Two resistances R1 and R2 are made of different materials- The temperature coefficient of the material of R1 is - and that of the material of R2 is - The resistance of the series combination of R1 and R2 does not change with temperature- then the ratio of resistances of the two wires R1R2 at 0-C will be -

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Resistance and Resistors

Two resistanc...

Question

Two resistances $R_{1}$ and $R_{2}$ are made of different materials. The temperature coefficient of the material of $R_{1}$ is $α$ and that of the material of $R_{2}$ is $- β$ . The resistance of the series combination of $R_{1}$ and $R_{2}$ does not change with temperature, then the ratio of resistances of the two wires $\frac{R_{1}}{R_{2}}$ at $0^{\circ} C$ will be :

\frac{α}{β}

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\frac{α + β}{α - β}

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\frac{α^{2} + β^{2}}{α β}

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\frac{β}{α}

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Solution

The correct option is D $\frac{β}{α}$

Let the resistances after an increase in temperature $Δ T$ be $R_{1}^{'}$ and $R_{2}^{'}$

The combination of resistances is independent of temperature rise.

So, $R_{1} + R_{2} = R_{1}^{'} + R_{2}^{'}$

$R_{1} + R_{2} = R_{1} (1 + α Δ T) + R_{2} (1 - β Δ T)$

$\Rightarrow R_{1} α Δ T = R_{2} β Δ T$

$\Rightarrow$ $\frac{R_{1}}{R_{2}} = \frac{β}{α}$

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Resistance and Resistors

Standard XII Physics

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The correct option is D βα Let the resistances after an increase in temperature ΔT be R′1 and R′2 The combination of resistances is independent of temperature rise. So, R1+R2=R′1+R′2 R1+R2=R1(1+αΔT)+R2(1−βΔT) ⇒R1αΔT=R2βΔT ⇒ R1R2=βα