A) 0.02
B) 0.1
C) 0.2
D) 1.2
Correct Answer: C
Solution :
[c] \[3=I\text{ }\!\![\!\!\text{ }20+10]\] \[I=\frac{1}{10}A\] \[V\text{ }\!\![\!\!\text{ }across\text{ }wire]=I{{R}_{w}}\] \[=\frac{1}{10}\times 20=2v\] Potential gradient \[\frac{V}{\ell }=\frac{2}{10}\Rightarrow 0.2V/cm\]You need to login to perform this action.
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