Electrical Engineering - Energy and Power - Discussion

I agree that the connection is not clear. If connected in series then option D otherwise option A.

It should be "B" because in both resisters power is same i.e, 0.5 w.

So 330 R will have high power loss i^2*r than the 33 R.

Heat is I^2R. This loss is more in 33 ohm resistor. Option A is correct.

33 ohm will draw more current compared to 330 ohm, since they are connected in parallel 12 remains fixed across both but current gets splitted according to the resistance opposed, obviously 330 will oppose more current hence will dissipate less while 33 will draw in more current thus dissipating heat more, if any mistake please clarify me.

Both load are resistive components. It dissipate power. For sure both of them will produce heat. So both of them.

Option A should be correct according to i^2r we see that more current flow through 33 ohm. So more heat should across it.

Question should be more clear. Whether the resistances are in series or in parallel? The answer differs D or A.

Because both the resistor has equal rating and heating of resistor depends on wattage only. Not on the type of connection.

Equal rating not given, why you people calling equal rating?

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.