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

Discussion:
59 comments Page 1 of 6.
Biswa said:
1 decade ago
i^2*r
=.025*.025*120=75mw
=.025*.025*120=75mw
Sumana dey said:
1 decade ago
i^2*r=.0258.025*120=75mw
Saleem said:
1 decade ago
i^2*r = 75mW....... 4.8W resistor carries max current of 200mA {i = root[4.8/120] = 200mA}. 150mW resistor carries max current of 35mA... 15mW resistor carries max current of 11.2mA... 480mW resistor carries max current of 63.2mA.....
Among those options 150mW resistor has nearer value 35mA max current. so, B is ryt ans.
Among those options 150mW resistor has nearer value 35mA max current. so, B is ryt ans.
Richa said:
1 decade ago
I^2*R=25*25*120*(10^-3)^2= 750mW
while this option is not available in the option.. and these are given.. please give me ful explanantion of this question
while this option is not available in the option.. and these are given.. please give me ful explanantion of this question
RAJESH BANOTH said:
1 decade ago
p=i2r
=2*(25*10-3)2*120
=150mw
=2*(25*10-3)2*120
=150mw
Sravanthi said:
1 decade ago
P=I^2*R is correct when current passing through it is given but they have given maximum current we known P=VI we known I 25ma. So the answer should be in multiples of 25.
So from the given options 150 is multiple of 25 so option b is correct I feel so the rating should contain maximum current and max voltage.
So from the given options 150 is multiple of 25 so option b is correct I feel so the rating should contain maximum current and max voltage.
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.
Akhila R said:
1 decade ago
Pankaj Ahuja is right
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
Jaswanth said:
1 decade ago
Sravanthi is correct.
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