Two plates of thermal conductivities $k_1$ and $k_2$,cross-sectional areas $A_1$ and $A_2$,and the same thickness $l$ are joined as shown in the figure. The equivalent thermal conductivity $k$ of the combination is:

  • A
    $k_1 A_1 + k_2 A_2$
  • B
    $\frac {k_1 A_1}{ k_2 A_2}$
  • C
    $\frac{k_1 A_1 + k_2 A_2}{A_1 + A_2}$
  • D
    $\frac{k_1 A_2 + k_2 A_1 }{k_1 + k_2}$

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