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
4.8 W
150 mW
15 mW
480 mW
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
59 comments Page 3 of 6.

THE MAN said:   1 decade ago
Rating of any device is such that it should carry current up to twice of the capacity. May be this reason so answer is doubled.

Rohit kumar said:   1 decade ago
P = i^2*R.

So that,

P = 25^2*120 = 25*25*120*(10^-3) = 75mW.

But is not exist so that what right answer of that question.

Sweta said:   1 decade ago
P = I^2*R.
P = (Imax)^2 * R.
P = (Irms*sqrt(2))^2*R.
P = (0.025*sqrt(2))^2* 120.
P = (0.035355)^2 *120.

P = 150mW.
(1)

Rojohn Ogalinola said:   7 years ago
Use a resistor rated for at least 2 times the required power dissipation.
Thus,
P=(25mA)^2 x 2(120ohm) = 150mW.
(1)

Eric said:   1 decade ago
p= i2r or p=vi so v=ir =25/1000 * 120=3v

p=vi =3*0.025 = 0.075= 75mv

OR

P=I2R =0.025Square *120 = 75mv

James said:   1 decade ago
75mw is correct but rating is alwaya more, so 150mw which is nerest to 75mw according to option is correct.

Rabi Sankar Samanta said:   1 decade ago
p = I2R (25mA=0.025A).

P = 0.025*0.025*120.

P = 0.075W.

P = 75 mW.

Answer did not maching.

R n trimukhe said:   1 decade ago
Simply go through ohms law,

V = IR.
V = 25*10-3*120.
V = 3 v.

P=VI.
P=3*25*10-3.
P=75 mWatt.

Ravi ranjan said:   8 years ago
P=i2r bit here I is the RMS and irms =I'm/√2 now apply 25ma square * 120 * 2 = 150mw.

Prakash said:   1 decade ago
p= 25*1.414*25*1.414*120*10*-3= 150mw

Vrms = Vm/1.414=25
its given in rms so convert vmax.


Post your comments here:

Your comments will be displayed after verification.