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I have been asked the following question:

The resistance, r, is not known in fig(1a), why does its value not affect potential at b?

By Thévenin's theorem, any arbitrary voltage source can be replaced by ideal voltage source with a resistor in series (without affecting the rest of the circuit - rendering those two equivalent). However, I don't think that this is the answer they are looking for, neither I can think of any other way to explain why this resistor does not affect value of \$V_b\$. I have tried to do mesh-analysis, but that failed miserably.

Any other ways?

Figure 1a

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  • \$\begingroup\$ r won't affect the voltage at a (unless it's a dead short). In practice, the supply will have some internal resistance, and the value of r will have an effect. \$\endgroup\$ – Leon Heller Feb 25 '13 at 14:19
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I'm not going to just give you the answer to your homework problem.

However, consider what it really means to be a voltage source. Let's say I have a 6V source that is good to 1 A. What voltage is it when nothing is connected? When I put a 60 Ω resistor accross it? A 30 Ω resistor accross it?

If I put the 6 V source and resistor inside a black box and brought out only the two leads, what difference could someone observe on the outside to distinguish the no load, 60 Ω and 30 Ω cases? Now consider this is the case you have. Your +6V and R are inside the box and the rest of the circuit on the outside.

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  • \$\begingroup\$ Thanks, it's clearer now :) Also, not a homework, just exercise sheets. \$\endgroup\$ – Bart Platak Feb 25 '13 at 16:58
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First, let's redraw the schematic in a more intuitive way.

schematic

What we have here is a series parallel circuit. Assuming \$r>0\$ and the voltage source can supply enough current to meet the demand, \$V_c\$ is connected to the positive terminal node, so it must stay at that potential, and \$V_a\$ is connected to the negative node and must stay at that potential. \$V_b\$ can be determined by the voltage divider rule.

If \$r=0\$, the supply is shorted, and all nodes will go to zero.

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If you want to throw Thévenin at this problem, then consider that the closest thing to an "ideal voltage source" is a power supply. That is, your Thévenin model of the 6v source would be modeled with a zero ohm series resistance.

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