Current Electricity Charging Discharging Of Capacitors Question 164
Question: The resistance of a wire of iron is 10 ohms and temp. coefficient of resistivity is $ 5\times {{10}^{-3}}/{}^\circ C $ . At $ 20{}^\circ C $ it carries 30 milliamperes of current. Keeping constant potential difference between its ends, the temperature of the wire is raised to $ 120{}^\circ C $ . The current in milliamperes that flows in the wire is
[MP PMT 1994]
Options:
20
15
10
40
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{R _{1}}{R _{2}}=\frac{(1+\alpha t _{1})}{(1+\alpha t _{2})} $
Therefore $ \frac{10}{R _{2}}=\frac{(1+5\times {{10}^{-3}}\times 20)}{(1+5\times {{10}^{-3}}\times 120)} $
Therefore $ R _{2}\approx 15\Omega $ Also $ \frac{i _{1}}{i _{2}}=\frac{R _{2}}{R _{1}} $
Therefore $ \frac{30}{i _{2}}=\frac{15}{10} $
Therefore $ i _{2}=20,mA $
BETA
