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I'm new to circuit analysis and I bumped into this problem that Im not sure how to solve. Originally the exercise had two ground nodes at the same ground level but I saw a thread that said that it could be modeled as it is shown in the figure below

Circuit

I don't know if the ground node affects the equivalent resistance calculation by making R3//R4//R5 or R4 is just in series with R5.

Thanks in advance!

Edit: I'm looking for the total equivalent resistance, from the beginning to the ground node.

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  • \$\begingroup\$ If all you mean is that on the other diagram, there was a ground symbol connected to R4 and a second ground symbol on R5, the two are identical. \$\endgroup\$ – Hearth Mar 17 at 13:31
  • \$\begingroup\$ You didn't mention between which points you are calculating equivalent resistance. \$\endgroup\$ – Mitu Raj Mar 17 at 14:14
  • \$\begingroup\$ Are you aware that R3 does nothing in your circuit with the values given? \$\endgroup\$ – Andy aka Mar 17 at 14:23
  • \$\begingroup\$ Hearth, Yes. Mitu Raj, the total one, from the upper left wire to the ground node. Andy aka, I'm not aware of that. Why doesn't it do nothing? \$\endgroup\$ – FelipeMedLev Mar 17 at 14:34
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What this problem probably wants you to notice is the symmetry of the circuit.

  1. If we consider the axis of symmetry running through the top node, the middle of R3, and the ground node, then the left and right sides of the circuit are exactly the same.
  2. Therefore, we can assume that the potentials at corresponding nodes are the same.
  3. Therefore, the left and right sides of R3 are at the same potential.
  4. Therefore, no current flows through R3 and we may at our convenience replace R3 with either an open or short circuit to simplify the analysis.

Once you perform either of these replacements, you will have a series-parallel arrangement that is easy to simplify into a single resistor.


This topology of circuit elements is known as a bridge circuit and it has many applications; most famously, in the Wheatstone bridge it is used to measure a resistance, by letting R5 be unknown, replacing R3 with a meter, and then adjusting a variable resistance R4 until R3 shows no voltage/current — indicating that the circuit is in the balanced condition discussed above.

(When the resistances are not symmetric, R3 has current, but we don't actually care most of the time about the magnitude of that current or the voltage across it; rather, the sign of that current or voltage tells us which direction to correct the variable resistance.)

If one is drawing a bridge circuit in a real schematic (except for the H bridge), it is conventionally shown in a diamond shape:

schematic

simulate this circuit – Schematic created using CircuitLab

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