A) \[260\,\Omega \]
B) \[760\,\Omega \]
C) \[960\,\Omega \]
D) \[1060\,\Omega \]
E) \[1160\,\Omega \]
Correct Answer: B
Solution :
Potential gradient along wire \[=\frac{potential\text{ }difference\text{ }along\text{ }wire}{length\text{ }of\text{ }wire}\] \[\therefore \] \[0.1\times {{10}^{-3}}=\frac{I\times 40}{1000}V/cm\] Current in wire, \[I=\frac{1}{400}A\] Or \[I=\frac{E}{R+R}\] \[\therefore \] \[\frac{2}{40+R}=\frac{1}{400}\] Or \[R=800-40=760\,\Omega \]
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