A) \[4\,\Omega \]
B) \[8\,\Omega \]
C) \[9\,\Omega \]
D) \[18\,\Omega \]
Correct Answer: A
Solution :
(a) \[4\,\Omega \] Total current through \[{{R}_{1}}\] \[{{l}_{a}}=\frac{{{R}_{2}}+{{R}_{3}}}{{{R}_{1}}+({{R}_{2}}+{{R}_{3}})}l\] \[=\frac{18}{36+18}=\frac{18}{54}l=\frac{l}{3}\] Also. \[{{l}_{4}}=l\] Since, \[{{P}_{1}}={{P}_{4}}\Rightarrow l_{a}^{2}{{R}_{1}}=l_{4}^{2}{{R}_{4}}\] \[\Rightarrow {{\left( \frac{l}{3} \right)}^{2}}\times 36={{l}^{2}}{{R}_{4}}\] which implies sign \[{{R}_{4}}=4\,\Omega \]
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