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I'm a newbie who just started studying EE and I'm presented with this problem as shown here.

enter image description here

I have to calculate all currents, meaning I1, I2, I3, I4, I5. I have the value of each resistor and the value of the current source I, which is 1A.

Now, for I1 I did the following: I added R2 and R3 (I think it's doable because there's an empty wire between them and R4...) and then I multiplied that value I got with R1 and divided it by the sum of those 2 components...

Basically it was 30 * 20 / 50 and I got 12 ohms.

Now I use the formula for calculating current. I1 = I * R4 / (12 + R4) So now I1 is 0.6A. This value is correct.

To get I2 I can use Kirchoff's current law and I get that 1 = 0,6 + I2, from which I get that I2 = 0.4. So far so good.

Now, how do I go about calculating the rest?

Since I5 is an empty wire, can I just say that the current there is the same as I2?

According to the solutions presented I5 is 0,4 too, I3 is -0,4 and I4 is 0. Now I know that I = I1 + I2 + ... + In... so the sum of all these Is must be 1. But why isn't I4 0.4 and I5 0?

What's the thinking behind it? I'm just not very clear as to how this works.

Thanks

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  • \$\begingroup\$ you can use the wire (call it R5) in your calculations ... it is a zero ohm resistor ... you have 0Ω in parallel with 20Ω ... from that you can calculate how much current flows in each \$\endgroup\$ – jsotola Feb 24 '18 at 5:13
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Since I5 is an empty wire, can I just say that the current there is the same as I2?

Yes it is a short circuit so I2 = I5

But why isn't I4 0.4 and I5 0?

R4 is shorted by the short circuit hence all the current passes through that short circuit and none passes through R4. Therefore I5 is the same magnitude as I3 and I2.

If the I5 short wasn't there I4 would equal I2 = -I3

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  • \$\begingroup\$ Oh yes, I thought that'd be the case! Thank you very much, helped a lot! \$\endgroup\$ – Aron Bara Feb 23 '18 at 17:54
  • \$\begingroup\$ By the way, when can I say that a R is shorted? I think I understand the intuition behind it but can't translate this to words or a general statement. \$\endgroup\$ – Aron Bara Feb 23 '18 at 18:01
  • \$\begingroup\$ If a resistor is bypassed in such a way as to prevent any current flowing through it, it can be said to be shorted-out. \$\endgroup\$ – Andy aka Feb 23 '18 at 18:07
  • \$\begingroup\$ @AronBara when both ends are connected to the same place/node. Note the top of 5 and the bottom of 5 is the same node. \$\endgroup\$ – Trevor_G Feb 23 '18 at 18:07
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    \$\begingroup\$ In the real world of course, the short circuit (I5) would have some resistance, maybe milliohms, so I4 wouldn't be zero, but tiny (inversely proportional to the ratio between the resistances) and the two currents would sum to I2. \$\endgroup\$ – Brian Drummond Feb 23 '18 at 18:27
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The trick is to realize the I5 line effectively bypasses R4.

Despite how it is drawn, there are actually only three nodes in this circuit.

enter image description here

Since both ends of R4 connect to the same node, it is shorted out.

If you redraw your circuit like this.... If it pretty obvious there is no current going through R4.

schematic

simulate this circuit – Schematic created using CircuitLab


PS: It is not a stupid question, lots of folks get tripped up by this one. Which is why teachers keep throwing them at you, so you get used to looking for the shorts.

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I think it can simply be proved that a current divider with one resistor shorted out passes all the current.

schematic

simulate this circuit – Schematic created using CircuitLab

The current through R1 and R2 is given simply as:

$$I_{R_1}=\frac{R_2}{R_1+R_2}I$$ $$I_{R_2}=\frac{R_1}{R_1+R_2}I$$

So now if you replace R1 with a short wire (R1=Rsc=0 ohms) you get: $$I_{R_1}=\frac{R_2}{0+R_2}I=I$$ $$I_{R_2}=\frac{0}{0+R_2}I=0$$

So as you see no current will pass through R2 and it all goes through R1.

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