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In the above original circuit,by node analysis,we see that \$R_3\$ and \$R_4\$ are in parallel combination. So we can replace the two with an equivalent resistance \$R_e\$. For that we need to remove the two resistances and use only a single resistance \$R_e\$ in between the two nodes. In the figure below,two equivalent circuits are drawn where the new resistance is \$R_p\$ in the first one and \$R_p'\$ in the second one.

According to the books,the first one is correct. But I want to know why we are placing \$R_p\$ in the place of \$R_3\$. Why aren't we placing \$R_p\$ in \$R_4\$'s wire just like in the second circuit?

In case of removing such resistances in parallel,how can we understand which resistances and wires to what extent can be erased? This issue is creating troubles for me in doing circuit analysis.

  • \$\begingroup\$ Tip: use \$ for inline MathJAX on this site. It's $ on most of the other sites but we don't understand how they manage to discuss cost! I fixed your post. \$\endgroup\$
    – Transistor
    Commented Dec 27, 2023 at 4:33
  • \$\begingroup\$ @Transistor In my experience, although electronic components can be obtained at reasonable prices, mathematical components are usually so cheap that the cost is not worth discussing! I got a box of 10,000 cosines from a friend for free over a decade ago, and I haven't even used half of them yet. \$\endgroup\$ Commented Dec 27, 2023 at 6:11

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

Figure 1. Circuit redrawn for clarity.

  • It should be clear from the equivalent circuit in Figure 1 that R3 and R4 are in parallel as you have correctly figured out.
  • It should now also be clear that R2 and R5 are in parallel so they too can be combined using the parallel resistor calculation.


simulate this circuit

Figure 2. Parallelling the resistors.

  • Now in Figure 2 it becomes clear that R1 is in series with the R2 || R5 combination so we can add those.
  • You will then be left with (R3 || R4) || (R1 + (R2 || R5)) which can again be solved with the parallel resistor rule.


  • You can parallel resistors that share the same nodes.
  • You can series resistors that share one node but have no alternate current branches.

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