A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.4 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this resistor dissipates during a 24-hour period, (b) the heat flux, and (c) the fraction of heat dissipated from the top and bottom surfaces.

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated (), measured in joules, by the cylindrical resistor is the power multiplied by operation time (), measured in hours. That is:

(1)

If we know that and , then the amount of heat dissipated by the resistor is:

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux (), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder (), measured in square meters:

(2)

(3)

Where:

- Diameter, measured in meters.

- Length, measured in meters.

If we know that , and , the heat flux of the resistor is:

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces (), no unit, is the ratio of the top and bottom surfaces to total surface:

(3)

If we know that and , then the fraction is:

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

it becomes positively charged by charging it