Because it has to be 1 when the phase angle is zero.
mw is the unit of real power and mvar is unit of reactive power. You should now the current and power factor angle to calculate the voltage.p=vi cos piq=vi sin piAnswerI think you mean MW, not mw -capital 'M' mega; lower-case 'm' milli!!! And the symbol for watt is a capital 'W', not a lower case 'w'. Also, I think you mean 'Mvar'(mega, not milli!).
Q = 3 Vph Iph sin(phase angle) = 31/2 Vline Iline sin(phase angle)
The defination of form factor is FF=RMS/avg(abs(f(t))) for sin waveform RMS=0.707*peak(f) avg(abs(f(t)))=2/pi*peak(f) so FF=0.707/(2/pi)=1.1106
This is a simple derivation problem. Take the derivative of the AC voltage waveform (I'll assume sin (t)), and set the derivative equal to the maximum value it can attain (which is 1): cos (t) = 1, t = 0 degrees, thus for an AC waveform of y = sin(t), the fastest increase in voltage occurs at 0, 360, etc. degrees. The fastest decrease in voltage will occur at 180, 180 + 360, etc. degrees.
kvar = kva*sin@
cos(phi - 1) = cos(phi)cos(1) + sin(phi)sin(1)
tan (phi)= (V* sin (theta) + Ia*Xs)/(V*cos (theta) +Ia*ra) theta is power factor angle torque angle= phi-theta
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
The expression for the unit vector r hat in spherical coordinates is r hat sin(theta)cos(phi) i sin(theta)sin(phi) j cos(theta) k.
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
sec x - cos x = (sin x)(tan x) 1/cos x - cos x = Cofunction Identity, sec x = 1/cos x. (1-cos^2 x)/cos x = Subtract the fractions. (sin^2 x)/cos x = Pythagorean Identity, 1-cos^2 x = sin^2 x. sin x (sin x)/(cos x) = Factor out sin x. (sin x)(tan x) = (sin x)(tan x) Cofunction Identity, (sin x)/(cos x) = tan x.
2
The equation cannot be proved because of the scattered parts.
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec