A) 2 V
B) 3 V
C) 1 V
D) 1.5 V
Correct Answer: C
Solution :
| [c] Total current drawn from the battery |
| \[i=\frac{E}{R+r}=\frac{6}{100+0}=0.06\,A\] |
| Resistance of 50 cm wire is |
| \[R'=\frac{\rho l'}{A}=\left( \frac{\rho }{A} \right)l'\] |
| \[=\left( \frac{R}{l} \right)l'\] \[\left( \because R=\frac{\rho l}{A} \right)\] |
| \[=\frac{100}{300}\times 50\] |
| So, \[R'=\frac{50}{3}\Omega \] |
| Hence, the potential difference between two points on the wire separated by a distance \[l'\] is |
| \[V=iR'=0.06\times \frac{50}{3}=1\,V\] |
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