Two resistors of the same material and cross section area but different lengths $L_{1}$ & $L_{2}$ have their resistances in the ratio

L1xL2/(L1+L2)

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L1+L2

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L1-L2

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1/L1 + 1/L2

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Solution

The correct option is B
L1+L2

$R = ρ \frac{L}{A}$

Hence, $R_{1} = ρ_{1} \frac{L_{1}}{A_{1}}$ and $R_{2} = ρ_{2} \frac{L_{2}}{A_{2}}$

Given that the material of the resistors is the same, and their cross-section areas are the same
Hence, $ρ_{1} = ρ_{2} = ρ$ and $A_{1} = A_{2} = A$
Thus, $R_{1} = ρ \frac{L_{1}}{A}$ and $R_{2} = ρ \frac{L_{2}}{A}$

Thus, $\frac{R_{1}}{R_{2}} = \frac{ρ \frac{L_{1}}{A}}{ρ \frac{L_{2}}{A}}$

$\frac{R_{1}}{R_{2}} = \frac{L_{1}}{L_{2}}$

Or, $R_{1} : R_{2} = L_{1} : L_{2}$

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Two resistors of the same material and cross section area but different lengths L1 & L2 have their resistances in the ratio

The correct option is B L1+L2 R=ρLA Hence, R1=ρ1L1A1 and R2=ρ2L2A2 Given that the material of the resistors is the same, and their cross-section areas are the same Hence, ρ1=ρ2=ρ and A1=A2=A Thus, R1=ρL1A and R2=ρL2A Thus, R1R2=ρL1AρL2A R1R2=L1L2 Or,R1:R2=L1:L2

Two resistors of the same material and cross section area but different lengths $L_{1}$ & $L_{2}$ have their resistances in the ratio