How Is Average Power Calculated in a Transformer Circuit?

[dwn]

Homework Statement

Please see the image attached. We are asked to find the average power (P) delivered to Z_L.

2. Relevant Questions

(My observation of the circuit) The circuit on the left is as follows: V + I1*(5 + jwL) = 0.
For the circuit on the right: V = I2*(Z + jωL)

Is a= N1/N2 or N2/N1?

The Attempt at a Solution

(a) Given that there is no source in the left circuit, we use the RE values from the right circuit:

[B]I[/B][SUB]2(max)[/SUB] = 2A    [B]I[/B][SUB]2(rms)[/SUB] = 2/2 = 1A       [B]P[/B] = 1[SUP]2[/SUP] *60 = 60 W

(b) Since we are given a V_1(rms)= 100∠60° V we can calculate V² by ( a = V₁/V₂ ).

a = N1/N2 = 0.5       V[SUB]2[/SUB]= 100/0.5 = 200V[SUB](rms)[/SUB] 
[B]P[/B] = V[SUP]2[/SUP]/Z    →   200[SUP]2[/SUP]/60 = 666.67 W

(c)

V[SUB]S(max)[/SUB] = 120∠0°   By finding V[SUB]1[/SUB], we are able to find V[SUB]2[/SUB].    V[SUB]S[/SUB]= I[SUB]1[/SUB]*5 + jωL

I thought this was the correct way of solving this since, jωL = V/I but we are not given the frequency ω. So obviously this is not the correct approach.

Can someone please critique and advise on everything. I do have a couple other questions regarding this circuit, but I will wait until the problem is completed. Thank you very much any help!

 

[rude man]

(a) Given that there is no source in the left circuit, we use the RE values from the right circuit:

[B]I[/B][SUB]2(max)[/SUB] = 2A    [B]I[/B][SUB]2(rms)[/SUB] = 2/2 = 1A       [B]P[/B] = 1[SUP]2[/SUP] *60 = 60 W

OK

(b) Since we are given a V_1(rms)= 100∠60° V we can calculate V² by ( a = V₁/V₂ ).

a = N1/N2 = 0.5       V[SUB]2[/SUB]= 100/0.5 = 200V[SUB](rms)[/SUB] 
[B]P[/B] = V[SUP]2[/SUP]/Z    →   200[SUP]2[/SUP]/60 = 666.67 W

No. Your V is right but your Z is not. Use the correct |Z| to get the current, then do I^2 R where R = 60 ohms.

(c)

What is the impedance of the load Z looking at the transformer input? Call that Z'. You have an ideal transformer. What is Z' as a function of Z and a? BTW a = N2/N1 = 2.

Then you can put R=5 ohms in series with the transformer input to compute I1, and you know I2/I1 = N1/N2. Then use |I2| to compute load power as in (b).

EDIT:
It may be simpler to write KVL for the inut and output circuits, realizing V2 = aV1 and I2=I1/a. So you get 2 equations in 2 unknowns: in I2 and V2. Then again P = |I2|^2 R, R = 60.

 

[dwn]

Part B:

Z[SUB]L[/SUB]= 60+j80        I[SUB]2[/SUB]= 200/(60+j80)  = 1.2 - j1.6   
Magnitude of I[SUB]2[/SUB]= 2A
P = 2[SUP]2[/SUP]*60 = 240 W

Part C:

a[SUP]2[/SUP]= Z[SUB]s[/SUB]/Z[SUB]p[/SUB]   Z[SUB]p[/SUB] = 15+j20

I[SUB]1[/SUB]= 120/(20+j20) = 3+j3       Magnitude of I[SUB]1[/SUB] = 4.24A 
I[SUB]2[/SUB]/I[SUB]1[/SUB] = a        2(4.24) = 8.48A = I[SUB]2[/SUB]
[B]P[/B]=8.48[SUP]2[/SUP]*60 = 4314.62W

How does it look now?

Thank you a lot for the guidance. I forgot about the Z_p : Z_s relationship.

 

[rude man]

Getting there.

For (c) you first need to reduce the 120V by sqrt(2).
That reduces your |I1|.
Then, it's a = I1/I2, not the other way around. Remember, a = N2/N1 = 2 here.
That reduces |I2| by a lot.
Then P = |I2|^2 * 60.