Why is effective resistance 4 ohm and not 4.4 ohm?

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Homework Statement


What is the effective resistance of this circuit?

Homework Equations


The answer given is 3A. so it means the total effective resistance is 4 ohm. But my answer is 4.4 ohm

The Attempt at a Solution


My attempt:
1/(1/6 + 1/4)+2 = 4.4 ohm
 

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andrevdh said:
diagram/circuit?
Thank you. I've just uploaded the picture.
 
Icy98 said:
1/(1/6 + 1/4)+2 = 4.4 ohm

Try to draw the circuit again, but "split up" the connection in point A. What happens with R3 and the connection at its right handside?
 

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It is "shorted out" by the wire forming the corner, so there is in effect "no resitance" due to the wire. That means that you can redraw the circuit without the R3 resistor.
 
Think of electric current flowing like water from the + terminal towards the - terminal of the battery.
 
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Icy98 said:
What is the reason that we can omit R3?

Complete the drawing from your 3rd post with the missing cable/connection. Then you will see.
 
Icy98 said:
What is the reason that we can omit R3?
Good thread ! You really adapted quickly to the PF culture :smile: !

In the upper right of the original picture, there is a ' 0 ##\Omega## resistor ' in parallel with the 2 ##\Omega## resistor. Does that help ?

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Thanks a lot to stockzahn, andrevdh and BvU for helping![emoji16][emoji5]️
 
It's a pleasure. The potential difference over a small resistor, or in this case the corner wire, is almost zero. That means that the potential difference over the R3 resistor is also almost zero so that almost no current flows through it. It is thus not contributing to the circuit and can be ignored or eliminated.