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:
61 comments Page 1 of 7.
Ankur said:
3 years ago
As per my knowledge, 75 MW is the correct answer as it is a DC problem.
(7)
Yogesh meena said:
7 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.
Then the average value will be √(2)*i.
So, P=[√(2)*i]^2*r,
=[1.414*25]^2*120.
=150mW.
(3)
Vishvajeet yadav said:
8 years ago
P = 75mW.
(3)
Santhu said:
8 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.
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.
(3)
Mustansar Zubair said:
11 months ago
P =VI
V = IR.
So,
P =I^2R.
V = IR.
So,
P =I^2R.
(2)
Ashhh said:
4 years ago
Why to take Irms instead of taking I max direactly?
(2)
KONDAVENI SANTHOSH said:
8 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)
Geethu said:
9 months ago
Good questions, Thanks all.
(1)
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.
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:
8 years ago
Use a resistor rated for at least 2 times the required power dissipation.
Thus,
P=(25mA)^2 x 2(120ohm) = 150mW.
Thus,
P=(25mA)^2 x 2(120ohm) = 150mW.
(1)
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