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?
half-watt resistor and a 330 half-watt resistor are connected across a 12 V source. Which one(s) will overheat?33
330
both resistors
neither resistor
Discussion:
58 comments Page 4 of 6.
Shekhar said:
1 decade ago
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.
Sunil hojage said:
1 decade ago
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*33;i=0.015
0.5=i*330;i=0.0015(not taking squaroot of i for simplicity)
so only current is change but out put is remain same
coz r=33 & r=330 & w=0.5
0.5=i*33;i=0.015
0.5=i*330;i=0.0015(not taking squaroot of i for simplicity)
Upasana sapra said:
1 decade ago
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.
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:
1 decade ago
Beacause they are connected in parallel.
Sana sahithi said:
1 decade ago
In both the resistors power dissipated is same.
Calculate currents and power in both resistors. Numerically they are equal.
Calculate currents and power in both resistors. Numerically they are equal.
Ankit Bhardwaj said:
1 decade ago
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...!
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:
1 decade ago
Circuit in series so power or heat dissipation will be same.
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
Gurbachan Dhiman said:
1 decade ago
Option D is correct because in // any type of load take current according its capacity ( Rating )
Jogi said:
1 decade ago
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.
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