question_answer

In the network shown below, the equivalent resistance between P and \[Q\] is \[\frac{4}{3}Q\]. Hence the value of r is

A)  \[3\,\,\Omega \]                                            

B)  \[4\,\,\Omega \]

C)  \[5\,\,\Omega \]                            

D)         \[6\,\,\Omega \]  

Correct Answer: D

Solution :

 Resultant of \[2\] and \[8\] in parallel (ratio\[=4\])                 \[=\frac{4}{5}\times 2=\frac{8}{5}=1\cdot 6\] Resultant of \[6\] and \[4\] in parallel (ratio\[=1\cdot 5\])                 \[=\frac{1.5}{1.6}=0.6\times 4=2.4\] As shown below, these two are in series. The equivalent circuit now consists of \[3,\,\,4,\,\,r\] in parallel.                 \[\frac{1}{3}+\frac{1}{4}+\frac{1}{r}=\frac{3}{4}\] or            \[\frac{1}{r}=\frac{3}{3}-\frac{1}{4}-\frac{1}{3}=\frac{2}{4}-\frac{1}{3}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\] So,          \[r=6\]