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$R=R_1=R_2=R_3=R_4=R_5=R_6= 10 \Omega$ $R_2 \quad and \quad R_3 \quad with \quad R_4 \quad and \quad R_5 \quad are\quad in \quad series$

Therefore, $\frac{1}{R_{tot}} = \frac{1}{R_1}+ \frac{1}{R_2+R_3}+\frac{1}{R_4+R_5}= \frac{1}{R}+\frac{1}{2R}+\frac{1}{2R}= \frac{2}{R}$

$R_{tot} = \frac{R}{2}$

What I don't get is, why we can ignore $R_6$?


marked as duplicate by Qmechanic Jan 5 '18 at 11:35

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  • $\begingroup$ is that solution correct @BasilHallaq $\endgroup$ – QuIcKmAtHs Jan 5 '18 at 11:33
  • $\begingroup$ I don't think the question is a duplicate since, because of the values of the resistors, some simplification to get the answer is possible. $\endgroup$ – jim Jan 5 '18 at 11:43
  • $\begingroup$ You should apply symmetry arguments. Mid point between R2/R3 is exactly the same as mid point between R4/R5. So no voltage drop on R6 and as per @jim answer, you can effectively ignore this resistor. $\endgroup$ – npojo Jan 5 '18 at 12:10

One approach to determining the effective resistance is to place a perfect source of emf, say E, across the terminals A and B. Then work out the current drawn by the battery, say I. If a battery of emf E draws a current I, then the effective resistance across the terminals is E/I. Because of the values of resistors you use, the potential difference across the resistor $R_6$ will be zero, so no current flows through this resistor and you can effectively remove this resistor from the circuit.


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