current-electricity Question 35
Question: Q. 13. The temperature coefficient of resistivity, for two materials $A$ and $B$, are $0.0031 /{ }^{\circ} \mathrm{C}$ and $0.0068 /{ }^{\circ} \mathrm{C}$ respectively.
Two resistors $R_{1}$ and $R_{2}$, made from materials $A$ and $B$, respectively, have resistances of $200 \Omega$ and $100 \Omega$ at $0^{\circ} \mathrm{C}$. Show on a diagram, the ‘colour code’, of a carbon resistor, that would have a resistance equal to the series combination of $R_{1}$ and $R_{2}$ at a temperature of $100^{\circ} \mathrm{C}$. (Neglect the ring corresponding to the tolerance of the carbon resistor).
A [Foreign Comptt. 2016]
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Solution:
Ans.
$$ R_{t}=R_{0}(1+\alpha \Delta t) $$
For Resistance $R_{1}$
$$ \begin{aligned} R_{1}^{\prime} & =R_{1}(1+\alpha \Delta t) \ & =200(1+0.0031 \times 100) \ & =262 \Omega \end{aligned} $$
For Resistance $R_{2}$
$$ \begin{align*} R_{2}^{\prime} & =100(1+0.0068 \times 100) \ & =168 \Omega \end{align*} $$
Hence, Total Resistance in series combination of $R_{1}$ and $R_{2}$ at $100^{\circ} \mathrm{C}$ :
$R=R_{1}{ }^{\prime}+R_{2}{ }^{\prime}=262+168$
$430 \Omega$
$1 / 2$
orange
[CBSE Marking Scheme 2016]
