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  • In the two circuits as shown in the figure  <div c

In the two circuits as shown in the figure

 

Option: 1

\mathrm{R}_{\mathrm{AB}}=\mathrm{R}_{\mathrm{CD}}=(2+\sqrt{3}) \Omega


Option: 2

\mathrm{R}_{\mathrm{AB}}=(\sqrt{3}+1) \Omega


Option: 3

R_{C D}=(\sqrt{5}+1) \Omega


Option: 4

 None of these
 


Answers (1)

best_answer

 

\mathrm{R_{A B}=1+\frac{x}{x+1}=x }
On solving we get

\mathrm{x=\left(\frac{\sqrt{5}+1}{2}\right) \Omega }

and

\mathrm{R_{C D}=\frac{1+y}{2+y}=y }

on solving we get

\mathrm{y=\frac{(\sqrt{5}-1)}{2} \Omega }
Hence,

\mathrm{R}_{\mathrm{AB}}>\mathrm{R}_{\mathrm{CD}}

Posted by

himanshu.meshram

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