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I am having difficulty calculating the Thevenin resistance of this circuit

Circuit

I know I need to replace the batteries with links but then I am confused about how to approach R1,2 & 3 in terms of being in series or parallel. Would I consider R2 and R1 in parallel and therefore 1*2/1+2 = 0.667 then add R3 as they are is series then treat R4 as parallel?

I'm looking for the Thevenin equivalent as seen across 'A' amd 'B'

Losing the will with this slightly...

Thanks

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  • \$\begingroup\$ I'm assuming you are looking for a Thevenin equivalent circuit as seen across 'A' and 'B' please confirm. \$\endgroup\$ – Warren Hill Sep 9 '17 at 13:01
  • \$\begingroup\$ Yes that's correct \$\endgroup\$ – T Ridge Sep 9 '17 at 13:07
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Assume R1 to be parallel to R2. This effective to be in series with R3 and this effective resistance in parallel with R4. This gives the effective Resistance across the terminals A and B, which is 625ohm.

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Step 1 - Yes. Because V1 = V2 (and common grounded), you can consider R1 and R2 to be in parallel, and in series with R3.

Redraw the schematic with this change and reconsider how R3 and R4 interact.

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  • \$\begingroup\$ Taking into account R1,2 & 3 I am getting 1.667k ohms then 1.667k * 1k(r4) / 1.667k + 1k I am getting 0.625k ohms \$\endgroup\$ – T Ridge Sep 9 '17 at 13:10
  • \$\begingroup\$ Redrawing I can see that 1.667k ohms would actually be in series with R4 from A-B therefore making the answer 1.667k ohms? I have watched a few youtube videos and there seems to be so many different approaches although I know logically the approach is the same \$\endgroup\$ – T Ridge Sep 9 '17 at 13:30
  • \$\begingroup\$ "because V1=V2": even if V1<>V2 the equivalent resistance is the same. \$\endgroup\$ – pasaba por aqui Sep 9 '17 at 15:24

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