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Apply a voltage of +v to point A and −v to point B. The voltage along the horizontal dividing line in the middle is 0V. Now look at the four resistors that connect the outer circle to the inner circle. The outer ends of these four resistors are all at a voltage of 0V. Hence, those four resistors, the four resistors of the inner circle, and the four resistors of the inner cross are all shorted out of the circuit.
Removing those twelve resistors yields a circuit with 3 sections in series, each of which has a resistance of (R+R)∥(R+R)=R. Therefore, the total resistance between points A and B is 3R.
There are two shortings in this circuit which will not allow any current to flow to the inner branches. Thus, simplifying it we have three resistances in series and thus the answer is 3R.
the Req of the 2 triangles above and below is R.
If you restructure the diagram you will sea that it a wheatstone bridge.So you can eliminate the circuits inside the main circle and evaluate the Req for the outer circle=R.
So , the final result will be R+R+R=3R
The 2 triangles have obviously r as eqln. resistance In case of circle setup inside the circle is shorted. Effectively outer perimeter of circle accounts for resistance and that also comes out to be r. So
++'R'effective=r+r+r=3r++
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Apply a voltage of +v to point A and −v to point B. The voltage along the horizontal dividing line in the middle is 0V. Now look at the four resistors that connect the outer circle to the inner circle. The outer ends of these four resistors are all at a voltage of 0V. Hence, those four resistors, the four resistors of the inner circle, and the four resistors of the inner cross are all shorted out of the circuit.
Removing those twelve resistors yields a circuit with 3 sections in series, each of which has a resistance of (R+R)∥(R+R)=R. Therefore, the total resistance between points A and B is 3R.
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Correct ! Those resistors will get eliminated , as the voltage will remain the same on the wires with no resistance between them . Voting you up !
aaa
This is the circuit transformation where each resistance are 'r'. If you can't view this image, go to bigger image
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now I have to convert the star-delta conversion for any one triangle....isn't it?
There are two shortings in this circuit which will not allow any current to flow to the inner branches. Thus, simplifying it we have three resistances in series and thus the answer is 3R.
the Req of the 2 triangles above and below is R. If you restructure the diagram you will sea that it a wheatstone bridge.So you can eliminate the circuits inside the main circle and evaluate the Req for the outer circle=R. So , the final result will be R+R+R=3R
Ahhh!!A classical case of 1)WHEATSTONE BRIDGE 2)STAR- DELTA TRANSFORM....
The 2 triangles have obviously r as eqln. resistance In case of circle setup inside the circle is shorted. Effectively outer perimeter of circle accounts for resistance and that also comes out to be r. So ++'R'effective=r+r+r=3r++