Question

The current through a conductor is $ a $ when the temperature is 0°C. It is $ b $ when the temperature is 100°C. The current through the conductor at 220°C is:

• A. \( \frac{5ab}{11b - 6a} \)
• B. \( \frac{5ab}{6a - 11b} \)
• C. \( \frac{5ab}{11a - 6b} \)
• D. \( \frac{11ab}{5a - 6b} \)

Hint:

When the current through a conductor changes with temperature, you can use the relation between temperature and resistance to calculate the change in current.

Answer 1

The Correct Option is C

The resistance of a conductor changes with temperature, and the current is inversely proportional to the resistance (Ohm's law: \( I = \frac{V}{R} \)). Let the resistance of the conductor at \( 0^\circ C \) be \( R_0 \). 
The resistance at a temperature \( T \) is given by the equation: \[ R_T = R_0 (1 + \alpha T), \] where \( \alpha \) is the temperature coefficient of resistance. At \( 0^\circ C \), the current is \( a \), so: \[ I_0 = a \quad \Rightarrow \quad R_0 = \frac{V}{a}. \] At \( 100^\circ C \), the current is \( b \), so: \[ I_{100} = b \quad \Rightarrow \quad R_{100} = \frac{V}{b}. \] Now, the resistance at \( 100^\circ C \) is: \[ R_{100} = R_0 (1 + \alpha \cdot 100) = \frac{V}{b}. \] Substituting \( R_0 = \frac{V}{a} \) into the equation: \[ \frac{V}{a} (1 + 100 \alpha) = \frac{V}{b}. \] Simplifying: \[ \frac{1}{a} (1 + 100 \alpha) = \frac{1}{b}. \] \[ \frac{1 + 100 \alpha}{a} = \frac{1}{b}. \] From this equation, we can solve for \( \alpha \): \[ \alpha = \frac{b - a}{100ab}. \] Now, to find the current at \( 220^\circ C \), we use: \[ R_{220} = R_0 (1 + \alpha \cdot 220) = \frac{V}{I_{220}}. \] Substituting \( R_0 = \frac{V}{a} \) and \( \alpha = \frac{b - a}{100ab} \): \[ \frac{V}{a} \left( 1 + 220 \cdot \frac{b - a}{100ab} \right) = \frac{V}{I_{220}}. \] Simplifying: \[ \frac{1}{a} \left( 1 + \frac{220(b - a)}{100ab} \right) = \frac{1}{I_{220}}. \] \[ I_{220} = \frac{5ab}{11a - 6b}. \] Thus, the current through the conductor at 220°C is \( \frac{5ab}{11a - 6b} \), which corresponds to option (C).