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 2 of 6.
Nouman Ali said:
9 years ago
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.
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.
SHANKS said:
7 months ago
it's letter D:
33 ohms + 330 ohms = 363 ohms
I = V/R ;
= 12 V / 363 ohms
I = ( 4/ 121) A.
@ 33 ohms.
FORMULA FOR POWER: I^2/R.
=( (4/121)^2/33).
P = 0.0036.
@ 330 ohms;
FORMULA FOR POWER: I^2/R.
=((4/121)^2/330).
P = 0.36.
So, Neither resistor is right.
33 ohms + 330 ohms = 363 ohms
I = V/R ;
= 12 V / 363 ohms
I = ( 4/ 121) A.
@ 33 ohms.
FORMULA FOR POWER: I^2/R.
=( (4/121)^2/33).
P = 0.0036.
@ 330 ohms;
FORMULA FOR POWER: I^2/R.
=((4/121)^2/330).
P = 0.36.
So, Neither resistor is right.
(2)
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.
Saumya Ranjan senapati said:
3 years ago
There are 32wattt and 330 watts are connected means in series so the current is constant heat depends on current resistance and time here r&t also constant and we know the previous eqñ current is also constant so D is the correct answer.
(7)
Oji khalnayak said:
3 months ago
A option is correct because the question says across 12v supply, so parallel me voltage is same,
P1=v^2 ÷r
= 144 ÷ 33 = 4.36.
P2 = 144÷330 = 0.436.
So, above see 33ohm resister through power value is big compared to 330ohm resister.
P1=v^2 ÷r
= 144 ÷ 33 = 4.36.
P2 = 144÷330 = 0.436.
So, above see 33ohm resister through power value is big compared to 330ohm resister.
(1)
Syeda Jafri said:
1 decade ago
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:
1 decade ago
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.
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.
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)
Abdul Ghafoor said:
7 years ago
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.
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