Use Norton’s Theorem to find the current in RL

  • Thread starter agata78
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  • #51
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I don't think this is a complete answer.You have left complex numbers unsolved.

Answer should be: 0.411∠-17.13°
 
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  • #52
gneill
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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°.
 
  • #53
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32.44∠-9.3°
__________ = 0.411∠(-9.3°-8.13°)= 0.411∠-17.13°
78.85∠8.13°
 

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  • #54
gneill
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You'll have to show where your numbers are coming from. Show your determinations of Ve and Ze.
 
  • #55
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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)
 

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  • #56
gneill
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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 ZE, 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
$$I = \frac{Ve}{Z_E + RL} = \frac{34.241 V~ ∠ -9.273°}{69.967 Ω ~∠ 9.188°} = 0.489 A ~∠ -18.461°$$
 

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