Use Norton’s Theorem to find the current in RL

[agata78]

Still don't know how to finish this one.

 

[gneill]

Go back to your expression:

(490100-406000j) / 410785

Now, with the denominator factor hint I gave, 410785/2833 = 145. So divide all the numerator numbers by 145.

In truth, your expression above is a perfectly good answer, too. In fact, in practical terms you could just use a calculator and present the result in decimal form:

1.193 - 0.988j

Add the appropriate units and you're done.

Sometimes you will get a problem that requests you to express the result in "reduced form", where you need to find the lowest common denominator as we've been doing.

 

[agata78]

Yes i know what you mean. Thank you for helping me with this.
My task is :

Use Thevenin's and Norton's theorems to find the current in Rl.

I checked out some good examples and it seems like Thevenin and Norton come out the same. Is it true? Am i right?

How to complete the task?

 

[gneill]

Yes i know what you mean. Thank you for helping me with this.
My task is :

Use Thevenin's and Norton's theorems to find the current in Rl.

I checked out some good examples and it seems like Thevenin and Norton come out the same. Is it true? Am i right?

How to complete the task?

Well, the current in the load should certainly come out the same. Thevenin and Norton simply replace a given network with sources with a single source (either voltage or current) and a single resistor. The equivalent circuit behaves just as the original sources and network as far as the load is concerned.

You've worked through finding the Norton model, and you can easily find the Thevenin model from that. The rest is just a matter of sticking your load onto one of the models and calculating the current.

 

[agata78]

It is getting a bit complicated now.

Voc= e

Isc= V/ R1= 50/ 20+j5= (1000 -j250) / 325

Rth= Voc/ Isc= ((980- 160j ) / 29 ) / ((1000- 250j ) /325 )

what do you think?

 

[gneill]

It is getting a bit complicated now.

Voc= e

Isc= V/ R1= 50/ 20+j5= (1000 -j250) / 325

Rth= Voc/ Isc=

Sorry, I don't follow what you're doing here.

The Thevenin impedance is identical in value to the Norton impedance. So if you've found one, you've found the other.

The Thevenin voltage is equal to the Norton current multiplied by the Norton impedance. That is,

$R_{th} = R_N$

$V_{th} = I_NR_N$

and of course this also means:

$R_N = R_{th}$

$I_N = V_{th}/R_{th}$

The Thevenin model is a voltage in series with an impedance. The Norton model is a current source in parallel with an impedance.

(Note that I use the term "impedance" here rather than resistance because we're working with an AC circuit with impedances)

 

[agata78]

At beginning i thought the same, but i just didnt understand your previous clue.

Vth= V/ R1

Thank you so much for your help.

 

[gneill]

At beginning i thought the same, but i just didnt understand your previous clue.
does it mean the question has been answered!

Thank you so much for your help.

It means that once you've slogged through the hard work of finding one of the equivalents, finding the other is very easy to do.

To find the current in the load RL, all you have to do is place RL on one of the equivalents and determine the current. Probably easiest to use the Thevenin model for that so you don't have to deal with current division with impedances.

 

[agata78]

I load= Vth / (Rth+ RL )
Am i right?

 

[gneill]

I load= Vth / (Rth+ RL )
Am i right?

Yup.

 

[agata78]

Vth= In x Rn

Vth= ((3380 -2800j ) / 2833 ) x( ( 553+324 j) / 29 )
Vth=(( 2776340-453280j) /82157 )

And then :

I load= Vth/ Rth+RL

Iload= (( 2776340-453280j) /82157 ) / ((553 +324j )/ 29) + 50

Before i go further am i correct?

 

[gneill]

That's fine. By the way, your Thevenin voltage can be reduced to $\frac{980}{29} - \frac{160}{29}j$ V.

 

[agata78]

I load = (1911100 - 638000j) / 4116985

Iload= 0.464-0.155j

Would you agree with me?

 

[gneill]

Yes I would agree. :thumbs:

 

[agata78]

Then it would be my current which i was looking for to answer my task.

Thank you so much!

 

[shaltera]

I don't think this is a complete answer.You have left complex numbers unsolved.

Answer should be: 0.411∠-17.13°

 

[gneill]

I don't think this is a complete answer.You have left complex numbers unsolved.

Answer should be: 0.411∠-17.13°

Polar or rectangular form for complex values are equally good if the problem doesn't specify that the answer must be given in one or the other.

I think you'll find that your polar value for the load current is not quite right. Should be more like 0.489 ∠-18.5°.

 

[shaltera]

32.44∠-9.3°
__________ = 0.411∠(-9.3°-8.13°)= 0.411∠-17.13°
78.85∠8.13°

 

[gneill]

You'll have to show where your numbers are coming from. Show your determinations of Ve and Ze.

 

[shaltera]

I=Ve/(Ze+Rl)

You'll have to show where your numbers are coming from. Show your determinations of Ve and Ze.

I=Ve/(Ze+Rl)

 

[gneill]

You should probably keep a couple of extra digits for intermediate values so that roundoff error doesn't creep into results.

Your polar value for R3 is okay. But the polar value you obtained for R1+R3 is a bit off. I see:
R1 + R3 = 60.208 Ω ∠ -4.764° (your angle is off a bit). As a result I'm seeing a value for Ve of

Ve = 34.241 V ∠ -9.273°

which is just a bit larger than your value.

I'm not sure where your numbers are coming from for Z_E, but assuming it's a Thevenin equivalent resistance it is given by R2 + R1 || R3 and should come to 19.069 + 11.172j Ω, or 22.101 Ω ∠ 30.366°.

Carrying on I find
$$I = \frac{Ve}{Z_E + RL} = \frac{34.241 V~ ∠ -9.273°}{69.967 Ω ~∠ 9.188°} = 0.489 A ~∠ -18.461°$$