The same mass of copper is drawn into two wir | vedclass

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Question

(d) $H = {i^2}RT = {i^2}\left( {\frac{{\rho l}}{A}} \right)\,t = \frac{{{i^2}\rho Vt}}{{{A^2}}}$ $(V =$ volume, $= Al)$

$ \Rightarrow $  $H \p

Vedclass · Accepted Answer

$16:1$

Vedclass · Answer

(d) $H = {i^2}RT = {i^2}\left( {\frac{{\rho l}}{A}} \right)\,t = \frac{{{i^2}\rho Vt}}{{{A^2}}}$ $(V =$ volume, $= Al)$

$ \Rightarrow $  $H \propto \frac{1}{{{r^4}}}$ $ \Rightarrow $ $\frac{{{H_1}}}{{{H_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^4} = {\left( {\frac{2}{1}} \right)^4} = \frac{{16}}{1}$.

The same mass of copper is drawn into two wires $1\, mm$ and $2\, mm$ thick. Two wires are connected in series and current is passed through them. Heat produced in the wire is in the ratio

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