A) 60
B) 15
C) 30
D) 45
Correct Answer: B
Solution :
\[{{\operatorname{R}}_{eff}}={{R}_{wire}}+{{R}_{resistor}}\] \[=10\Omega +990\Omega =1000\Omega \] 'V drop across wire \[{{\operatorname{V}}_{w}}=I\times Rwire\] \[{{\operatorname{V}}_{w}}=\frac{V\,\,battery}{{{\operatorname{R}}_{eff}}}\times 10\Omega \] \[{{\operatorname{V}}_{w}}=\frac{2}{1000}\times 10=2\times {{10}^{-2}}\operatorname{volt}\] \[\frac{\operatorname{V}}{\ell }=0.001\,\operatorname{volt}\]
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