A current of 5 A flows through a conductor `25 ^@C`, when connected to a power supply of 100 V . At what temperature the current will drop to 2A , if it remains connected to the power supply (Temperature coefficient of conductor is `0.005 ^@C^(-1)`.)

Answer

To solve the problem, we need to determine the temperature at which the current flowing through a conductor drops from 5 A to 2 A while connected to a constant voltage supply of 100 V. The temperature coefficient of resistance for the conductor is given as \( \alpha = 0.005 \, ^\circ C^{-1} \).

### Step-by-Step Solution:

1. **Determine the Initial Resistance**:
Using Ohm's Law, we can find the initial resistance \( R_1 \) when the current is 5 A.
\[ V = I_1 \cdot R_1 \implies R_1 = \frac{V}{I_1} = \frac{100 \, V}{5 \, A} = 20 \, \Omega \]

2. **Determine the Final Resistance**:
When the current drops to 2 A, we can find the new resistance \( R_2 \) using the same voltage.
\[ V = I_2 \cdot R_2 \implies R_2 = \frac{V}{I_2} = \frac{100 \, V}{2 \, A} = 50 \, \Omega \]

3. **Relate the Resistances Using Temperature**:
The resistance of a conductor changes with temperature according to the formula:
\[ R = R_0 (1 + \alpha (T - T_0)) \]
where \( R_0 \) is the original resistance at temperature \( T_0 \). Here, \( R_0 = R_1 = 20 \, \Omega \) at \( T_0 = 25^\circ C \). We can express \( R_2 \) as:
\[ R_2 = R_1 (1 + \alpha (T - 25)) \]

4. **Substituting Values**:
We know \( R_1 = 20 \, \Omega \) and \( R_2 = 50 \, \Omega \), so we can substitute these values into the equation:
\[ 50 = 20 (1 + 0.005 (T - 25)) \]

5. **Solving for Temperature**:
Dividing both sides by 20:
\[ 2.5 = 1 + 0.005 (T - 25) \]
Subtracting 1 from both sides:
\[ 1.5 = 0.005 (T - 25) \]
Now, divide both sides by 0.005:
\[ T - 25 = \frac{1.5}{0.005} = 300 \]
Finally, adding 25 to both sides gives:
\[ T = 300 + 25 = 325 \, ^\circ C \]

### Conclusion:
The temperature at which the current will drop to 2 A is **325 °C**.