Electrical Engineering - Alternating Current and Voltage - Discussion
Discussion Forum : Alternating Current and Voltage - General Questions (Q.No. 19)
19.
Sine wave A has a positive-going zero crossing at 45°. Sine wave B has a positive-going zero crossing at 60°. Which of the following statements is true?
Discussion:
24 comments Page 1 of 3.
Abdul Majid said:
7 years ago
Hi all.
Option B is correct. Here is the explanation:
That signal leads which crosses a point before the other signal.
δ _t=∠/360*f.
Where Δ_t=tB-tA.
And ∠ is phase difference given by:
∠=∠B-∠A.
And it should be cleared here if Δ t is +ve then signal A is lagging and Vice versa
In our problem.
∠=60-45=15=+ve so Δ t is also +ve and hence wave A is lagging wave B by 15 degrees.
I hope this helped you.
Thanks
Option B is correct. Here is the explanation:
That signal leads which crosses a point before the other signal.
δ _t=∠/360*f.
Where Δ_t=tB-tA.
And ∠ is phase difference given by:
∠=∠B-∠A.
And it should be cleared here if Δ t is +ve then signal A is lagging and Vice versa
In our problem.
∠=60-45=15=+ve so Δ t is also +ve and hence wave A is lagging wave B by 15 degrees.
I hope this helped you.
Thanks
Eoj said:
2 years ago
Get the power factor of both waves.
Wave A = 45 degrees.
Wave B = 60 degrees.
P.F. (Wave A) = cos (45) = 0.707.
P.F (Wave B) = cos (60) = 0.5.
The implication of this is that Wave A is already at the peak, while Wave B is about the get the peak. Thus, Wave A is leading Wave B by 15 degrees.
Wave A = 45 degrees.
Wave B = 60 degrees.
P.F. (Wave A) = cos (45) = 0.707.
P.F (Wave B) = cos (60) = 0.5.
The implication of this is that Wave A is already at the peak, while Wave B is about the get the peak. Thus, Wave A is leading Wave B by 15 degrees.
Trone said:
1 decade ago
Yes I agree. The answer must be option B.
Because they didn't indicate the peak of the pulse. If they have peak we can clearly say if A lags B or B lags A since it just states sine waves so I assume that the wave have the same wave length and peak. So B is leads wave A by 15 degree.
Because they didn't indicate the peak of the pulse. If they have peak we can clearly say if A lags B or B lags A since it just states sine waves so I assume that the wave have the same wave length and peak. So B is leads wave A by 15 degree.
Harish said:
1 decade ago
Option A is the correct option. Because if you plot the graph as angles on x-axis a and respective variable(voltage or current) on y-axis, you will observe that wave A is 15 degrees before that of wave B. Therefore you can say wave A leading wave B by 15 degrees.
Fayaz Ahmed Memon said:
1 decade ago
Option A is a wrong ansswer.
The correct answer is option B because when wave B is at 60 degree then the wave A is at 45 degree so we can say that wave B is leading wave A by 15 degree (60-45) or wave A lags wave B by 15 degree.
The correct answer is option B because when wave B is at 60 degree then the wave A is at 45 degree so we can say that wave B is leading wave A by 15 degree (60-45) or wave A lags wave B by 15 degree.
STUTI KUSHWAHA said:
10 years ago
Answer is A. Because if we think with the help of graph then we find that a start from 45 and after 15 means at 60 b will start then it is clear that a is lead with 15.
Arslan zahoor said:
1 decade ago
45-60 = -15.
-15 value of wave A, means that's leading the other (B) wave form, such as in capacitor case -90 degree means that current lead the voltage to 90 degree.
-15 value of wave A, means that's leading the other (B) wave form, such as in capacitor case -90 degree means that current lead the voltage to 90 degree.
Dhaval said:
8 years ago
Answer should be B because lagging and leading does not depend on Cross zero first. It depends upon the angle of the wave. So right answer should be B.
(1)
Muralidhar jangid said:
1 decade ago
Answer A is correct answer because when we plot graph, than we observe that wave A lead wave B by 15 degree.
Kamran said:
6 years ago
As meter move from left to right. So as respect to this 45 angle will lead to 60 degree angle. A is correct.
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