A) 3 ohms
B) \[\sqrt{39}\,ohms\]
C) \[\sqrt{69}\,ohms\]
D) 10 ohms
Correct Answer: C
Solution :
The given circuit can be simplified as follows \[R=3+\frac{10\times (3+R)}{10+3+R}\]\[=3+\frac{30+10R}{13+R}\] \[R=\frac{39+3R+30+10R}{13+R}=\frac{69+13R}{13+R}\] \[13R+{{R}^{2}}=69+13R\] Þ \[R=\sqrt{69}\Omega \].You need to login to perform this action.
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