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Kirchoffs law

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¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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No, there is no simple way.

And there is no reason to need to have all your currents positive (or indeed, all of your voltages). If you discover the signs of voltage and current differ in a resistor then you have problems :) Not so for a voltage or current source.

If you now note the voltage drops across each resistor, and after deciding that a (any) node is nominally 0V, you might that the current (conventional or otherwise) does not flow in the direction you think it does.

This might make you wonder how your equations work, given that they were (apparently) predicated on current going in a particular direction.

The answer for you would lie in doing this calculation again, using a method of creating the equations that does not rely on anything more than the basic Kirchoff's laws. I know I'm banging on about this, but maybe there's a reason.
 

tom6123

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ok i calculated the voltage drop across each resistor by dividing the r1,r3 and r4 by 5v, i got r1 at 0.4237v, r3 at 1.39821v and r4 at 3.17775v
is this what you meant
 

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R3 should be around 8 or 9 volts from my calculations (I left the piece of scrap paper at home)
 

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Yes, that looks very much like what I had written down on a piece of paper when I was working this out in bed last night :D
 

Harald Kapp

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is there a simple way of working out the direction of a current in this circuit assuming conventional current flow

Mark the currents in the branches by little arrows as I did for I1 (through R1):
bla-png.13835

Do this for the other currrents. Do this for voltages, too. It doesn't mattter which way the arrows point. For the calculations we assume the direction of the arrow to be the positive current (conventional current flow from "+" to "-" assumed). Likewise voltages across resistors are accordingly positive when the arrow points in the same direction as the current arrow.
This is a convention for doing the math only
After doing the math it may turn out that some voltages and currents are negative, i.e. negative with respect to the direction drawn into the schematic. This only meand that the current doesn't flow as drawn, but in the opposite direction. Kirchhoff's laws are still valid.
 

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Laplace

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is there a simple way of working out the direction of a current in this circuit assuming conventional current flow

The simplest way is to work out the node voltage equations. Current flows from a more positive node to a less positive node.

The node equations sum the currents at each node in terms of the node voltages & resistance to adjacent nodes. I like to write the node equations so that each term (i.e., current) in the equation begins with the same node voltage for that node. Nodes on either side of a voltage source are treated as the same node, while still accounting for the voltage offset. If current sources are involved, currents flowing out of the node are positive -- think of current flowing out of a positive node to ground. The method of node equations is really much simpler and less error prone than using mesh or loop currents. See node equations below:
NodeEquations.png
 

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tom6123

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ok i have managed to work out the direction of the currents
- R₄⋅I₁ + 5 - R₁⋅I₁ - R₃⋅(I₁ - I₂) = 0
- R₃⋅(I₂ - I₁) - R₂⋅I₂ - 21 - R₅⋅I₂ = 0

(-R₁ - R₃ - R₄)⋅I₁ + R₃⋅I₂ = -5
R₃⋅I₁ + (-R₂ - R₃ - R₅)⋅I₂ = 21


which came out to I₁=-0.0919505577, I₂=-0.211335544 which are both anticlockwise
how do i work out the current
or am i able to presume that

R₁ is 91.95mA
R₂ 211.34mA
R₃ 119.39mA
R₄ 91.95mA
R₅ 211.34mA
 
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Harald Kapp

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What is I1, I2, I3,...? You haven't labeled these in the schematic, so it is difficult to guess which ones you mean.
Help us understand:
  1. label the currrents and mark the arrows (as I suggested) in the diagram.
  2. for ease of understanding, name the currents such that I1 flows through R1, I2 flows through R2 etc. This is mathematically or physically not necessary, but makes it much easier connecting the equations with the circuit.
  3. label the currents through the voltage sources Iv1 and Iv2 to distinguish them from currents through resistors.
 

tom6123

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ok is this what you meant?
 

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Harald Kapp

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Halfways. You've indicated the current direction by arrows, that's o.k. You've indicated the voltages by + and -, that's o.k., too (although personally I'd use arrows, too, but that is part of how I was trained).
What's missing is the labels. We still don't know which of the currents is I1, I2, ... and which od the voiltages would be V1, V2... - although the latter is probably not necessary as you can express any voltage by V=Ixx*Rxx.
 

tom6123

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would this circuit suffice ?


also for the current through resistors
would it be correct to write:
IR₁ is 91.95mA
IR₂ 211.34mA
IR₃ 119.39mA
IR₄ 91.95mA
IR₅ 211.34mA
, or would i have to state it as

IR₁ is-91.95mA
IR₂ -211.34mA
IR₃ 119.39mA
IR₄ 91.95mA
IR₅ 211.34mA
 

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Are you certain about the signs on the voltages across R1 and R4?

What is the voltage across R3? What does that tell you about the voltages across R1 and R4?

Sure the arrows you have drawn indicate the way you assumed the current was flowing for the purposes of creating the equation, but what does a negative current for R1 and R4 then imply? Look at the total current, not the results from each individual equation.

As Laplace and I have suggested (to no effect it appears), go back and use the basic identities given by Kirchoff's laws, and not the shortcut method you have used. I'll relent and let you know that you will eventually get the same or equivalent equations, however you'll have a set of more basic equations that will help you in determining (fairly easily) what direction the current is flowing in.
 

tom6123

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Ok i have just realised by looking at the circuit i posted above was wrong and that both the currents flow in an counterclock wise direction.
the voltage across R3, i calculated it at 8.954V
 

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So which way does the current flow in R3?
 

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So what is the counter clockwise path that includes R1?
 

tom6123

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from the -5v to r4,r3 then r1 back to the cell
is this what you meant?
 
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