Question 48: Current Electricity

Question

The resistance of a wire at room temperature $$20^{\circ} \mathrm{C}$$ is found to be $$10 \Omega$$. If resistance of the wire increases by $$10 \%$$, then the temperature of the wire will be (The temperature coefficient of the material of the wire is $$0.002 /{ }^{\circ} \mathrm{C}$$)

Accepted Answer

We can solve the problem using the concept of temperature dependence of resistance, which is given by the relation:
$$ R = R_0 (1+\alpha T) $$
Here,
Ã¢ÂÂ¢Ã¢ÂÂ$$R$$ is the resistance at temperature $$T$$ (measured from 0ÃÂ°C),
Ã¢ÂÂ¢Ã¢ÂÂ$$R_0$$ is the resistance at 0ÃÂ°C, and
Ã¢ÂÂ¢Ã¢ÂÂ$$\alpha$$ is the temperature coefficient.
However, the wireÃ¢ÂÂs resistance is given at room temperature (20ÃÂ°C) as 10 ÃÂ©. So first, we find the resistance at 0ÃÂ°C:
Write the relation at 20ÃÂ°C:
$$    R(20) = R_0 (1+ \alpha 	imes 20)    $$
Substitute the known values:
$$    10 = R_0 (1+0.002	imes20)= R_0 (1+0.04)= 1.04 \, R_0    $$
Solve for $$R_0$$:
$$    R_0 = \frac{10}{1.04}\approx 9.615 \, \Omega    $$
Now, when the resistance increases by 10%, the new resistance becomes:
$$ R' = 10\,\Omega 	imes 1.10 = 11 \, \Omega $$
This new resistance at an unknown temperature $$T$$ (measured from 0ÃÂ°C) is given by:
$$ 11 = R_0 (1+\alpha T) $$
Substitute $$R_0 \approx 9.615 \, \Omega$$ and $$\alpha = 0.002$$:
$$ 11 = 9.615 \, (1+0.002T) $$
Now, solve for $$T$$:
Divide both sides by 9.615:
$$    1+0.002T = \frac{11}{9.615}\approx 1.144    $$
Subtract 1 from both sides:
$$    0.002T = 1.144-1 = 0.144    $$
Solve for $$T$$:
$$    T = \frac{0.144}{0.002} = 72\, ^\circ 	ext{C}    $$
Thus, the new temperature of the wire is $$72^\circ 	ext{C}$$, which corresponds to Option D.

Answer

$$52^0 \mathrm{C}$$

Answer

$$22^0 \mathrm{C}$$

Answer

$$62^0 \mathrm{C}$$

Answer

$$72^0 \mathrm{C}$$

Current Electricity

The resistance of a wire at room temperature $20^{\circ} \mathrm{C} is found to be 10 \Omega. If resistance of the wire increases by 10 \%, then the temperature of the wire will be (The temperature coefficient of the material of the wire is 0.002 /{ }^{\circ} \mathrm{C}$)

Choose the correct answer: