Q. The resistance of a wire of iron is $10\ ohm$ and temp. coefficient of resistance is $5 \times 10^{- 3}/℃.$ At $20℃$ it carries $30\ milliampere$ of current. Keeping constant potential difference between its ends, the temperature of the wire is raised to $120℃.$ The current in $milliampere$ that flows in the wire is
A) 20
B) 15
C) 10
D) 40
Explanation
(a) $\frac{R_{1}}{R_{2}} = \frac{(1 + \alpha t_{1})}{(1 + \alpha t_{2})} \Rightarrow \frac{10}{R_{2}} = \frac{(1 + 5 \times 10^{- 3} \times 20)}{(1 + 5 \times 10^{- 3} \times 120)} \Rightarrow R_{2} \approx 15\Omega$
Also $\frac{i_{1}}{i_{2}} = \frac{R_{2}}{R_{1}} \Rightarrow \frac{30}{i_{2}} = \frac{15}{10} \Rightarrow i_{2} = 20\ mA$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.