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Schematic showing multiple series and parallel resistors!

I know how to find resistance, current, and voltage through series/parallel circuits. The problem I'm having is figuring which part of this circuit to simplify first. What is the correct order of operations for each branch? Or how can I redraw this circuit to make more sense?

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    \$\begingroup\$ A hint: what is the voltage across R3, the one in the middle. It does not help that there are several resistors called R3. \$\endgroup\$ – Bimpelrekkie Oct 1 '15 at 9:59
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enter image description here

Like the users before me pointed out, you can solve it just by redrawing the circuit, and noticing a balanced wheatstone bridge.

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The problem is that using standard series parallel transformations does not work in this case. You have 3 options. Kirchhoff's current rule, Kirchhoff's voltage rule, Y/Delata transform.

https://en.wikipedia.org/wiki/Y-%CE%94_transform

https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

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  • \$\begingroup\$ Instead of getting all the laws out already, why not simply look at the circuit and examine what's actually happening ? Maybe no laws are needed ? \$\endgroup\$ – Bimpelrekkie Oct 1 '15 at 10:02
  • \$\begingroup\$ I'm going to look at some examples with these rules and be back later. \$\endgroup\$ – srp009 Oct 1 '15 at 10:18
  • \$\begingroup\$ What if I say: there is NO NEED to use any of the laws or rules ??? Why does no one try to "see" what this circuit is doing ? I must have a special gift to see that the voltage across the resistor in the middle is 0V. So on the left there's 2 x 4 = 8 ohms and on the right there's 4 x4 = 16 omhs. 8 and 16 in parallel gives: 5 1/3 ohm. 12V / 5.33 = 2.25 A There, solved no Kirchoff, Thevenin or Wheatstone needed. \$\endgroup\$ – Bimpelrekkie Oct 1 '15 at 12:50
  • \$\begingroup\$ Thanks. So the R3 in the middle is a dead end with 0V and 0I ? and you just have to calculate the series/parallel like that middle R3 resister inst there? \$\endgroup\$ – srp009 Oct 1 '15 at 16:00

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