# Electrical Engineering - Energy and Power - Discussion

Discussion Forum : Energy and Power - General Questions (Q.No. 1)

1.

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

Discussion:

56 comments Page 1 of 6.
Vikas said:
8 years ago

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.

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.

Girma said:
1 decade ago

The question has two key words: half-watt, and (connected) across. The resistors are designated to work up to half-watt power, more than it make them overheat. If the word across is used, it means the resistors are connected in parallel. Therefore,

R1 = 33 ohm, and its power dissipation is P1 = V*I1 = 12*12/33 = 4.3W.

R2 = 330 ohm, and its power dissipation is P2 = V*I2 = 12*12/330 = 0.43.

P1 = 4.3W>0.5W, and P2 = 0.43W<0.5W,

Hence, R1 = 33ohm is overheat, option A is the correct answer.

R1 = 33 ohm, and its power dissipation is P1 = V*I1 = 12*12/33 = 4.3W.

R2 = 330 ohm, and its power dissipation is P2 = V*I2 = 12*12/330 = 0.43.

P1 = 4.3W>0.5W, and P2 = 0.43W<0.5W,

Hence, R1 = 33ohm is overheat, option A is the correct answer.

Anai relish said:
1 decade ago

Yeah its connected in parallel so we have different currents

so I(1)=v/r1=12/33=0.3636

P(1)=I^2*r1=4.3636>0.5W (so it gets heated up......)

I(2)=v/r2=12/330=o.o3636

P(1)=I^2*r2=0.43636<0.5W (so it is not heated.....)

Therefor the answer is op:A

If it is connected in series then we have

R=r1+r2=363

I=v/R=12/363=0.033

P1=I^2*r1=0.0359<0.5W (not heated....)

P2=I^2*r2=0.0359<0.5W (not heated....)

So answer is Option D is correct answer.

so I(1)=v/r1=12/33=0.3636

P(1)=I^2*r1=4.3636>0.5W (so it gets heated up......)

I(2)=v/r2=12/330=o.o3636

P(1)=I^2*r2=0.43636<0.5W (so it is not heated.....)

Therefor the answer is op:A

If it is connected in series then we have

R=r1+r2=363

I=v/R=12/363=0.033

P1=I^2*r1=0.0359<0.5W (not heated....)

P2=I^2*r2=0.0359<0.5W (not heated....)

So answer is Option D is correct answer.

(1)

Junaid Ahmed Hassan said:
6 years ago

By Ohm's law, V=I * R.

Power dissipated across the resistor, P=I^2 x R.

Let R1=33 ohm, R2=330 ohm.

So, let's calculate the current through each resistor. We know V = 12V.

Current through the resistor R1, I(R1)=V/R1= 12/33= 0.36 A.

Power dissipated by R1, P(R1)= I(R1)^2 x R1 = 0.36^2 x 33 = 4.27 Watts.

Current through resistor R2, I(R2)=V/R2=12/330=0.036 A.

Power dissipated by R2, P(R2)= I(R1)^2 x R2= 0.036^2 x 330 = .42 Watts.

Power dissipated across the resistor, P=I^2 x R.

Let R1=33 ohm, R2=330 ohm.

So, let's calculate the current through each resistor. We know V = 12V.

Current through the resistor R1, I(R1)=V/R1= 12/33= 0.36 A.

Power dissipated by R1, P(R1)= I(R1)^2 x R1 = 0.36^2 x 33 = 4.27 Watts.

Current through resistor R2, I(R2)=V/R2=12/330=0.036 A.

Power dissipated by R2, P(R2)= I(R1)^2 x R2= 0.036^2 x 330 = .42 Watts.

(3)

HAris said:
4 years ago

All.

Here the value of resistances is given just to confuse us, cause.

"The rate at which heat is dissipated is called power" by formula : Power = heat/time.

So power is directly proportional to heat so if power rating is the same of each resistor so heat dissipation will also be the same. So neither of the resistor will be heated up at the power rating of 0.5 watts.

Here the value of resistances is given just to confuse us, cause.

"The rate at which heat is dissipated is called power" by formula : Power = heat/time.

So power is directly proportional to heat so if power rating is the same of each resistor so heat dissipation will also be the same. So neither of the resistor will be heated up at the power rating of 0.5 watts.

(5)

Arti said:
1 decade ago

Heat generated = (square of current)* Resistance * time

This means higher the current, more is the heat generated.

For 33ohms resistance across 12 V source, current is 0.36A.Hence heat generated per second will equal to 4.28 joules.

For 330 ohms resistance across 12V source, current is 0.036A. So heat generated per second= 0.436 joules.

Hence Option A is correct.

This means higher the current, more is the heat generated.

For 33ohms resistance across 12 V source, current is 0.36A.Hence heat generated per second will equal to 4.28 joules.

For 330 ohms resistance across 12V source, current is 0.036A. So heat generated per second= 0.436 joules.

Hence Option A is correct.

Sunnieth said:
5 years ago

1) What is a half watt resisting?

2) What does it mean to connect resistors across a potential difference

3) What is the influence of series or parallel connection of the resistors themselves in this question?

4) If a half watt is actually 0.5watt fix rating, does the half watt rating cancels out the effect of the nature of resistance arrangement?

2) What does it mean to connect resistors across a potential difference

3) What is the influence of series or parallel connection of the resistors themselves in this question?

4) If a half watt is actually 0.5watt fix rating, does the half watt rating cancels out the effect of the nature of resistance arrangement?

(2)

Kooba said:
4 years ago

The word across means parrallel,

ie. The two resistance are parallel to the voltage (12v) source,

Then, both resistance have 12 volts resistance across then,

Then the power source across 33ohm is using v^2/R = 12^2/33 = 4.36 watts (overheat) ,

With 330 ohm, power = 12*12/330 =0.436 watts (won't over heart).

Therefore, the answer is "A".

ie. The two resistance are parallel to the voltage (12v) source,

Then, both resistance have 12 volts resistance across then,

Then the power source across 33ohm is using v^2/R = 12^2/33 = 4.36 watts (overheat) ,

With 330 ohm, power = 12*12/330 =0.436 watts (won't over heart).

Therefore, the answer is "A".

(8)

Arun said:
8 years ago

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.

Pandiaraj.S said:
1 decade ago

I am sorry to say that Answer Option D is wrong and correct Answer is Option A.

Given:

Both R are connected in parallel and 0.5W capacity.

Solution:

I1 = V / R1 = 12 / 33 = 0.364A

W1 = I1^2*R1 = 0.364^2*33 = 4.37W - Overheat

I2 = V / R2 = 12 / 330 = 0.0364A

W2 = I2^2*R2 = 0.0364^2*330 = 0.0437W - Good design

Answer:

Option A

Given:

Both R are connected in parallel and 0.5W capacity.

Solution:

I1 = V / R1 = 12 / 33 = 0.364A

W1 = I1^2*R1 = 0.364^2*33 = 4.37W - Overheat

I2 = V / R2 = 12 / 330 = 0.0364A

W2 = I2^2*R2 = 0.0364^2*330 = 0.0437W - Good design

Answer:

Option A

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