Resistance of a resistor at temperature t- C is Rt-R01- t- t2- where R0 is the resistance at 0-C- The temperature coefficient of resistance at temperature t- C is

The correct option is C α +2β t/(1+α t+β t^2) Given, #160;R_t=R_0(1+α t+β t^2) ⇒dR_t/dt=R_0(α+2β t) Temperature coefficient of resistance at t^o ...

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Byju's Answer

Standard XII

Physics

Dependence of Resistance on Temperature

Resistance of...

Question

Resistance of a resistor at temperature $t^{\circ} C$ is $R_{t} = R_{0} (1 + α t + β t^{2})$ , where $R_{0}$ is the resistance at $0^{\circ} C$ . The temperature coefficient of resistance at temperature $t^{\circ} C$ is

\frac{(1 + α t + β t^{2})}{α + 2 β t}

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(α + 2 β t)

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\frac{α + 2 β t}{(1 + α t + β t^{2})}

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\frac{α + 2 β t}{2 (1 + α t + β t^{2})}

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Solution

The correct option is C $\frac{α + 2 β t}{(1 + α t + β t^{2})}$
Given, $R_{t} = R_{0} (1 + α t + β t^{2}) \Rightarrow \frac{d R_{t}}{d t} = R_{0} (α + 2 β t)$
Temperature coefficient of resistance at $t^{o} C$ is
$α = \frac{1}{R_{t}} \frac{d R_{t}}{d t} = \frac{1}{R_{0} (1 + α t + β t^{2})} . R_{0} (α + 2 β t)$ $= \frac{α + 2 β t}{1 + α t + β t^{2}}$

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Resistance of a resistor at temperature t∘C is Rt=R0(1+αt+βt2), where R0 is the resistance at 0∘C. The temperature coefficient of resistance at temperature t∘C is

The correct option is C α+2βt(1+αt+βt2)Given, Rt=R0(1+αt+βt2)⇒dRtdt=R0(α+2βt)Temperature coefficient of resistance at toC is α=1RtdRtdt=1R0(1+αt+βt2).R0(α+2βt) =α+2βt1+αt+βt2

Resistance of a resistor at temperature $t^{\circ} C$ is $R_{t} = R_{0} (1 + α t + β t^{2})$ , where $R_{0}$ is the resistance at $0^{\circ} C$ . The temperature coefficient of resistance at temperature $t^{\circ} C$ is