Current Electricity Charging Discharging Of Capacitors Question 144
Question: Two wires of resistance R1 and R2 have temperature coefficient of resistance $ {\alpha _{1}}\text{ and }{\alpha _{2}} $ , respectively. These are joined in series. The effective temperature coefficient of resistance is
[MP PET 2003]
Options:
A) $ \frac{{\alpha _{1}}+{\alpha _{2}}}{2} $
B) $ \sqrt{{\alpha _{1}}{\alpha _{2}}} $
C) $ \frac{{\alpha _{1}}R _{1}+{\alpha _{2}}R _{2}}{R _{1}+R _{2}} $
D) $ \frac{\sqrt{R _{1}R _{2}{\alpha _{1}}{\alpha _{2}}}}{\sqrt{R _{1}^{2}+R _{2}^{2}}} $
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Answer:
Correct Answer: C
Solution:
$ {R _{t _{1}}}=R _{1}(1+{\alpha _{1}}t) $ and $ {R _{t _{2}}}=R _{2}(1+{\alpha _{2}}t) $ Also $ {R _{eq.}}={R _{t _{1}}}+{R _{t _{2}}}\Rightarrow R _{eq}=R _{1}+R _{2} $
$ +(R _{1}{\alpha _{1}}+R _{2}{\alpha _{2}})t $
Therefore $ R _{eq}=(R _{1}+R _{2}){ 1+( \frac{R _{1}{\alpha _{1}}+R _{2}{\alpha _{2}}}{R _{1}+R _{2}} ).t } $ So $ {\alpha _{eff}}=\frac{R _{1}{\alpha _{1}}+R _{2}{\alpha _{2}}}{R _{1}+R _{2}} $
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