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I'm pretty stuck trying to solve the below circuit. I need to find the node voltages and branch currents. I've tried it using mesh, nodal and superposition analyses, but I just can't get the right answer. I'm really not sure where I'm going wrong.

Using the mesh analysis, I took the clockwise left-hand loop as x and the clockwise right-hand loop as y.

Then: $$10V=10I_x+30(I_x-I_y)$$ $$10V=40I_x-30I_y$$ And: $$6V=40I_y+30(I_y-I_x)+20I_y$$ $$6V=90I_y-30I_x$$ Solving these gives $$I_x=0.4A, I_y=0.2A$$ But I've been told that $$I_x=0.267A, I_y=-0.022A$$ I know my solution is wrong anyway, because I end up with a resistor supplying power to the circuit. Where am I going wrong?

circuit

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3 Answers 3

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If you consider the conventional current (current flows from positive node to negative node) your equations are wrong. At the 30Ω resistor the currents actually sum up. So the correct equations are:

$$ 10V=40 I_x + 30 I_y $$

and

$$ 6V=30 I_x + 90 I_y $$

which give the correct answer.

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  • \$\begingroup\$ Even if I've defined the I_x and I_y loop to both be running clockwise? Wouldn't they be pointing in opposite directions in the middle? \$\endgroup\$ Nov 2, 2019 at 4:55
  • \$\begingroup\$ But why would you do that? From the signs in the sources the currents should sum in the middle, regardless of how you choose the path. \$\endgroup\$ Nov 2, 2019 at 4:57
  • \$\begingroup\$ I guess I'd do that because I don't really know what I'm doing. My lecturer kind of rushed over a lot of this stuff. Do you consider the direction of current flow for all the branches? \$\endgroup\$ Nov 2, 2019 at 4:59
  • \$\begingroup\$ You have to be consistent throughout the analysis, if in your first equation the current goes from positive to negative, in the second equation you have to do the same, otherwise it will always fail. Before writing the equations you have to define the current paths, in this case they will sum in the middle no matter what. \$\endgroup\$ Nov 2, 2019 at 5:03
  • \$\begingroup\$ Thanks for all your help, and I'm sorry I'm asking so many questions. On the diagram I've got, say I define the right hand loop as clockwise. Do I then make the $$I_4$$ branch negative, because it's in the opposite direction? Or do I just treat it as positive because it's a voltage drop? \$\endgroup\$ Nov 2, 2019 at 5:07
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schematic

simulate this circuit – Schematic created using CircuitLab

You can use this diagram to understand the current direction and accordingly get correct loop equations.

  1. Decide the loop current direction.
  2. Then mark voltage drop according to the current direction (current entering at a point is considered as positive).
  3. Apply KVL, in the direction of loop current.

It can be seen from the diagram that currents flowing through R2(30 ohm) are actually adding up. Now you can find out all currents and voltages at every node.

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  • \$\begingroup\$ Thanks for your help. If I do this, I get equations $$10=40I1+30I2$$ and $$6=30I1+10I2$$, which have solutions I1 = 0.16A and I2 = 0.12A. Is this correct? The equations Salatiel had give different solutions. \$\endgroup\$ Nov 2, 2019 at 7:57
  • \$\begingroup\$ @JamesOswald: Your second loop equation is not correct. It should be, 6= 30*I1 + 90*I2, which is same as given by Salatiel. Please check it again. \$\endgroup\$ Nov 2, 2019 at 9:31
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At Mesh 1:

$$ 10 - 10i_x - 30(i_x - i_y) = 0 $$ $$ 40i_x - 30i_y = 10$$

At Mesh 2:

$$-20i_y - 6 - 40i_y - 30(i_y - i_x) = 0$$ $$ 30i_x - 90i_y = 6$$

Arranging these equations in Matrix format

$$\begin{bmatrix}40 & -30\\\ 30 & -90\end{bmatrix} \begin{bmatrix} i_x \\\ i_y\end{bmatrix} = \begin{bmatrix}10 \\\ 6\end{bmatrix}$$

Solving these two equations produces

$$i_x = 0.2667 A$$ $$i_y = 0.0222 A$$

Try double checking your loop equations. You might’ve made a mistake in assigning signs.

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