# Load resistance, dissipated power (AC)

Discussion in 'Electronics Homework Help' started by evol_w10lv, Apr 1, 2013.

1. ### evol_w10lv

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0
Feb 19, 2013
Hi!
I hope that someone will help me to calculate few tasks.

1. We know alternating voltage = 120V, Resistance of wire = 0.3 Omh as we can see in scheme.
I have to solve voltage in load resistance, dissipated power in wires and dissipated power in load resistance.

2. The same situation, but there are added two ideal transformers and relation 1:5 and 5:1 as in scheme. Need to solve voltage in load resistance, dissipated power in wires and dissipated power in load resistance.

Which forumulas I have to use? How to start this tasks?

2. ### duke37

5,364
772
Jan 9, 2011
Starting with 1.
The wire reistance is not 0.3R, it is 0.6R.

You can calculate the total resistance, 120V, 100A
You can therefore find the load resistance.
You can then calculate the dissipation in each component.

3. ### Laplace

1,252
184
Apr 4, 2010
Because these are ideal transformers, given the value of current in one loop you know the value of current in all loops. Look at the middle loop -- What is the value of the voltage source on the other side of the transformer? What is the value of the load resistance reflected back through the transformer? (hint: reflected resistance is related to the square of the turns ratio) The resistance seen in the middle loop is the wire resistance (Rw) plus the reflected resistance (Rr). So the formula is E/I=2Rw+Rr and from that calculate the value of the load resistance (Rl) from Rr. The simplest way to get the voltage in the load resistance is V=IRl=100Rl (hint: it is somewhat less than 120 V).

4. ### evol_w10lv

73
0
Feb 19, 2013

Rtotal = V/I = 120/100 = 1.2 Ohms
Rl = Rtotal - 2*Rw = 1.2 - 2*0.3 = 0.6 Ohms
Vl = I * Rl = 100 * 0.6 = 60 V
Ptotal = V^2/Rtotal = 120^2/1.2 = 120000 W
Pl = Vl^2/Rl = 60^2/Rl = 60^2/0.6 = 6000 W
Vw1 = Vw2 = I * Rw = 100 * 0.3 = 30 V
Pw = Pw1 + Pw2 = (Vw1^2/Rw1) + (Vw2^2/Rw2) = (30^2/0.3)/(30^2/0.3) = 6000 W
Efficiency = Pl/Ptotal = 6000/120000 = 0.5

Is it correct? And internal resistance is it something simular wire resistance? When it's need to use formula Pl = (V^2 + Rl) / (Rl + r)^2 ?
Later I will try to solve second task.

Last edited: Apr 5, 2013
5. ### Laplace

1,252
184
Apr 4, 2010
Internal resistance might be considered similar to wire resistance when the internal portion of a device comprises wire; however, when internal resistance is due to leakage factors it would be modeled as a parallel resistance rather than being in series with other elements of the model.

It seems (to me) that the purpose of this exercise is to demonstrate that high voltage power transmission is more efficient than low voltage transmission over the same wires due to I^2R losses, so it may be more appropriate to think of Rw as transmission line rather than internal resistance.

I believe your calculations are correct (not 120 kW) but there is a more direct method to calculate power:

Ptotal=ExI=120x100= 12,000 W
Pw=I^2xR=100x100x0.6= 6,000 W
Pl=Ptotal-Pw=12,000-6,000= 6,000 W

6. ### evol_w10lv

73
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Feb 19, 2013
It means that there Ptotal = 12000 W and Pw = 6000 W as in the first task, because in the all three loops are same current I = 100 A?

Then between first and second loop Rr = 0.04*Rl, using formula [(1/n)^2 * Rl] and between second and third loop Rr = 25*Rl. Where or how can I get connection between formulas to get Rl for all scheme?

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,838
Jan 21, 2010
Start from the end.

If the output current of the last transformer is 100A, what is the input current?

8. ### evol_w10lv

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Feb 19, 2013
If we are using ratio, that current of the first transformer should be 20 A
And If the current is increased as we can see, I guess, then the voltage in the load is decreased.
How I have to use the square of the turns ratio to get Rreflected?

9. ### Laplace

1,252
184
Apr 4, 2010
If your textbook doesn't cover how to do that, then we should not do it here. So consider another approach. As before, the total power is the power dissipated in the wire plus the power dissipated in the load. But you already know the current through the wires and the current through the load, so calculate the I^2R power dissipation.

That will get you the value of the load resistance. But do you need the value of the load resistance, or just the power dissipated in the load?

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,838
Jan 21, 2010
Or perhaps just the power lost in the resistors which simulate long wires?

11. ### evol_w10lv

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0
Feb 19, 2013
Actually, I have to calculate dissipated power in the load (Pl), dissipated power in the wires (Pw), total dissipated power (Pt) and voltage in load (Vl).

Il = 100A
Iw = Il/5 = 100/5 = 20A
Pw = Iw^2 * Rw = 20^2 * 0.6 = 240W
I(at start) = Iw/0.2 = 20/0.2 = 100A
Pt = I * U = 100*120 = 12000 W
Pl = Pt - Pw = 12000 - 240 = 11760 W
Rl = Pl/I^2 = 11760/100^2 = 1.176
Vl = Rl * Il = 1.176 * 100 = 117.6 V
Efficiency = Pl/Ptotal = 11760/12000 = 0.98

Is it correct?

12. ### Laplace

1,252
184
Apr 4, 2010
You got the same answer as myself. Does that make it correct?

However, I calculated the load resistance a different way from the loop voltage equation for the inner loop using the reflected load resistance and source voltage of 600 volts at 20 amps.

13. ### evol_w10lv

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Feb 19, 2013 In electronics I believe you more than I believe myself.
Thank you for helping me.  