Question

A wire of length 'l' has a resistance 'R'. If half of the length is stretched to make the radius half of its original value then find the final resistance of wire.

• A. 2R
• B. $\frac{2R}{17}$
• C. $\frac{17R}{2}$
• D. $\frac{2}{17R}$

Answer 1

Answer: (C)

$\frac{17R}{2}$

Answer 2

R¢ =$\left( \frac{r}{r/2} \right)^{4}$ × $\frac{R}{2}$ $\left\lbrack R_{final} = \left( \frac{r_{original}}{r_{final}} \right)^{4} \times R_{original} \right\rbrack$R¢ =16 ×

R/2 = 8R; \ R_Net = R¢ + R¢¢ = 8R + R/2 =$\frac{17R}{2}$