Resistance of a resistor at temperature t- C is R t - R 01- at - bt 2- where R 0 is the resistance at 0- C- The temperature coefficient of resistance at temperature t- C isA- a-2 b tB- 1-a t-b t2-a-2 b tC- a-2 b t-1-a t-b t2D- a-2 b t-21-a t-b t2

The correct option is C a + 2bt(1 + at + bt^2)R_t=R_o(1+α t) α =R_t R_oR_o t lim Δ t→ 0 α =1R_tdRdT α =1R_o(1+a t+bt^2)dR_o(1+a t+bt^2)dt α =1R_o(1+a t+bt^2)R_ ...

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Byju's Answer

Standard XII

Physics

Idea of Temperature

Resistance of...

Question

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

\frac{(1 + a t + b t^{2})}{a + 2 b t}

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(a + 2 b t)

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\frac{a + 2 b t}{(1 + a t + b t^{2})}

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\frac{a + 2 b t}{2 (1 + a t + b t^{2})}

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Solution

The correct option is C $\frac{a + 2 b t}{(1 + a t + b t^{2})}$
$R_{t} = R_{o} (1 + α t)$
$α = \frac{R_{t} - R_{o}}{R_{o} t}$
$l i m Δ t \to 0$
$α = \frac{1}{R_{t}} \frac{d R}{d T}$
$α = \frac{1}{R_{o} (1 + a t + b t^{2})} \frac{d R_{o} (1 + a t + b t^{2})}{d t}$
$α = \frac{1}{R_{o} (1 + a t + b t^{2})} R_{o} (a + 2 b t)$
$α = \frac{a + 2 b t)}{1 + a t + b t^{2}}$

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Resistance of a resistor at temperature t∘C is Rt=R0(1+at+bt2); where R0 is the resistance at 0∘C. The temperature coefficient of resistance at temperature t∘C is

The correct option is C a+2bt(1+at+bt2)Rt=Ro(1+αt) α=Rt−RoRot lim Δt→0 α=1RtdRdT α=1Ro(1+at+bt2)dRo(1+at+bt2)dt α=1Ro(1+at+bt2)Ro(a+2bt) α=a+2bt)1+at+bt2

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