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 1 of 6.

Ankur said:   5 months ago
As per my knowledge, 75 MW is the correct answer as it is a DC problem.
(1)

Ashhh said:   2 years ago
Why to take Irms instead of taking I max direactly?
(1)

Sam said:   4 years ago
Yes it should be at least greater than the rated value.

Yogesh meena said:   5 years ago
The given current value is in rms.

Then the average value will be √(2)*i.
So, P=[√(2)*i]^2*r,
=[1.414*25]^2*120.
=150mW.
(2)

Bila said:   5 years ago
@All.

The correct answer is the one where we have to measure Irms (= Ief = effective current) first. Hence, 150 mW.

Other who points out that the answer is 150 mW but NOT calculating Irms first - e.g. just jumped into the P=I^2*R - their answer just so happens to be the right answer (150 mW). You should calculate Irms first.
(1)

Vishvajeet yadav said:   5 years ago
P = 75mW.
(3)

Santhu said:   6 years ago
The resistor is power dissipating element.

Then current take in RMS value,
The problem is given in maximum value
so we convert the current maximum value into RMS value,
current rms=max current/sqr(2).
p=(I^2)R.
after calculation
p=2*75mw=150mw.

The power rating of resistor depends on RMS value.
(1)

KONDAVENI SANTHOSH said:   6 years ago
p=(i^2)r,
p=(25*25*10^-6)*120,
p=75000*10^-6,
p=75*10^-3,
p=75mw.
(1)

Bibek said:   6 years ago
I think 75 mW is the correct answer.

Rojohn Ogalinola said:   6 years ago
Use a resistor rated for at least 2 times the required power dissipation.
Thus,
P=(25mA)^2 x 2(120ohm) = 150mW.


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