Question 361: Current Electricity

Question

A material $$'B'$$ has twice the specific resistance of $$'A'.$$ A circular wire made of $$'B'$$ has twice the diameter of a wire made of $$'A'$$. Then for the two wires to have the same resistance, the ratio $${l \over B}/{l \over A}$$ of their respective lengths must be

Accepted Answer

$${ho _B} = 2{ho _A}$$
$${d_B} = 2{d_A}$$
$${R_B} = {R_A} \Rightarrow {{{ho _B}{\ell _B}} \over {{A_B}}} = {{{P_A}{\ell _A}} \over {{A_A}}}$$
$$	herefore$$ $${{{\ell _B}} \over {{\ell _A}}} = {{{ho _A}} \over {{ho _B}}} 	imes {{d_B^2} \over {d_A^2}}$$ $$ = {{{ho _A}} \over {2{ho _A}}} 	imes {{4d_d^2} \over {d_A^2}} = 2$$

Answer

$$1$$

Answer

$${l \over 2}$$

Answer

$${l \over 4}$$

Answer

$$2$$

Current Electricity

A material $'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of a wire made of 'A'. Then for the two wires to have the same resistance, the ratio {l \over B}/{l \over A}$ of their respective lengths must be

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