# AC Circuit; find average and reactive power to load (and load V and I)

1. Jun 6, 2013

### Color_of_Cyan

1. The problem statement, all variables and given/known data

In the circuit shown, a load having an impedance of 39 + j26Ω is fed from a voltage source through a line having an impedance of 1 + j4Ω. The rms value of the source voltage is 250V.

[Broken]

a) Calculate the load current IL and voltage VL

b) Calculate the average and reactive power delivered to the load

2. Relevant equations

V = IR

I = V/R

voltage division

Polar form conversion

j*j = -1

1/j = -j

average power = VIcos∅

cos∅ = power factor

calculating power factor

reactive power = (apparent power)sinθ

3. The attempt at a solution

Simplify to this first:

[Broken]

and then voltage division:

VL = (250∠0 deg)*[ (39Ω + 26Ωj)/(40Ω + 30Ωj) ]

and in complete polar form:

VL = (250∠0 deg)*[ (46.8∠33.6)/(50∠36.8) ]

VL = (234∠-3.25)V

since the drop VL on the diagram is in the opposite direction though

VL = (-234∠-3.25)V

IL = (250∠0 V) / (50∠36.8 Ω)

IL = 5∠-36.8 A

Now, not sure how to get part b from here. Any hints on how to calculate the angle to get power factor? What is meant by the angle between voltage and current to get power factor (and then being able to find average power)?

Last edited by a moderator: May 6, 2017
2. Jun 6, 2013

### Staff: Mentor

You have the angle of the load voltage relative to the source voltage, and you have the angle of the load current relative to that same reference, so you should be able to determine the angle of the load current relative to the load voltage.

I think your diagram illustrates the wrong idea, or maybe it is just misleading? The power line is represented by 1 + j4. The load voltage is the voltage across the 39 + j26.

I don't see it.

3. Jun 6, 2013

### Color_of_Cyan

Sorry for asking, but how exactly do you do this? Any hints? Subtract the angles? It seems confusing.

That's right. Were you referring to my simplification though? The simplification was just to find the impedance / voltage across.

And I mean there is a + and - for the VL on the question diagram while the battery is the "other way" (isn't it?)

Last edited: Jun 6, 2013
4. Jun 7, 2013

### Staff: Mentor

If you draw the phasor diagram there should be no confusion. Simply subtract the angles.

The + and - are exactly how I'd expect them. No reversal necessary. The + of VL corresponds to the + of the source, so the potential divider equation gives VL as required. (To get current flowing in the load in the direction of the IL arrow, the top of the load must be + with respect to the bottom of the load.)

5. Jun 7, 2013

### Color_of_Cyan

So VL stays positive then.

Is the angle you are giving me the same for also determining power factor?

And would the angle go: -36.8 - (-3.25) ?

Last edited: Jun 7, 2013
6. Jun 8, 2013

### Staff: Mentor

Yes, current relative to voltage gives the pf. The question does not ask you for the power factor angle, though. All it asks for is load voltage (relative to source, presumably), and load current (again, presumably relative to souce voltage).

7. Jun 8, 2013

### Color_of_Cyan

So if

average power = VIcos∅

cos∅ = power factor

Then can you calculate the average power as:

Pavg 5*-234(cos(-3.25°)) ?

8. Jun 8, 2013

### Staff: Mentor

V I cosɸ, but using the power factor angle ɸ that you determined earlier. (As a check, your answer should be the same as if you'd used I².R )

9. Jun 9, 2013

### Color_of_Cyan

That angle was a typo, so

Pavg = 5*-234*cos(-33.55)

Pavg = -975W

=975W supplied?