Electrical Engineering - Energy and Power - Discussion

Both the resistors have same power rating so no one will be heat up.

The heat is cause by current flowing through it so these two resistors are connected in series and current is same across both so neither resistor is overheated because value of current is very less.

The Key point is;

Less resistance value resistor will dissipate more heat.
The connection is a parallel circuit.
Formulae is Power = V*V/R.

The resistor will overheat because the capacity of both resistors are 5watt And due to 12v supply no one will take higher current than it's rated capacity. (respectively i1=0. 3636, i2=. 03636).

Power of both are same so only current will be change. The output of the resistor in the form of heat that heat output is remains same for both for same wattage.

In question, if both resistors are connected in series across 12v supply, then answer is D, either the A correct answer.

According to power loss formula, 330 ohm resistor will produce overheating because both are connected in series.

33 Ω resistor will overheat as the connection is parallel it's not specific. So, the Answer is A.

Option A is the right answer. Power dissipation depends on the voltage and resistances of both the resistors.
Power dissipation in 33 Ohm resistor is 4.363 W as 12^2/33 = 4.363 W and in 330 Ohm resistor, it is 12^2/330 = 0.436. So resistor with 33 Ohm resistance will over heat.

The value is the maximum wattage the part can dissipate without damage. The value specified is usually at room temp (25C), and the rating will need to be de-rated for higher temperatures.

Some types of resistors may need heatsinking to meet the stated power specs. The large 5, 10, 25, and 50-watt resistors with aluminum cases usually need a fairly large heat sink to achieve their ratings.

So, if P = I^2 * R or P = V^2/R for the resistor in your application is less than the rated power (taking into consideration any derating at higher temps) then your part will be fine.

So, answer A is right one as it was a wrong design by using half a watt 33-ohm resistor this circuit is not right.