0

schematic

simulate this circuit – Schematic created using CircuitLab

I have really gotten stuck on this question where I need to calculate I1, I2, I3 and I4. The numbers for the circuit is: V(a)=2 V, V(b)=6 V, V(c)=8 V, R1=1 kiloOhm, R2=2 kiloOhm, R3=4 kiloOhm, R4=4 kiloOhm and R5=10 kiloOhm. I see four meshes and my go-to solution is the mesh current method with four equations where I create four mesh-currents consisting Ia, Ib, Ic, and Id. The equations I have created are:

  1. -6 + 2(Ia- Ib) + 1(Ia-Ic) = 0
  2. 4(Ib -Id) + 2(Ib - Ia) + 10(Ib) = 0
  3. -2 + 1(Ic-Ia) + 4(Ic - Id) = 0
  4. 8 + 4(Id -Ic) + 4(Id-Ib) = 0

These four equations are not adding up and I cannot understand why. My goal is to solve for Ia, Ib, Ic and Id.

FYI, the correct answer is I1 = -2 mA, I2 = 2 mA, I3 = 1 mA, I4 = -1 mA.

I would be so grateful if any of you could explain what I should do or think because I have been stuck on this for days.

Picture of the circuit

DiodeLight97
  • 7
  • 3
  • 1
    Tip: add the component values and known voltages and currents to the schematic. It saves an awful lot of looking back and forward. There's a CircuitLab schematic editor button on the editor toolbar. Drag and drop components, R to rotate, H and V to flip. Double-click to edit. When you are finished click Save and Insert to create an editable and copyable schematic. No CircuitLab account required. – Transistor Jul 29 '23 at 15:36
  • Thanks for the advice! Followed your tip, and wasn't able to put out the currents but I guess you get what I am asking for in the second picture :) – DiodeLight97 Jul 29 '23 at 15:54
  • why are R1 and R3 separate in your attempted solution circuit diagram? ... same with R2 and R4 – jsotola Jul 29 '23 at 17:43
  • What do you mean? Because they are separate resistors? :) – DiodeLight97 Jul 29 '23 at 17:45
  • they are series resistors – jsotola Jul 29 '23 at 17:47
  • Why is that? I don't understand... – DiodeLight97 Jul 29 '23 at 17:54
  • are you not familiar with series connected resistors and that they can be replaced by one resistor? ... the four resistors are not connected in the center – jsotola Jul 29 '23 at 23:58

2 Answers2

0

HINT

You are making a meal of this when you should just mark up the voltages on the main nodes and see this: -

enter image description here

R5 has 8 volts on each node by simple inspection.

FYI, the correct answer is I1 = -2 mA, I2 = 2 mA, I3 = 1 mA, I4 = -1 mA.

That cannot be true if your main image schematic is to be believed.

If the crossing wires at the centre are in fact joined then maybe it can be the right answer. Use Millman's Theorem to find the voltage at the central node in that case.

enter image description here enter image description here

Then it's dead easy after that.

Andy aka
  • 434,556
  • 28
  • 351
  • 777
  • Hi! Thanks for your hint! I unfortunately still don't understand how you get the voltages over each resistor. Could you explain that? And also why my equations are wrong? If you have the time... thanks again! – DiodeLight97 Jul 29 '23 at 16:07
  • Voltage sources in series add up hence a 6 volt source (Vb) stacked on the top of a 2 volt source (Va) produces 8 volts end-to-end. Read my answer again on why I think your proposed answer may be incorrect. – Andy aka Jul 29 '23 at 16:09
  • Thanks for your response! You do not need Millman's theorem to solve the for the currents. Only Mesh-corrent method is enough or Node-voltage method. The currents are correct that I have written. – DiodeLight97 Jul 29 '23 at 16:57
  • Then you have drawn the main circuit diagram incorrectly. There is no option here as I explained in my answer. – Andy aka Jul 29 '23 at 17:18
0

Your lower diagram is very confusing and could be the cause of your difficulties. Where wires cross are they connected or not. Assuming there is a junction where the wires from R1,R2,R4, and R5 cross, then then upper diagram is incorrect. A dot should be placed to symbolize there is an electrical connection. The upper diagram shows that R2 is connected to R5 only, and R1 is connected to R1 only. There should never be a crossing junction. See this question on schematic drawing.

This is a correct way of drawing the diagram. Notice the dots where there is to be a wire connection. Note also that the four-branch connection is made into two T-connections.

I prefer that the mesh currents all have the same clock wise orientation sothe the analysis of each mesh is consistent.

schematic

simulate this circuit – Schematic created using CircuitLab

Apply KVL to the outer perimeter mesh to find that the voltage across \$R_5=0\$. So the mesh current through R5, \$I_d=0\$. So mesh D can be ignored, resulting in \$I_2 = I_b\$ and \$I_3 = -I_c\$

$$\left(I_a-I_b\right)R_1+\left(I_a-I_b\right)R_4=2\text{V}\tag{Mesh A}$$

Since \$I_a-I_b=I_1\$ and \$I_a-I_c=-I_4\$

$$I_1 R_1-I_4 R_4=2\text{V}\tag{Equ A}$$

Repeat (methodically) for mesh b and mesh c to get a set of three simultaneous equations.The equation for mesh d is $$0=0\tag{Equ D}$$

Labeled diagrams help maintain clarity and help getting the polarity of voltages and currents correct. KVL must be applied consistently. That is why the mesh currents are all drawn either all clockwise or all counter clockwise.

RussellH
  • 12,496
  • 2
  • 9
  • 34
  • Your schematic is wrong - there's a jumper in the middle of OP's schematic, while in yours there's a node, essentially connecting between \$R_1\$ and \$R_3\$ to \$R_2\$ and \$R_4\$. – uriyabsc Jul 30 '23 at 09:35
  • 1
    The equations obtained in the OP indicate a junction in the centre. So did the OP make a mistake when adding the upper schematic diagram? The lower diagram in the question can be interpreted either way, hence the first part of my answer. The mesh loops are not labeled on either diagram. So the schematic diagram in my answer **might be** wrong @uriyabsc . This lack of clarity in the question could result in the question being closed. – RussellH Jul 30 '23 at 14:47
  • Hi! Yes, I drawed the schematic wrong. Sorry about that. The correct method is to solve it with node-voltage method after removing R5 which worked. Thanks a lot for your answers! – DiodeLight97 Aug 04 '23 at 05:53