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I'm trying to find E Thevenin between points A and B. I found the Thevenin resistance, but I'm having trouble finding the voltage.

Note that there were wires that connected the load resistor, but since it's removed, I moved the nodes all the way to the 110 resistor.

I started out like this: since 165 and 55 are in series, their equivalent resistance is 220 ohms. Therefore, the current in this branch is I = 110V / 220Ω = 0.5A. Is that correct? Obviously the voltage U(AB) is the same as the voltage across the 110 resistor. The 110 resistor is a voltage divider, am I right? So how do I apply the voltage divider formula to find the voltage across the 110 resistor (the one in the middle)?

enter image description here

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  • \$\begingroup\$ Try this: combine the 165R and 55R below it. Then do a source transformation. Then you can combine the left 110R, and do another source transformation. I'll leave the rest to you. \$\endgroup\$ – uint128_t Jan 20 '16 at 16:41
  • \$\begingroup\$ I end up with a current source of 0.5 A on the left, and a 55 ohm resistor on the right. That gives me a voltage of 27.5V, which isn't correct according to my simulator... Maybe the 165 and 55 are combined parallel? (I combined them like they were in series) \$\endgroup\$ – Quant Jan 20 '16 at 16:54
  • \$\begingroup\$ You'll have to post your work, something's going wrong somewhere. Remember to keep doing source transformations and combining resistors until you have only a Thevenin source. Incidentally, what does your simulator say? \$\endgroup\$ – uint128_t Jan 20 '16 at 17:05
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One way is to do the thevenin twice.

  1. the circuit with 110V source, 165R, 110R and 55R. Output R = 220 in parallel with 110 = 73.33R Offload V = 36.67V

  2. Add the remaining circuit. We have 36.67V source in series with 73.33 + 55 + 55 = 183.33ohms. This is potted down by the 110R Output R is 183.33 in parallel with 110 = 68.75R Vout offload = 36.67 / (183.33 + 110) * 110 = 13.75V

Answer: Equivalent = 13.75V + 68.75ohms

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I agree with step 1 from @user1582568

However, the equivalent is 36.667 with a series R of 183. So, all the Rs are in series for step 2 (36.667 is applied across 183.333 + 110 = 293.333 ohms). Thus the current is simply 36.667 / 293.3333 = 0.125 A.

0.125 A times 110 ohms = 13.75 V.

Does that agree with your simulation?

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  • \$\begingroup\$ Yes it does, thank you! But one question, to see if I got it right: at the end we have a 36.67V voltage source in series with 73.33 + 55 + 55 + 110? How did we get this new voltage? \$\endgroup\$ – Quant Jan 20 '16 at 17:24
  • \$\begingroup\$ @MikeP I think you got the same answer as me, did you think my answer was incorrect? \$\endgroup\$ – user1582568 Jan 20 '16 at 17:36
  • \$\begingroup\$ @Quant the voltage is obtained as in my answer. We pot down 36.67V by a resistances 183.33oms and 110ohms, so the current in the divider would be 36.67 / (183.33 + 110). then multiply this by 110 to get 13.75V. \$\endgroup\$ – user1582568 Jan 20 '16 at 17:40
  • \$\begingroup\$ @user1582568, sorry I must have misunderstood. I divided 68.75 by 13.75 and got 0.200 A, so I thought something wasn't exactly right. \$\endgroup\$ – MikeP Jan 20 '16 at 18:04
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As you already know, computing a Thevenin Equivalent involves 2 parts:

  1. Computing \$Z_{TH}\$
  2. Computing \$V_{TH}\$

Obtaining \$Z_{TH}\$

As you have already computed it I am going to discuss briefly how I would get it to see if we agree.

After shorting out the voltage source we find that: $$Z_{TH} = (165 + 55)//110 + 55 + 55 = \frac{(165 + 55)·110}{165 + 55 + 110} + 55 + 55 [Ω]$$

Where (//) denotes the parallel of those resistors and (+) the series equivalent. Note how the // operation takes precedence. Carrying it out shows \$Z_{TH} = 183,83 Ω\$

Obtaining \$V_{TH}\$

As you pointed out, having an open circuit between terminals A and B "removes" all the resistors in the right mesh from the circuit (namely both 55 and the right-most 110 ohm resistors). We also see how \$V_{AB} = V_{TH}\$ is indeed the same as \$V_{R=110}\$. The only thing we have to do is obtaining that \$V_{R=110}\$. To do so we will solve the left-most mesh. Obtaining the current follows Ohm's law and it is given by: $$I =\frac{V_{Source}}{R_{165} + R_{110} + R_{55}} = \frac{110V}{165Ω + 110Ω + 55Ω} = 0,33A = 333,33mA$$

Then: $$V_A = V_B + I·R_{110} \to V_A - V_B = V_{TH} = I·R_{110} = 0,33A·110Ω = 36,67V$$

If you want to see the "equation" for the voltage divider you can just substitute I in the second equation for the expression we obtained in the first one and you would get: $$V_{TH} = \frac{V_{Source}}{R_{165} + R_{110} + R_{55}}·R_{110}$$

Finally, here is the circuit with the current I used:

schematic

simulate this circuit – Schematic created using CircuitLab

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