A) \[R\]
B) \[\frac{2}{3}R\]
C) \[\frac{3}{7}R\]
D) \[\frac{8}{15}R\]
Correct Answer: D
Solution :
[d] The equivalent circuit is as shown in figure. The resistance of arm AOD (=R+R) is in parallel to the resistance R of arm AD. Their effective resistance \[{{R}_{1}}=\frac{2R\times R}{2R\times R}=\frac{2}{3}R\] The resistance of arms AB, BC and CD is \[{{R}_{2}}=R+\frac{2}{3}R+R=\frac{8}{3}R\] The resistance \[{{R}_{1}}\] and \[{{R}_{2}}\]are in parallel. The effective resistance between A and D is \[{{R}_{3}}=\frac{{{R}_{1}}\times {{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{\frac{2}{3}R\times \frac{8}{3}R}{\frac{2}{3}R+\frac{8}{3}R}=\frac{8}{15}R\]You need to login to perform this action.
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