Two rectangular rods of thermal resistance $5 k w^{- 1}$ and $10 k w^{- 1}$ are joined in parallel combination their equivalent thermal resistance will be

15 k w^{- 1}

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3.33 k w^{- 1}

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7.5 k w^{- 1}

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6.67 k w^{- 1}

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Solution

The correct option is B $3.33 k w^{- 1}$
In case of parallel combination ( similar to case of resistance in current electricity)
$\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$∴ R_{e q} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
$= \frac{5 \times 10}{5 + 10} = 3.33 k w^{- 1}$

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Two rectangular rods of thermal resistance 5kw−1 and 10kw−1 are joined in parallel combination their equivalent thermal resistance will be

The correct option is B 3.33kw−1In case of parallel combination ( similar to case of resistance in current electricity)1Req=1R1+1R2∴Req=R1R2R1+R2=5×105+10=3.33kw−1

Two rectangular rods of thermal resistance $5 k w^{- 1}$ and $10 k w^{- 1}$ are joined in parallel combination their equivalent thermal resistance will be

The correct option is B $3.33 k w^{- 1}$
In case of parallel combination ( similar to case of resistance in current electricity)
$\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$∴ R_{e q} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
$= \frac{5 \times 10}{5 + 10} = 3.33 k w^{- 1}$