A) 24 ohm
B) \[\frac{44}{9}\]ohm
C) 26.4 ohm
D) 18.7 ohm
Correct Answer: C
Solution :
\[\frac{P}{Q}=\frac{R}{{{S}'}}\] (For balancing bridge) Þ \[{S}'=\frac{4\times 11}{9}=\frac{44}{9}\] Þ \[\frac{1}{{{S}'}}=\frac{1}{r}+\frac{1}{6}\] Þ \[\frac{9}{44}-\frac{1}{6}=\frac{1}{r}\] Þ \[r=\frac{132}{5}=26.4\,\Omega \]You need to login to perform this action.
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