# Use Norton’s Theorem to find the current in RL

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

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

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°.

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

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

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