A) 0.1
B) 0.2
C) 0.4
D) None of these
Correct Answer: B
Solution :
\[Current\,\,i\,\,=\,\,\frac{V}{R}=\frac{3}{10+20}\,=\,\,0.1\,\,A\] \[\therefore \] Potential difference across \[20 \Omega = 0.1 \times 20 = 2 V\] Potential gradient \[=\,\,\,\frac{Potential\text{ }difference\text{ }across\,\,20\,\,\Omega }{length\text{ }of\text{ }20\,\Omega \text{ }wire}\,~~~~~~~~~~~\text{ }200\mathbf{2}.~~~~~~~~~~~length\text{ }of\text{ }20Q\text{ }wire\mathbf{1}.~~~~~~~~~~~Potential\text{ }difference\text{ }a\] \[=\,\,\,\,\frac{2}{10}=0.2\,V-{{m}^{-1}}\]You need to login to perform this action.
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