45(voltage equal in parallel).

Then,

Real power =Ir^2 * 45 = 12.8.

Then,

Real power =Ir^2 * 45 = 12.8.

Rajan said:
6 years ago

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).

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).

Sudhi. said:
8 years ago

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.

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

ARUN said:
9 years ago

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.

I = V/Z = 0.534 A.

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

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

Nasim said:
10 years ago

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

Lochana.vyas said:
10 years ago

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.

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

Sindhu priya said:
1 decade ago

I=V/R=24/45=0.533

P=VI=24*0.533=12.8

P=VI=24*0.533=12.8

(2)

Hareshwar said:
1 decade ago

Good answer.

C.P. BORKAR said:
1 decade ago

First calculate

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

THEN P=VI

P=24*0.533

P=12.8

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

THEN P=VI

P=24*0.533

P=12.8

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