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If we have three resistors, two resistors on two separate nets with a third resistor connecting these two branches together, how do you calculate their total resistance?

I thought the three resistances were in parallel and therefore 1/R1 + 1/R2 + 1/R3 would calculate the correct total resistance but this does not seem to be the case.

What also puzzles me is why the two branches might have been connected together in the first place. I understand that the voltage across R2 will simply be double the voltage of the positive line but by doing this, what does it acieve?

Parallel resistances in 'C' configuration

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    \$\begingroup\$ Before you can think about calculating the "total resistance" you need to decide which 2 points in the circuit you're looking at. The "total resistance" looking "in" from the De-Mux IC will be different to the "total resistance" looking in from RX_P & RX_N. \$\endgroup\$
    – brhans
    Nov 14, 2022 at 17:38

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They are not in parallel. To be in parallel, both ends have to be connected together. That isn't the case here.

If you don't have a load attached between RX_P and RX_N, there's no current through R1 or R3, so the only load on the "De-miltiplexer" is R2, and the total resistance it sees is R2.

If you have a load between RX_P and RX_N, call it RL, then you have a series combination R1 + RL + R3, and this combination is in parallel with R2. You should be able to calculate the parallel combination of R2 and this series combination to get the total load on the "De-miltiplexer".

More likely this circuit is intended to drive a transmission line, which makes everything more complicated. In that case, you will need to know the characteristic impedance of the strip line, and if it isn't perfectly terminated, what the termination impedance is and the electrical length of the line, before you can find the total load on the "De-miltiplexer".

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