This is one of those problems where it helps to "visualize" what you have and what you need to find. Think about Ohmic resistance in the + x-axis direction and inductive reactance in the + y-axis direction. You are told R and V(R). Nice. You are told only a combined V(rL), though. That part is a vector you don't actually know the direction of, yet. But whatever it is, you are told it has a resistive part and an inductive part.

So. Draw out the vector for V(rL) at some angle -- try something like 40° or so as a random guess. Label the length of that vector 86V. Mark the end of it with an arrow head. Note that it includes a resistive part (which is the COS() part of that vector) and an inductive part (which is the SIN() part of that vector.) Now, starting at the arrow head of your 86V vector, draw out another vector about 30V long pointing due right (along the + direction of the x-axis.) That's your V(R) vector. Mark the end of that one with an arrow head, as well, and label the length of it as 30V.

Finally, draw a vector going from (0, 0) to the last arrow head (end of the 30V vector.) Label that one as 110V. Guess why? Because the 110Vrms must be the magnitude of the resulting vector.

So now you have three sides for a triangle. It's not a nice right triangle. It's one of those cosine law things. A pain. But at least you know all three sides!!

What you need to find out is the length of the resistive part of the V(rL). But to get there you need the angle for your V(rL) vector. Look back at your drawing. Starting at the arrow tip of the V(rL) vector (or start of the V(R) vector) and draw a dotted line in the opposite direction of your V(R) vector. The angle made between this dotted line and your V(rL) vector is ALSO the same as the angle that V(rL) makes with the x-axis (which is what you want.) So you just need to compute that somehow.

To get it, you need to compute the angle between V(rL) and V(R) and then subtract that from π. To get the angle, you need the law of cosines:

1. ? ? ?cos θ = (a2 + b2 - c2) ? (2?a?b)

or,

2. ? ? ?θ = arccos( (a2 + b2 - c2) ? (2?a?b) )

Once you have θ, you have the angle you want (?) from:

3. ? ? ?? = π - θ

With that you can find the resistive voltage vector part by:

4. ? ? ?V(r) = V(rL) ? cos ?

Almost home, now. You know that the current in a series circuit is everywhere the same. So the current in R must also be the current in r. So all you need to do now is add up the resistive parts of the voltage vector and multiply that by the current, I = V(R) ? R.

5. ? ? ?P(total) = P(r) + P(R) = V(r)?V(R) ? R + V(R)2 ? R

6. ? ? ?P(total) = [ V(R)?V(rL)?cos ? ] ? R + V(R)2 ? R

That's the general equation for the total power dissipated in the entire circuit. You can also extract the individual parts from (6), as well. P(r) is the first term and P(R) is the second term of equation (6).

Now to supply values:

7. ? ? ?θ = arccos( (302 + 862 - 1102) ? (2?30?86) ) = 2.4 radians

8. ? ? ?V(r) = V(rL) ? cos( ? ) = 86 ? cos( π - θ ) = 63.41586 V

9. ? ? ?P(total) = 30?63.41586 ? 50 + 302 ? 50 ≈ 38.05 W + 18 W = 56.05 W

Remember that (9) is the TOTAL power. The first term, 38.05 W, is P(r). The second term, 18 W, is P(R).

By the way, you can double check a few things. You have V(r) and V(R) which sum up to provide the resistive voltage component. You can also get V(L), the inductive reactive component. That's just:

10. ? ? V(L) = V(rL) ? sin( ? ) = 86 ? sin( π - θ ) = 58.08983 V

Now, it should be the case that:

11. ? ? V = √[V(L)2 + ( V(r) + V(R) )2]

12. ? ? 110 ≈ √[58.089832 + ( 63.41586 + 30 )2]

13. ? ? 110 ≈ √[58.089832 + 93.415862]

14. ? ? 110 ≈ √[12100.9512]

Which works out close enough. The inductance, by the way, works out to about 256.89mH and r is 105? Ω

Set the sum of the Voltage drops equal to zero and solve for i total.

50i + [86 (30/86)]V + 110V = 0

50i = 80V

i = 1.6A

Power lost in the internal resistance = (30V)*(1.6A) = 48 Watts

Edit add on: I support Jonathan's answer also. I put the 30 Volts across the wrong resistance and was using the wrong approach for solving the problem anyway. My apologies to the asker.

I support Mr. Jonathan.

I= E/R= 30/50= 0.6A

z= 110/0.6= 183.333Ω

z^2= (50+r)^2+XL^2 ---(1)

zrL= 86/0.6= 143.333Ω

zrL^2= r^2+XL^2 ----(2)

(1)-(2)

(50+r)^2+XL^2-r^2-XL^2 = 183.333^2-143.333^2

2500+100r= 13066.64

r= 105.6664[Ω]

P=I^2R = 0.6^2*105.6664= 38.04[W]

XL^2=zrL^2-r^2

XL=√(143.333^-105.6664^2)

XL=96.845Ω

L= 96.845/(2*3.14*60)= 256.89[mH]

Here's the figure to further understand how to solve this problem:





http://s1324.photobucket.com/user/jerome...

A 110VRMS source at 60 Hz in series with a resistor R = 50Ω and an inductor L with an internal resistance r. The voltage across both r and the inductance L is VrL = 86 V and the voltage across the second resistor R is VR = 30 V.





What is the power loss in the internal resistance?