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well i thought that Both currents are counter-clockwise in direction, the opposite of what I assumed in writing out my previous equations. that the current in the 5V source is the reverse of normal assumptions (due to the overwhelming voltage present in the 21V source.
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OK, let's consider the way current is flowing.
We know it (conventional current) flows from the 21V source upward then right to left through R2, then something happens and it eventually flows left to right through R5 and back up into the 21V voltage source.
We agree on that, right?
The current in R2 and R5 is equivalent, and equivalent to the current through the 21V source (it has to be because these are in series)
The current likewise is equivalent through R1, R4, and the 5V source because they are in series.
If you draw just those arrows, what do you see?
Now examine R3. We've agreed that the (conventional) current flows from top to bottom through it.
Now, pretend you're being swept along by conventional current. From the left side of R4 (you're going left to right here) what path could you follow (remembering to only go in the direction of the arrows) to get back to this point?
We know it (conventional current) flows from the 21V source upward then right to left through R2, then something happens and it eventually flows left to right through R5 and back up into the 21V voltage source.
We agree on that, right?
The current in R2 and R5 is equivalent, and equivalent to the current through the 21V source (it has to be because these are in series)
The current likewise is equivalent through R1, R4, and the 5V source because they are in series.
If you draw just those arrows, what do you see?
Now examine R3. We've agreed that the (conventional) current flows from top to bottom through it.
Now, pretend you're being swept along by conventional current. From the left side of R4 (you're going left to right here) what path could you follow (remembering to only go in the direction of the arrows) to get back to this point?
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And then downward through the 5V voltage source to the node at the bottom left where we started.
Correct.
There are two important issues here,
1) current can only flow in one direction through a wire.
2) current can flow either way through a voltage source.
Correct.
There are two important issues here,
1) current can only flow in one direction through a wire.
2) current can flow either way through a voltage source.
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If you look at the node equation for current at the top central node (the junction of R1, R2, and R3) and the currents, then the practical consequence would be fairly obvious.
This is one reason to start analysing circuits from Kirchoff's laws, not from things that are derived from them -- it insulates you from some of the understanding you would otherwise pick up very naturally.
This is one reason to start analysing circuits from Kirchoff's laws, not from things that are derived from them -- it insulates you from some of the understanding you would otherwise pick up very naturally.
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The current through R3 is the difference between 2 values, it does not mean that one current flows one way and another current flows the other way through that resistor.
If you had taken the basic step of describing the currents at the nodes, it would be clear to you that the current flowing in to any node is balanced by currents flowing out of the node.
Instead you took a shortcut which makes it appear that current is flowing both ways, and doesn't explicitly deal with it as a single current.
The current from the 5V source cannot flow through R3 because the current is flowing the other way through this resistor! In this circuit, the 5V source is effectively slowing down, or acting against the current flow.
If you had taken the basic step of describing the currents at the nodes, it would be clear to you that the current flowing in to any node is balanced by currents flowing out of the node.
Instead you took a shortcut which makes it appear that current is flowing both ways, and doesn't explicitly deal with it as a single current.
The current from the 5V source cannot flow through R3 because the current is flowing the other way through this resistor! In this circuit, the 5V source is effectively slowing down, or acting against the current flow.
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Yes.
Look at the junction of R1, R2, and R3.
Through R2 211 mA enters the node. 92 mA leaves via R1, and 119 via R3 <-- That's an example of Kirchoff's law for node currents
Add up the currents entering and the currents leaving the node. They are equal. (And note that if you were to call currents entering the node positive and those leaving it negative, the sum would be zero).
The complicating factor is that a negative current (leaving a node) will still be a negative current at the next node it encounters, but it's entering, so it should be positive. The problem is solved by not ascribing meaning to the sign, and knowing that it will simply fall out in the end. Another solution is to name your nodes (say) A, B, C, D, E, F, and your currents around node A (assuming it links to nodes B, C, & D) would be IAB, IAC, & IAD. These would sum to zero. at node D, you would talk about IDA, and others. Note that IAD = -IDA.
If you start from this point, you then say that the component between A and D (let's call it R1) has a current IR1 being equal to IAB. Thus IDA = -IR1. Once you've solved for the various currents, you can substitute back to the node currents with the sign telling you the direction, and more importantly summing to zero.
That's complex and long, but it will (hopefully) give you an understanding of the current law and how it works, as well as giving you insight into the actual current flows.
Later you can ignore all of that, jump straight to the currents through the components and then (from experience) determine how the currents flow at a node, even though the currents through the components connected to the node may appear not sum to zero until you fiddle the signs.
Look at the junction of R1, R2, and R3.
Through R2 211 mA enters the node. 92 mA leaves via R1, and 119 via R3 <-- That's an example of Kirchoff's law for node currents
Add up the currents entering and the currents leaving the node. They are equal. (And note that if you were to call currents entering the node positive and those leaving it negative, the sum would be zero).
The complicating factor is that a negative current (leaving a node) will still be a negative current at the next node it encounters, but it's entering, so it should be positive. The problem is solved by not ascribing meaning to the sign, and knowing that it will simply fall out in the end. Another solution is to name your nodes (say) A, B, C, D, E, F, and your currents around node A (assuming it links to nodes B, C, & D) would be IAB, IAC, & IAD. These would sum to zero. at node D, you would talk about IDA, and others. Note that IAD = -IDA.
If you start from this point, you then say that the component between A and D (let's call it R1) has a current IR1 being equal to IAB. Thus IDA = -IR1. Once you've solved for the various currents, you can substitute back to the node currents with the sign telling you the direction, and more importantly summing to zero.
That's complex and long, but it will (hopefully) give you an understanding of the current law and how it works, as well as giving you insight into the actual current flows.
Later you can ignore all of that, jump straight to the currents through the components and then (from experience) determine how the currents flow at a node, even though the currents through the components connected to the node may appear not sum to zero until you fiddle the signs.
Here is a solution using mesh currents for two different cases. In the first case the mesh currents reinforce each other in R3. In the second case the mesh currents oppose each other in R3. Note how the loop equations sum the voltage drops around the loop, and how the direction of the mesh current affects the sign of the voltage drop. Also note that in the setup the only thing which changed is the direction of the I2 mesh, while in the solution to the loop equations the only thing which changed is the sign of the I2 current.
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