Question

If the radius of a current carrying wire is less than the critical radius, then the addition of electrical insulation will enable the wire to carry a higher current because:

• A. The thermal resistance of the insulation is reduced.
• B. The thermal resistance of the insulation is increased
• C. The heat loss from the wire would increase
• D. The heat loss from the wire would decrease

Hint:

Remember the behavior around the critical radius of insulation: - If \( r_o<r_c \), adding insulation increases heat transfer. - If \( r_o>r_c \), adding insulation decreases heat transfer. - At \( r_o = r_c \), heat transfer is maximum for a given insulation thickness.

Answer 1

The Correct Option is C

Step 1: Understand the concept of critical radius of insulation.
The critical radius of insulation (\( r_c \)) for a cylindrical object like a wire is the radius at which the total thermal resistance to heat transfer from the wire to the surroundings is minimum. For insulation with thermal conductivity \( k \) surrounding a wire with outer radius \( r_o \), the critical radius is given by \( r_c = \frac{k}{h} \), where \( h \) is the convective heat transfer coefficient from the outer surface of the insulation to the ambient air. Step 2: Analyze the case when the wire radius is less than the critical radius (\( r_o<r_c \)).
In this scenario, adding insulation (increasing the outer radius of the insulated wire) will decrease the total thermal resistance. This is because the increase in conductive resistance due to the added insulation is less than the decrease in convective resistance due to the increased outer surface area available for heat transfer. Step 3: Relate thermal resistance to heat loss.
Heat loss \( Q \) is related to the overall temperature difference \( \Delta T \) and the total thermal resistance \( R_{total} \) by the equation \( Q = \frac{\Delta T}{R_{total}} \). If the thermal resistance decreases, for a constant temperature difference between the wire and the ambient air, the heat loss from the wire will increase. Step 4: Understand why increased heat loss allows for higher current carrying capacity.
A current-carrying wire generates heat due to its electrical resistance (\( I^2R_{electrical} \)). The maximum current a wire can safely carry is limited by the maximum permissible operating temperature of the wire's insulation. If the heat generated exceeds the rate at which heat can be dissipated to the surroundings, the temperature of the wire and its insulation will rise. By adding insulation in the case where \( r_o<r_c \), we increase the heat loss from the wire for a given temperature difference. This means that for the same maximum permissible temperature, the wire can dissipate a larger amount of heat, which in turn implies that it can carry a higher current (\( I^2R_{electrical} \) can be larger while maintaining the temperature limit). Step 5: Evaluate the given options.
Option 1 is incorrect because adding insulation when \( r_o<r_c \) reduces the thermal resistance.
Option 2 is incorrect because adding insulation when \( r_o<r_c \) reduces the thermal resistance.
Option 3 is correct because adding insulation when \( r_o<r_c \) increases the heat loss from the wire.
Option 4 is incorrect because adding insulation when \( r_o<r_c \) increases the heat loss from the wire.
Step 6: Select the correct answer.
If the radius of a current carrying wire is less than the critical radius, then the addition of electrical insulation will enable the wire to carry a higher current because the heat loss from the wire would increase.