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Is the 2nd step in the solution sheet wrong?

I don't know how they can just ignore the node where the 9 ohm resistor and the 12 ohm resistor were located. Wouldn't current be branched/divided when it flows from either the a or the b terminal through both the 18 and 9 ohm; and 12 and 6 ohm resistors, thus making the 2nd picture not a series circuit?

enter image description here

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  • \$\begingroup\$ Redraw the net from left to right, start with node (a), then resistors, then the red node, then resistors, then node (b). Sketch it out, and post it here for feedback if you still have questions. \$\endgroup\$ – P2000 Jun 23 '20 at 18:07
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Look harder at the schematic and you will see this: -

  • The 9 ohm and 18 ohm are in parallel
  • The 6 ohm and 12 ohm are in parallel

Just keep looking is my advice and follow the red enclosed node around from top to bottom.

It's a teaching aid to make you realize how things are connected together.

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Re-arranging circuit components in an effort to aid circuit simplification is a skill that engineers use so very often that we often do it mentally. Re-arranging can make clear that nodes are common (as in this case), or that components are in parallel or in series. This re-arranging skill is particularly useful when applying Thevenin or Norton equivalents.

schematic

simulate this circuit – Schematic created using CircuitLab


Satisfy yourself that a test current applied from Node A -to- Node B would measure the same in the red-coloured wires in any of the diagrams shown.
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