Question

The resistance of an iron wire is $10\Omega$ and its temperature coefficient of resistance is $5\times {10}^{-3}{/}^{o}C$. A current of 30 mA is flowing in it at ${20}^{o}C$. Keeping potential difference across its ends constant, if its temperature is increased to ${120}^{o}C$ then the current flowing in the wire will be (in mA)

• A. $20$
• B. $15$
• C. $10$
• D. $40$

Answer 1

Answer: (A)

$20$

Given:

Temperature coefficient of resistance, $\alpha =5\times {10}^{-3}{/}^{\circ }C$

Initial temperature, $T_0={20}^{\circ }$

Initial resistance at $T_0$, $R_0=10\Omega$

Initial current at $T_0$, $I_0=30mA$

Increased temperature, $T_1={120}^{\circ }$

Increased resistance is,

$R_1=R_0[1+\alpha (T_1-T_0\left)\right]$

$R_1=10[1+5\times {10}^{-3}(120-20\left)\right]$

$R_1=10[1+5\times {10}^{-1}]$

$R_1=10[1+0.5]$

$R_1=10\left[1.5\right]=15\Omega$

Since, potential difference across the ends of wire is constant,

$V=I_0{R}_0=I_1{R}_1$

$\Rightarrow I_1=\frac{I_0{R}_0}{R_1}$

$\therefore I_1=\frac{\left(30\right)\left(10\right)}{15}$

$\therefore I_1=\frac{300}{15}=20mA$