For the circuit shown in the figure,calculate | vedclass

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(B) To find the equivalent resistance between $A$ and $B$,we simplify the circuit from the rightmost end.$1$. The last two resistors (a series $R$ and

Vedclass · Accepted Answer

$R$

Vedclass · Answer

(B) To find the equivalent resistance between $A$ and $B$,we simplify the circuit from the rightmost end.$1$. The last two resistors (a series $R$ and a vertical $R$) are in series: $R_{	ext{eq1}} = R + R = 2R$.$2$. This $2R$ is in parallel with the vertical $2R$ resistor: $R_{	ext{eq2}} = \frac{2R 	imes 2R}{2R + 2R} = R$.$3$. Now,this $R$ is in series with the next horizontal $R$: $R_{	ext{eq3}} = R + R = 2R$.$4$. This $2R$ is in parallel with the next vertical $2R$: $R_{	ext{eq4}} = \frac{2R 	imes 2R}{2R + 2R} = R$.$5$. Continuing this pattern,the next horizontal $R$ is in series with $R$: $R_{	ext{eq5}} = R + R = 2R$.$6$. This $2R$ is in parallel with the next vertical $2R$: $R_{	ext{eq6}} = \frac{2R 	imes 2R}{2R + 2R} = R$.$7$. Finally,this $R$ is in series with the first horizontal $R$: $R_{	ext{eq7}} = R + R = 2R$.$8$. This $2R$ is in parallel with the first vertical $2R$: $R_{	ext{eq8}} = \frac{2R 	imes 2R}{2R + 2R} = R$.Thus,the equivalent resistance between $A$ and $B$ is $R$. The correct option is $B$.

For the circuit shown in the figure,calculate the resistance between the points $A$ and $B$.

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