Question

The potential difference between the ends of a wire at all temperatures is constant and is \(240\,\text{V}\). When the temperature of the wire is increased from \(0^\circ\text{C}\) to \(1000^\circ\text{C}\), its resistance increases by \(25\,\Omega\). If the temperature coefficient of resistance of the material of the wire is \(1.25\times10^{-4}\,^\circ\text{C}^{-1}\), then the current through the wire at \(0^\circ\text{C}\) is

• A. \(0.6\,\text{A}\)
• B. \(1.2\,\text{A}\)
• C. \(1.8\,\text{A}\)
• D. \(2.4\,\text{A}\)

Hint:

For metallic conductors, \[ \boxed{ R_t=R_0(1+\alpha t). } \] If the potential difference is constant, \[ \boxed{ I=\frac{V}{R}. } \] As temperature increases, the resistance increases and the current decreases.

Answer 1

The Correct Option is B

Step 1: Use the temperature dependence of resistance. The resistance at temperature \(t\) is \[ R_t=R_0(1+\alpha t), \] where \[ R_0 \] is the resistance at \[ 0^\circ\text{C}. \]

Step 2: Determine the resistance at \(0^\circ\text{C}\). The increase in resistance is \[ R_{1000}-R_0 = R_0\alpha(1000). \] Given, \[ R_{1000}-R_0=25\,\Omega, \] and \[ \alpha=1.25\times10^{-4}\,^\circ\text{C}^{-1}. \] Thus, \[ 25 = R_0(1.25\times10^{-4})(1000), \] \[ 25 = 0.125R_0, \] \[ R_0 = 200\,\Omega. \]

Step 3: Calculate the current. Using Ohm's law, \[ I=\frac{V}{R}. \] Therefore, \[ I = \frac{240}{200} = 1.2\,\text{A}. \] Hence, \[ \boxed{I=1.2\,\text{A}.} \] Therefore, the correct option is \(\boxed{(B)}\).