A) \[\frac{4}{3}r\]
B) \[\frac{10}{3}r\]
C) \[\frac{5}{3}r\]
D) \[10r\]
Correct Answer: B
Solution :
Resistance of wires CD and ED are in series, \[{{R}_{1}}=2r+2r=4r\] \[\therefore \]Resistance of wire CE is in parallel with\[{{R}_{1}}\] \[\frac{1}{{{R}_{2}}}=\frac{1}{4r}+\frac{1}{2r}=\frac{1+2}{4r}\] \[\frac{1}{{{R}_{2}}}=\frac{3}{4r}\Rightarrow {{R}_{2}}=\frac{4}{3}r\] \[\therefore \]Resistance of wire AC, \[{{R}_{2}}\] and resistance of wire EB are in series \[=r+\frac{4r}{3}+r=\frac{3r+4r+3r}{3}=\frac{10}{3}r\]
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