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# Physics/circuit

Question

ckt
in the circuit of part b
I understand that the two 12's are parallel for first step but after that ,we can join the 4 and the top 6 in parallel(if connect the equivalent of two 12's of first step in place of the right most 12 k resistor), and then make the equivalent with remaining 6,why did'nt do this ?

Hello Ali,

The circuit is drawn in a way that makes it look more complicated than necessary. If you redraw the circuit, it will be easier to see the sense of doing the analysis their way, and perhaps also show the problem with what you suggested doing.

1. Draw the upper of the 2 circles.
2. From that circle, draw a fairly long horizontal line. Put the 6K in the middle of it.
3. Draw a vertical line downward from the right end of the previous line. Put one of the 12 K resistors in the middle of this line.
4. Draw another vertical line, with 12K, downward from just to the left of the previous vertical line.
5. Draw another horizontal line, with no resistor in it from the bottom of the 2 vertical lines over to the lower of the 2 circles.
6. Draw a third vertical line, starting from just to the left of the 6K, down to where it meets the horizontal line of #5. Include the 4K in the middle of this vertical line.

Compare the drawing resulting from these steps and the drawing the problem came with until you see that they show equivalent connections. I think you will then see that the analysis the problem gave (yielding an equivalence resistance of 3K) makes sense with this drawing.

I hope this helps,
Steve

Physics

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#### Steve Johnson

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