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I have this problem during the exam and for some reason, I haven't asked my teacher yet. I hope that someone can clarify it for me.

Here is the question:

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On the left, Voltage between AB is given and we need to find voltage between CD. After I have U_CD, I use it and try to find back U_AB on the right but I cannot get the same number. I don't know what's wrong with the point that U_AB in the right side should be equal to the one on the left.

Maybe I lack of basic knowledge and it's time to study it again. Thank you.

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  • \$\begingroup\$ Find the voltage C first as potential divider, then for D and compare the results, Ucd = Potential at C - Potential at D \$\endgroup\$
    – Triak
    Commented Dec 18, 2015 at 13:07
  • \$\begingroup\$ If you have a balanced Wheatstone bridge with, say, 10V applied between the two vertical nodes, the voltage measured across the other two, horizontal, nodes is zero. But applying 0V across the horizontal nodes ... \$\endgroup\$
    – Chu
    Commented Dec 18, 2015 at 13:26
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    \$\begingroup\$ @Chu I do not see how your suggestion helps us closer to a solution \$\endgroup\$
    – Bimpelrekkie
    Commented Dec 18, 2015 at 13:28
  • \$\begingroup\$ What Triak means: make life easier on yourself by choosing a reference point so let's make B ground so Vb = 0V, then Va = 10 V. Now we have 2 voltage dividers: 1) Z1 with Z2 and 2) Z3 with Z4. Now calculate Vcb and Vdb. Since B is ground I use: Vcb = Z2/(Z1+Z2)*Vab similar for Vcd = Z4/(Z3+Z4) *Vab, then Vcd = Vcb - Vdb \$\endgroup\$
    – Bimpelrekkie
    Commented Dec 18, 2015 at 13:30
  • \$\begingroup\$ There's a source missing from the circuit, which we assume is across AB in the first circuit. But in the 2nd circuit that source is not shown. \$\endgroup\$
    – Chu
    Commented Dec 18, 2015 at 13:33

2 Answers 2

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The two circuits are not the same. In your first circuit there's a voltage source across AB and no voltage source across CD. But the second circuit is different - there's now a voltage source across CD, and no voltage source across AB.

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  • \$\begingroup\$ Ok Now I see the problem. I was confused about voltage source and measured voltage. Therefore, if I want to have two equivalent circuit, voltage source should be connected to A. V_AD and V_BD should be calculated by voltage divider when voltage source is in A. \$\endgroup\$
    – James
    Commented Dec 18, 2015 at 15:05
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You cannot calculate back Vab from Vcd like that ! For example: If Z1 = Z2 = Z3 = Z4 and then when you apply Vab = 10 V, Vcd will be 0 V.

But applying Vcd = 0 V does not mean Vab must be 10 V, Vab can be anything !

Also assume Z1 - Z4 are not all equal but have such a value that when Vab = 10 V, Vdc becomes 1 V. Now calculate back and assume Vcd = 1 V, Vab can only be lower than 1 V, I mean how could you (physically) end up with 10 V when applying 1 V to a passive network ? You can't !

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