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I know that Kirchoff's law says that the incoming current is equal to outgoing current.

But I want to understand his: suppose we have 4 wires like the one in diagram.

enter image description here

So let us consider current i1, comes to the junction and gets split into three direction towards i2, i3, i4 and similar is the case with current i2 but then why we imagine only the current i3 and i4 as outgoing current whereas current is also going to i1 and i2.

May be I am missing something but I am stuck on this problem for very long.

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  • \$\begingroup\$ Your statement of KCL is a bit vague. Better is: the sum of the currents towards a junction is equal to the sum of currents away from that junction. So in your case I1+I2=I3+I4 \$\endgroup\$ – Chu Feb 19 '17 at 13:12
  • \$\begingroup\$ Yeah that's the definition given in the book. But consider it as follows the current i1 at reaching the node has three path to go and similarly for i2 not only two path as shown. \$\endgroup\$ – Shivam Kumar Feb 19 '17 at 13:15
  • \$\begingroup\$ It's just an example, same here: obrazki.elektroda.pl/4446431200_1487510279.png Current entering the node must be equal the current that is leaving the node. That's all. What is important is that the current cannot magically disappear. What's enter the node must leave the node. \$\endgroup\$ – G36 Feb 19 '17 at 13:19
  • \$\begingroup\$ Once drawn on paper, you cannot imagine other ways again. The currents i1 and i2 are entering the nodes and those currents have to leave the nodes. Only path available is other two paths. So, they take those paths. One is named i3 and the other i4. i2 now cannot take path of i1 because, it is the net current which is being represented. \$\endgroup\$ – Umar Feb 19 '17 at 13:22
  • \$\begingroup\$ Yes . I saw that. But I Just want to understand that when i1 reaches the node it has three paths to go but why is outgoing current only in two paths \$\endgroup\$ – Shivam Kumar Feb 19 '17 at 13:23
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It's just an example. You may imagine a different case or example if you want.

Any of those current may be inflowing or outflowing, as long as total current inflowing equals total current outflowing.

Also, currents inflow or outflow depending of the conditions in other nodes. If there is a negative voltage difference with respect to another node and an impedance low enough (i.e., finite) to it, then there will be an inflow of current.

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  • \$\begingroup\$ So what are you saying is that i1 may the incoming current minus the outgoing one. \$\endgroup\$ – Shivam Kumar Feb 19 '17 at 13:13
  • \$\begingroup\$ That way to look at it would be more close to kirchoff loop rule. \$\endgroup\$ – Enric Blanco Feb 19 '17 at 13:18
  • \$\begingroup\$ What I am trying to understand is that you consider i1 , after it reaches the node it has three paths to go isn't it ? \$\endgroup\$ – Shivam Kumar Feb 19 '17 at 13:19
  • \$\begingroup\$ Well, in a narrow sense, yes. If i1 is coming from, say, a power supply and you connect it to three loads, then is has to split in three. What comes in must go out. The node itself doesn't sink any current. \$\endgroup\$ – Enric Blanco Feb 19 '17 at 13:23
  • \$\begingroup\$ So i1 is the net current of what we originally gave and what flown in from i2 ? And similarly with i2? \$\endgroup\$ – Shivam Kumar Feb 19 '17 at 13:25
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You cannot suppose two current directions at the same time. If you consider i1 and i2 enter, and i3 and i4 are outgoing currents then you cannot consider current is outgoing i1 and i2 aswell...

The best thing you can do to proof it is to draw a simple circuit and write down the KCL equations. You can assume whatever currents direction you want, but once you have resolved the equations you will see which are the real "directions".

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  • \$\begingroup\$ Could you please elaborate a bit. \$\endgroup\$ – Shivam Kumar Feb 19 '17 at 13:17
  • \$\begingroup\$ I am not supposing two directions. After i1 has reached the node it has three ways to go. Isn't it? \$\endgroup\$ – Shivam Kumar Feb 19 '17 at 13:18
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Here is circuit with 2 inputs and 2 outputs

schematic

simulate this circuit – Schematic created using CircuitLab

Here is circuit with 1 input and 3 outputs.

schematic

simulate this circuit

Here is circuit with 4 inputs.

schematic

simulate this circuit

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