Electronics - Alternating Current and Voltage - Discussion
Discussion Forum : Alternating Current and Voltage - General Questions (Q.No. 7)
7.
Calculate the positive-going slope of the waveform in the given circuit.
Discussion:
10 comments Page 1 of 1.
Vasu said:
1 decade ago
As wahab said its the slope formula.
slope =Y2-Y1/(X2-X1);
Here y2=5V and y1=0V and x2=4ms , and x1=2ms as the wave is from centre[consider only +ve going waveform that goes from (2,0)to (4,5)]
slope =Y2-Y1/(X2-X1);
Here y2=5V and y1=0V and x2=4ms , and x1=2ms as the wave is from centre[consider only +ve going waveform that goes from (2,0)to (4,5)]
(2)
Jagan said:
1 decade ago
How please explain it clearly.
S.wahab said:
1 decade ago
5-0/2(v/ms)=2.5v/ms
Jai parkash said:
1 decade ago
I am not clear from above. Plz more explain sir plz.
Bana said:
1 decade ago
y=mx, so, m=(y/x),
m=((y2-y1)/(x2-x1))
m=((y2-y1)/(x2-x1))
Amal bal komath said:
1 decade ago
Slope = (Y2-Y1)/(X2-X1)
=(5-(-5))/(4-0)
=10/4
=2.5 is the answer.
=(5-(-5))/(4-0)
=10/4
=2.5 is the answer.
Goutham .B said:
10 years ago
Slope = tan (a).
tan = 5/2 = 2.5.
tan = 5/2 = 2.5.
Gaurav said:
8 years ago
10/2 = 5 D is the right answer
-5 and 2..........and 4, 5.
-5 and 2..........and 4, 5.
Matvel said:
8 years ago
Why do you use the slope formula here?
Kasi said:
2 years ago
Very good explanation. Thanks @Vasu.
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