Use Norton’s Theorem to find the current in RL

  • Thread starter Thread starter agata78
  • Start date Start date
  • Tags Tags
    Current Theorem
AI Thread Summary
Norton’s Theorem is applied to find the current in the load resistor (RL) using the equation for Norton resistance (RN). The calculations involve complex numbers, where the load resistor should not be included in the Norton equivalent circuit to accurately determine the current through it. The discussion highlights the importance of correctly simplifying complex expressions and using the complex conjugate to eliminate imaginary components from the denominator. Participants emphasize the need for careful arithmetic and suggest using online calculators for verification. Ultimately, the correct approach involves calculating the open circuit voltage at terminals AB and dividing it by the Norton resistance to find the current.
  • #51
I don't think this is a complete answer.You have left complex numbers unsolved.

Answer should be: 0.411∠-17.13°
 
Last edited:
Physics news on Phys.org
  • #52
shaltera said:
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
32.44∠-9.3°
__________ = 0.411∠(-9.3°-8.13°)= 0.411∠-17.13°
78.85∠8.13°
 

Attachments

  • web.gif
    web.gif
    6.9 KB · Views: 534
  • #54
You'll have to show where your numbers are coming from. Show your determinations of Ve and Ze.
 
  • #55
I=Ve/(Ze+Rl)

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

I=Ve/(Ze+Rl)
 

Attachments

  • ve.gif
    ve.gif
    15.9 KB · Views: 359
  • #56
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
$$I = \frac{Ve}{Z_E + RL} = \frac{34.241 V~ ∠ -9.273°}{69.967 Ω ~∠ 9.188°} = 0.489 A ~∠ -18.461°$$
 
Back
Top