An aluminium rod and copper rod of equal length `1 m` and cross-sectional area `1 cm^(2)` are welded as shown. `K_(A1) = 200 W//m^(@)C, K_(Cu) = 400 W//m^(@)C`. Find heat current

Explanation

`R_(A1) = (l)/(K_(A1)A) = (l)/(200 xx 10^(-4)) = 50^(@)C//W`
`R_(Cu) = (l)/(K_(Cu) A) = (1)/(400 xx 10^(-4)) = 25^(@)C//W`
(a) `R_(eq) = R_(A1) + R_(Cu) = 50 + 25 = 75^(@)C//W`
Heat current `i= (100 - 25)/(R_(eq)) = (75)/(75) = 1 W`
(b) `R_(eq) = (R_(A1) . R_(Cu))/(R_(A1) + R_(Cu)) = (50 xx 25)/(50 + 25) = (50)/(3) .^(@)C//W`
Heat current `i = (100 - 25)/(50//3) = 4.5 W`