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Consider the circuit below:

enter image description here

The task is to find the node voltages V1, V2 and V3 (labelled 1, 2 and 3 above). In order to solve the problem I apply the node voltage method; summing all the currents going out of each node. However, I am having difficulties expressing the current going from node 1 to node 3 and vice versa (the one passing through the dependent voltage source). This is how I attempted to express the current from node 1 to 3:

enter image description here

However, the mark scheme states, without any explanation, the following: enter image description here

I am confused by the - sign as in my head the current is moving from 1 to 3, and the difference in potentials across the resistor must be: V1 - (V3 - 8Vb). I have tried searching for similar problems, but the problems which I have encountered have only considered the dependent voltage source on its own (no resistor etc. connected in series). It would be deeply appreciated if anyone could explain how to express the current in terms of the node and dependent voltages.

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Let's call the 4 Ohm resistor R1.

We can express the current through R1 (in the left-to-right direction) as

\$I(\mathrm{R1}) = \frac{1}{\mathrm{R1}}\left(V_1 - \left(V_3 - 8V_b\right)\right)\$,

which is just what you've already done.

Now the only thing I think you're missing is that Vb can be expressed in terms of V2 and V3, so now

\$I(\mathrm{R1}) = \frac{1}{\mathrm{R1}}\left(V_1 - \left(V_3 - 8\left(V_2-V_3\right)\right)\right)\$,

which of course you can simplify further using the usual rules.

I don't know who Mark Scheme is, but I believe he made a sign error in his solution.

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  • \$\begingroup\$ Thank you for your response. If we were to look at the current from node 3 to 1 (the opposite direction) we would use the same expression, but with a negative sign? Am I correct? \$\endgroup\$ – Joey Dec 4 '13 at 18:06
  • \$\begingroup\$ @Joey, yes, that's right. \$\endgroup\$ – The Photon Dec 4 '13 at 18:36

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