Six wires each of cross-sectional area A and length l are combined as shown in figure. thermal

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10-2.Transmission of Heat

normal

Six wires each of cross-sectional area $A$ and length $l$ are combined as shown in the figure. The thermal conductivities of copper and iron are $K_1$ and $K_2$ respectively. The equivalent thermal resistance between points $A$ and $C$ is

$\frac{{l({K_1} + {K_2})}}{{{K_1}{K_2}A}}$

$\frac{{2l({K_1} + {K_2})}}{{{K_1}{K_2}A}}$

$\frac{l}{{({K_1} + {K_2})A}}$

$\frac{{2l}}{{({K_1} + {K_2})A}}$

Solution

Let $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ be the thermal of Wheat stone's bridge. the point $\mathrm{B}$ and $\mathrm{D}$ must be at the same temperature when the bridge is balanced. Therefore. thermal resistance of arm $BD$ becomes ineffective.

Now the equivalent circuit at balance is

The effective resistance between $A$ and $C$ is

$\mathrm{R} =\frac{\left(2 \mathrm{R}_{1}\right)\left(2 \mathrm{R}_{2}\right)}{2 \mathrm{R}_{1}+2 \mathrm{R}_{2}}$

$=\frac{2 \mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}} $

${\rm{R}} = \frac{{2\frac{1}{{{{\rm{K}}_1}{\rm{A}}}} \cdot \frac{l}{{{{\rm{K}}_2}{\rm{A}}}}}}{{\frac{l}{{{{\rm{K}}_1}{\rm{A}}}} + \frac{l}{{{{\rm{K}}_2}{\rm{A}}}}}} = \frac{{2l}}{{\left( {{{\rm{K}}_1} + {{\rm{K}}_2}} \right){\rm{A}}}}$

Standard 11

Physics

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normal

Six wires each of cross-sectional area $A$ and length $l$ are combined as shown in the figure. The thermal conductivities of copper and iron are $K_1$ and $K_2$ respectively. The equivalent thermal resistance between points $A$ and $C$ is

Solution

Similar Questions

The intensity of radiation emitted by the Sun has its maximum value at a wavelength of $510\,\, nm$ and that emitted by the North Star has the maximum value at $350\,\, nm$. If these stars behave like black bodies then the ratio of the surface temperature of the Sun and the North Star is

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