SATHEE: current-electricity Question 35

Current Electricity

current-electricity Question 35

Question: Q. 13. The temperature coefficient of resistivity, for two materials $A$ and $B$ , are $0.0031 /^{\circ} C$ and $0.0068 /^{\circ} C$ respectively.

Two resistors $R_{1}$ and $R_{2}$ , made from materials $A$ and $B$ , respectively, have resistances of $200 Ω$ and $100 Ω$ at $0^{\circ} C$ . 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 $100^{\circ} C$ . (Neglect the ring corresponding to the tolerance of the carbon resistor).

A [Foreign Comptt. 2016]

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Solution:

Ans.

$R_{t} = R_{0} (1 + α Δ t)$

For Resistance $R_{1}$

$\begin{aligned} R_{1}^{'} & = R_{1} (1 + α Δ t) & = 200 (1 + 0.0031 \times 100) & = 262 Ω \end{aligned}$

For Resistance $R_{2}$

$\begin{aligned} R_{2}^{'} & = 100 (1 + 0.0068 \times 100) & = 168 Ω \end{aligned}$

Hence, Total Resistance in series combination of $R_{1}$ and $R_{2}$ at $100^{\circ} C$ :

$R = R_{1}^{'} + R_{2}^{'} = 262 + 168$

$430 Ω$

$1 / 2$

orange

[CBSE Marking Scheme 2016]

Current Electricity

current-electricity Question 35

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Current Electricity

current-electricity Question 35

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

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Other Resources

Syllabus

Test Platform

Sample Mock Test

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