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Step-by-Step Solution
Step 1: Identify the physical principles involved
We have a conductor connected across a constant voltage source. The current changes with temperature because the resistance changes with temperature. At 0°C, the current is 2 A, and at 100°C, the current is 1.2 A. We want to find the current at 50°C.
Step 2: Express the relationship between current and resistance
Because the voltage across the conductor remains constant, using Ohm’s Law (V = iR), current (i) is inversely proportional to resistance (R):
i \propto \frac{1}{R}
Step 3: Use the temperature dependence of resistance
For a metal conductor, the resistance can be approximated to vary linearly with temperature T as:
R_T = R_0 \bigl(1 + \alpha T \bigr)
where R_0 is the resistance at 0°C, and \alpha is the temperature coefficient of resistance.
Step 4: Determine the temperature coefficient of resistance, \alpha
Let the current at 0°C be i_0 = 2\text{ A} and the corresponding resistance be R_0 . At 100°C, the current is i_{100} = 1.2\text{ A} and the corresponding resistance is:
R_{100} = R_0 \bigl(1 + 100\,\alpha \bigr).
Since the voltage is constant:
i_0 \, R_0 = i_{100} \, R_{100} = i_{100} \bigl[R_0 (1 + 100\alpha)\bigr].
Substituting i_0 = 2\text{ A} and i_{100} = 1.2\text{ A} gives:
2\, R_0 = 1.2\, R_0 \bigl(1 + 100\,\alpha \bigr).
Dividing both sides by 1.2\,R_0 :
1 + 100\,\alpha = \frac{2}{1.2} = \frac{5}{3}.
Hence,
100\,\alpha = \frac{5}{3} - 1 = \frac{2}{3}, \quad \alpha = \frac{1}{150}.
Step 5: Calculate the resistance at 50°C
At 50°C, the resistance is:
R_{50} = R_0 \bigl(1 + 50\,\alpha \bigr).
Substituting \alpha = \frac{1}{150} ,
R_{50} = R_0 \left(1 + 50 \times \frac{1}{150}\right) = R_0 \left(1 + \frac{1}{3}\right) = \frac{4}{3}\,R_0.
Step 6: Determine the current at 50°C
Let the current at 50°C be i_{50} . Since i \, R is constant for a given voltage:
i_{50}\, R_{50} = i_0 \, R_0 \quad \Longrightarrow \quad i_{50} = \frac{i_0\, R_0}{R_{50}}.
Substitute i_0 = 2 \text{ A} and R_{50} = \frac{4}{3} R_0 :
i_{50} = \frac{2\,R_0}{\frac{4}{3}\,R_0} = \frac{2}{\frac{4}{3}} = \frac{2 \times 3}{4} = 1.5\,\text{A}.
Step 7: Convert the result to the desired format
The current 1.5 A is equivalent to 1500 mA. In terms of 10^2\text{ mA} :
1.5\,\text{A} = 1500\,\text{mA} = 15 \times 10^2\,\text{mA}.
Therefore, the final answer is:
\boxed{15 \times 10^2\ \text{mA}}.
