Which of the resistors will overheat?

Problem : A 33 Ω half-watt resistor and a 330 Ω half-watt resistor are connected across a 12 V source. Which one(s) will overheat?

Choices: A.) 33 Ω B.) 330 Ω C.) Neither resistors D.) Both will overheat.

My solution: Since the phrase is "the resistors are connected across the voltage source, I assumed that the resistors are in parallel. So, same voltage. V = 12V

Power dissipated in 33 Ω = V^2 / R1 = 12^2 / 33 = 4.36 W
Power dissipated in 330 Ω = V^2 / R2 = 12^2 / 330 = 0.436 W

The rating of the 33 Ω resistor is half-watt and the power dissipated in it is 4.36 which is more than the rating. Hence, 33 Ω resistor will overheat.

BUT the answer on the book is neither resistor will overheat. There is no solution. Now I am confused which is the correct answer as others say the answer should be (A) which is 33 Ω, saying that the resistors is connected in series and not int parallel.

Or maybe, it's because of the half-watt rating in both of the resistors. It is designed to keep the heat at half-watt, avoiding both to overheat... not sure.

What do you think guys?

Your power dissipation calculation is correct for parallel connection. The only possible reason for a different answer is that the resistors are not parallel but series connected.

Hello Jayce,
Your calculations are correct.
Here are mine:

Find the total resistance if the circuit:

Rt = R1 x R2/R1 + R2

Rt = 330 x 33/330 + 33
Rt = 10,890/363
Rt = 30Ω

Next, determine the total circuit current:

Itotal = V/R

Itotal = 12V/30Ω
Itotal = 0.4A
Itotal = 400mA

Now determine the current through each resistor

Ibranch = It x /(Rtotal ÷ Rbranch)

Ibranch1 = 0.4 x /(30 ÷ 330)
= 0.4 x 0.091
= 0.036A
= 36mA

Ibranch2 = 0.4 x /(30 ÷ 33)
= 0.4 x 0.91
= 0.364
= 364mA

Itotal = Ibranch1 + Ibranch2 = 0.4
Itotal = 0.036 + 0.364
Itotal = 0.4
Itotal = 400mA

Determine the power dissipated by each resistor:

Pr1 = R1 x I²
Pr1 = 330 x 0.0013
Pr1 = 0.429
Pr1 = 429mW

Pr2 = R2 x I²
Pr2 = 33 x 0.13245
Pr2 = 4.27
Pr2 = 4.27W

Even before going through most of the calculations above, we jest needed to
determine the total power dissipated in the circuit and that would have already
given us excessive power dissipation.

i.e Ptotal = V²/R
Ptotal = 12²/30
Ptotal = 144/30
Ptotal = 4.8W

I have to agree with FvM that the resistors must be in series and not parallel.
I have to admit that the question asked was a bit obscure, as it could have been
either configuration.
Can you hassle a teacher over this matter. To me its just too deceptive.
Regards,
Relayer

Assume series connection of resistors
Effective resistance : 33Ω + 330Ω = 360Ω
Maximum current : 12V/360Ω = 33.3mA
Power dissipation on 33Ω resistor : (33.3mA)^2 x 33Ω = 1.1W
Power dissipation on 330Ω resistor : (33.3mA)^2 x 330Ω = 11W
My answer is bot resistor will overheat because both resistors are half watt rated
Suppose you ask “Which resistor dissipate more heat” the answer is 330Ω because it will dissipate 11W

Power dissipation on 33Ω resistor : (33.3mA)^2 x 33Ω = 1.1W
Power dissipation on 330Ω resistor : (33.3mA)^2 x 330Ω = 11W

Click to expand...

No. My math says 37 mW and 367 mW. Apparently the square operation got lost.

Agree with FvM.

0.033 ^ 2 = 0.001 * 33 = 0.033 or 33mW.

Bob