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10-2.Transmission of Heat
normal
Five wires each of cross-sectional area $A$ and length $l$ are combined as shown. The thermal conductivity of copper and steel are $k_1$ and $k_2$ respectively. The equivalent thermal resistance between $A$ and $C$ is
A
$\frac{l}{{\left( {{k_1} + {k_2}} \right)\,A}}$
B
$\frac{{2l}}{{\left( {{k_1} + {k_2}} \right)\,A}}$
C
$\frac{{l\,\left( {{k_1} + {k_2}} \right)}}{{{k_1}{k_2}A}}$
D
$\frac{{l{k_1}{k_2}}}{{k_1^2 + k_2^2}}$
Solution
$\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \Rightarrow \frac{1}{R_{e q}}=\frac{k_{1} A}{2 \ell}+\frac{k_{2} A}{2 \ell}$
$R_{e q}=\frac{2 \ell}{\left(k_{1}+k_{2}\right) A}$
Standard 11
Physics
