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31 Mar 2009 12:35:16 IST

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Resistance Temperature Coefficient

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If two resistances ${R}_{1}\;and\;{R}_{2}$ are connected in parallel, then find the equivalent value of $\alpha\;\;(which\;is\;the\;R.T.C.)$ .

If you haven't understood the question, this is how it's like:-
If the resistances would have been connected in series, then the equivalent value of $\alpha$ is given by $\alpha\;=\;\frac{{R}_{1}{\alpha}_{1}\;+\;{R}_{2}{\alpha}_{2}}{R}$ .

I know the approach but I'm not getting the answer independent of $\Delta T$ .

Please give the solution also, along with the answer.

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Comments (6)

Minna

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31 Mar 2009 12:43:54 IST

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temperature co efficient of resistance is the ratio of increase in resistance to the resistance at 273K per

degree rise in temp.

R(t)=R₀{1+(temp coefft)(t)}

where R₀ is resistance at 273K

Minna

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31 Mar 2009 12:45:46 IST

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Im sorry...............

my answer above is completely irrelevant...............

XYZ

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31 Mar 2009 12:47:02 IST

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Yaar, I'm not asking what temperature co-efficient of resistance is???

I'm asking to find the resultant value of temperature coefficient when there are two resistors R1 and R2, connected in parallel.

Bipin Dubey

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31 Mar 2009 15:15:29 IST

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$\frac{R_{1}R_{2}}{R_{1}+R_{2}}(1+\alpha \Delta T)=\frac{R_{1}(1+\alpha_{1}\Delta T)R_{2}(1+\alpha_{2}\Delta T)}{R_{1}(1+\alpha_{1}\Delta T)+R_{2}(1+\alpha_{2}\Delta T)}$

Taking reciprocal on both the sides :

$\left( \frac{1}{R_{1}}+\frac{1}{R_{2}}\right)\left( \frac{1}{1+\alpha\Delta T}\right)=\frac{1}{R_{1}\left( 1+\alpha_{1}\Delta T\right)}+\frac{1}{R_{2}\left( 1+\alpha_{2}\Delta T\right)}$

As we know $\frac{1}{1+x}=(1+x)^{-1}\approx 1-x$ for x<<1..

Same can be applied to above expressions in LHS and RHS since coefficients are very small.

$\left( \frac{1}{R_{1}}+\frac{1}{R_{2}}\right)(1-\alpha\Delta T)=\frac{1-\alpha_{1}\Delta T}{R_{1}}+\frac{1-\alpha_{2}\Delta T}{R_{2}}$

$\alpha = \frac{\frac{\alpha_{1}}{R_{1}}+\frac{\alpha_{2}}{R_{2}}}{\frac{1}{R_{1}}+\frac{1}{R_{2}}}$

$\alpha = \frac{\alpha_{1}R_{2}+\alpha_{2}R_{1}}{R_{1}+R_{2}}$

XYZ

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31 Mar 2009 16:09:32 IST

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I think, without using approximation, we can't get the value of RTC independent of T.

Also, since the linear relationship of resistance with temperature is valid for a small temperature range, so we can use the approximation here.

Thanks a lot for the reply.

s faiyaz

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31 Mar 2009 22:30:54 IST

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