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 1 of 6.
Sam said:
1 decade ago
Both the resistors have equal ratings.
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
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)
Gurbachan Dhiman said:
1 decade ago
Option D is correct because in // any type of load take current according its capacity ( Rating )
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