Discussion :: Energy and Power - General Questions (Q.No.1)
|Sam said: (Dec 19, 2010)|
|Both the resistors have equal ratings.|
|Sunil Hojage said: (Mar 5, 2011)|
|w=i^2r..........but w=0.5watt is given
so only current is change but out put is remain same
coz r=33 & r=330 & w=0.5
0.5=i*330;i=0.0015(not taking squaroot of i for simplicity)
|Upasana Sapra said: (May 5, 2011)|
|Both resistance are connected in series. Therefore R(eq.)=363.
Therefore I=V/R =12/363 = 0.033
Now calcuate power across each resistor. p=I^2*R.
Both power does not exceed by 0.5W. Therefore,neither res. get heated as it is half-Watt resistance.
|Rachana said: (Jun 7, 2011)|
|Beacause they are connected in parallel.|
|Sana Sahithi said: (Aug 3, 2011)|
|In both the resistors power dissipated is same.
Calculate currents and power in both resistors. Numerically they are equal.
|Ankit Bhardwaj said: (Sep 14, 2011)|
|No need to calculate anythin as it's
given in the que. that both the resistors are of same
ratings i.e 1/2 watts....
thats why heat dissipation shall be same...!
|Raj said: (Sep 21, 2011)|
|Circuit in series so power or heat dissipation will be same.|
|Pandiaraj.S said: (Oct 6, 2011)|
|I am sorry to say that Answer Option D is wrong and correct Answer is Option A.
Both R are connected in parallel and 0.5W capacity.
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
|Anai Relish said: (Nov 6, 2011)|
|Yeah its connected in parallel so we have different currents
P(1)=I^2*r1=4.3636>0.5W (so it gets heated up......)
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
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.
|Gurbachan Dhiman said: (Nov 24, 2011)|
|Option D is correct because in // any type of load take current according its capacity ( Rating )|
|Jogi said: (Dec 8, 2011)|
|Actually depends on understanding that how both the resisters are connected. It is not clear how they are connected ie series or parell. It depends upon indually and its answer is different in both condition.|
|Himanshu said: (Dec 20, 2011)|
|Option d is correct as both are taking same power and are connected in parallel (as given in question they are conected across 12 we supply. Across means parallel).|
|Arti said: (Jan 7, 2012)|
|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.
|Biswajit said: (Jan 31, 2012)|
|What is half-watt resistor ?|
|Latha said: (Feb 22, 2012)|
|@Arti then how do you consider the wattage of resistance?|
|Shekhar said: (Mar 2, 2012)|
|When the resistance of the conductor is more then current drop will be more then heat will generat more as compare to low resistance of the conductor.|
|Suresh Maniar said: (Nov 11, 2012)|
|Both Resistances are Connected across same voltage of 12V.
Wattage=V square/R since V=Const or same for both smaller are value shall draw more power and heat up more.
|Dayo said: (Feb 26, 2013)|
|Both resistors are connected in series, therefore same voltage flow across each of them & there s also balance in heat dissipated by each.|
|Akhilesh Pandit said: (Mar 1, 2013)|
|Power consumed by each of them is same so both of them get equally heated.|
|Sagar Singh said: (Mar 23, 2013)|
|Since they are connected in parallel and 33 ohm is very less than 330 ohm more current will flow through 33 ohm, and therefore more heat will dissipate across 33 ohm and it gets over heated.|
|Syeda Jafri said: (Aug 15, 2013)|
|Heating depends upon the amount of current flowing through each resistance, not on power. If considered in parallel arrangement, 33 ohm resistance has higher current flow so it should get overheated & option A should be correct.|
|Majid Khan said: (Dec 5, 2013)|
|Heated beyond a safe or desirable point is over heating so just pay attention on words of question when both are connected to same source and no external power conditions are changing then why are you so confuse Option D is ok.|
|Ankush Chauhan said: (Feb 14, 2014)|
|Less resistance more current and more current more heating.|
|Ofri said: (May 13, 2014)|
|Since both resistors are connected across the voltage is in parallel so the option D is correct.|
|Girma said: (May 20, 2014)|
|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.
|Neelam said: (Jun 2, 2014)|
|Current is same through resisters connected in series.|
|Manikonda Malakondaiah said: (Jul 3, 2014)|
|As per the given question both are equal wattage so given two resistors never get over heated. So the answer is option D.|
|Kirthika.S said: (Jul 6, 2015)|
|Answer is wrong. Because we doing in project side in small value of resistors are easily access the heat.|
|Munawar said: (Jul 9, 2015)|
|It is [A] 33 W will overheat, because its wattage VI = 12*12/33 = 4.36 W against the allowed 0.5 W. For 330 W, wattage = 12*12/330 = 0.436 W within the 0.5 W.|
|Akhil said: (Aug 26, 2015)|
|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.
|Oorja said: (Oct 27, 2015)|
|I agree that the connection is not clear. If connected in series then option D otherwise option A.|
|Pratap Kumar said: (Dec 20, 2015)|
|Heat is I^2R. This loss is more in 33 ohm resistor. Option A is correct.|
|Arun said: (Jan 19, 2016)|
|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.|
|Marshall said: (Feb 13, 2016)|
|Both load are resistive components. It dissipate power. For sure both of them will produce heat. So both of them.|
|Stuti Kushwaha said: (Feb 16, 2016)|
|Option A should be correct according to i^2r we see that more current flow through 33 ohm. So more heat should across it.|
|S K Pillai said: (Mar 2, 2016)|
|Question should be more clear. Whether the resistances are in series or in parallel? The answer differs D or A.|
|Sonu Kumar said: (Apr 30, 2016)|
|Because both the resistor has equal rating and heating of resistor depends on wattage only. Not on the type of connection.|
|Chavitina Ramakrishna said: (Aug 2, 2016)|
|Equal rating not given, why you people calling equal rating?|
|Vikas said: (Sep 12, 2016)|
|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.
|Nouman Ali said: (Nov 8, 2016)|
|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.
|Presley C said: (Jul 27, 2017)|
|33 Ω resistor will overheat as the connection is parallel it's not specific. So, the Answer is A.|
|Subhrajit Jena said: (Feb 7, 2018)|
|According to power loss formula, 330 ohm resistor will produce overheating because both are connected in series.|
|Vinny said: (May 5, 2018)|
|In question, if both resistors are connected in series across 12v supply, then answer is D, either the A correct answer.|
|Akshay Jadhav said: (Jun 28, 2018)|
|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.|
|J D D said: (Jul 18, 2018)|
|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).|
|Swetha said: (Aug 7, 2018)|
|The Key point is;
Less resistance value resistor will dissipate more heat.
The connection is a parallel circuit.
Formulae is Power = V*V/R.
|Abdul Ghafoor said: (Aug 27, 2018)|
|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.|
|Sher Shah said: (Aug 28, 2018)|
|Both the resistors have same power rating so no one will be heat up.|
|D Shwetha said: (Sep 6, 2018)|
|Here, it is given that, the resistors are 0.5W & also a parallel connection.
Heat is directly proportional to power (H=P*t). So, power rating 0.5W & Heat is also both resistors are same.
|Junaid Ahmed Hassan said: (Oct 31, 2018)|
|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.
|Vishal said: (Mar 20, 2019)|
|Since power is given heat = I^2 x R x t, since I^2 R is the same for both resistors, neither will overheat.|
|Sunnieth said: (May 31, 2019)|
|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?
|Haris said: (Jan 7, 2020)|
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.
|Saikiran said: (Feb 7, 2020)|
|What is a half watt resistor?|
|Kooba said: (Jun 23, 2020)|
|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".
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