Electrical Engineering - Energy and Power - Discussion
Discussion Forum : Energy and Power - General Questions (Q.No. 18)
18.
A 3.3 k
resistor dissipates 0.25 W. The current is
resistor dissipates 0.25 W. The current isAnswer: Option
Explanation:
W = I2R
I = square root(W/R)
= square root( 0.25/(3.3x1000) )
= 0.0087038828
= 8.7 mA
Discussion:
5 comments Page 1 of 1.
Rajkumar said:
8 years ago
Answer:
V=IR, therefore I=V/R
W=V*I ----> V=W/I ----->(1)
sub (1)
I = (W/I)/R ----> I^2=W/R,
Finally, I = SquareRoot (W/R).
Given W=0.25, R = 3.3 Kilo Ohm.
Convert Kilo Ohm into Ohm, (1000 Ohm = 1 kilo Ohms).
Now R = 3300 Ohms, W = 0.25 Watt, Current (I) = ?.
I = √ (W/R)
= √(0.25/3300)
= √(0.0000757).
= 0.0087 A.
= 8.7 mA (1000 milli Amp = 1 Amp).
V=IR, therefore I=V/R
W=V*I ----> V=W/I ----->(1)
sub (1)
I = (W/I)/R ----> I^2=W/R,
Finally, I = SquareRoot (W/R).
Given W=0.25, R = 3.3 Kilo Ohm.
Convert Kilo Ohm into Ohm, (1000 Ohm = 1 kilo Ohms).
Now R = 3300 Ohms, W = 0.25 Watt, Current (I) = ?.
I = √ (W/R)
= √(0.25/3300)
= √(0.0000757).
= 0.0087 A.
= 8.7 mA (1000 milli Amp = 1 Amp).
(2)
Suresh said:
8 years ago
How to calculate this square root? Please explain in detail.
(1)
Sagar said:
8 years ago
The correct calculation is 0.25/3.3 =0.757.
Hassan Ahmed Eid said:
8 years ago
The right answer is 8.7A not 8.7mA because after calculation we didn't do any unit conversion the for there will no be any coefficient before the ampere.
Zabi Ullah said:
5 years ago
P=vi.
v=ir.
And p=i^2R.
v=ir.
And p=i^2R.
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