Electrical Engineering - Energy and Power - Discussion
Discussion Forum : Energy and Power - General Questions (Q.No. 9)
9.
A 120 
resistor must carry a maximum current of 25 mA. Its rating should be at least
resistor must carry a maximum current of 25 mA. Its rating should be at leastDiscussion:
59 comments Page 4 of 6.
Biswa Bhusan Pradhan said:
1 decade ago
I think the answer is 75 mW but it is not in d option.
I2R= 0.025*0.025*120 = 75 mW.
I2R= 0.025*0.025*120 = 75 mW.
Pankaj Ahuja said:
1 decade ago
I= 25mA=25/1000 = .025A
R=120 ohm
P= I2R= .025*.025*120 = 3/40 W = 75mW.
R=120 ohm
P= I2R= .025*.025*120 = 3/40 W = 75mW.
Ankur said:
2 years ago
As per my knowledge, 75 MW is the correct answer as it is a DC problem.
(7)
KONDAVENI SANTHOSH said:
7 years ago
p=(i^2)r,
p=(25*25*10^-6)*120,
p=75000*10^-6,
p=75*10^-3,
p=75mw.
p=(25*25*10^-6)*120,
p=75000*10^-6,
p=75*10^-3,
p=75mw.
(2)
Shyam said:
1 decade ago
Current should be in RMS value. So answer will be (75/2).
Sam said:
6 years ago
Yes it should be at least greater than the rated value.
Avd said:
1 decade ago
0.025*0.025*120 = 0.075 W
0.075*1000 = 75 mW Correct
0.075*1000 = 75 mW Correct
Ashhh said:
3 years ago
Why to take Irms instead of taking I max direactly?
(2)
Sathish said:
1 decade ago
They are not that one is RMS value its 75 mW only.
Amit said:
1 decade ago
Answer should be 74mW not 150mW as V=IR and P=VI.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers