cos(phi - 1) = cos(phi)cos(1) + sin(phi)sin(1)
sec(x)=1/cos(x) - (hint: look at the third letter: sec->(1/)cos, cosec->(1/)sin, cot->(1/)tan)
Yes.If the angle between them is 90 degrees. As we know that A.B=|A| |B| cos (phi). When phi=90 degree,cos 90=0. Hence A.B= |A| |B| *0 =0.
(a+b)/a=a/b=phi (the golden ratio, as defined) (a+b)/a=phi (we'll solve this equation) 1+b/a=phi (just changing the form of the left side a little) 1+1/phi=phi (a/b=phi so b/a=1/phi) phi+1=phi2 (multiply both sides by phi) phi2-phi-1=0 (rearrange) From here, we can use the quadratic equation to find the positive solution: phi=(-b+√(b2-4ac))/(2a) phi=(1+√(1+4))/2 phi=(1+√5)/2≈1.618
Manipulate normally, noting:cot x = cos x / sin xcos² x + sin² x = 1 → sin²x = 1 - cos² xa² - b² = (a + b)(a - b)1 = 1²ab = baa/(bc) = a/b/c(1 + cot x)² - 2 cot x = 1² + 2 cot x + cot² x - 2 cot x= 1 + cot² x= 1 + (cos x / sin x)²= 1 + cos² x / sin² x= 1 + cos² x / (1 - cos² x)= ((1 - cos² x) + cos² x)/(1 - cos² x)= 1/(1² - cos² x)= 1/((1 + cos x)(1 - cos x))= 1/(1 - cos x)/(1 + cos x)QED.
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
Because it has to be 1 when the phase angle is zero.
tan (phi)= (V* sin (theta) + Ia*Xs)/(V*cos (theta) +Ia*ra) theta is power factor angle torque angle= phi-theta
The phase angle phi in the cosine function cos(wtphi) affects the horizontal shift of the graph of the function. A positive phi value shifts the graph to the left, while a negative phi value shifts it to the right.
sec(x)=1/cos(x) - (hint: look at the third letter: sec->(1/)cos, cosec->(1/)sin, cot->(1/)tan)
Equation for power in an AC circuit is Power (P) = root3 Voltage (V) x Current (I) x power factor (cos phi) P = root3 V I Cos Phi. Value of cos phi is maximum 1. Normally 0.95 and above is considered to be good. root 3 and V assuming to be constant, if the cos phi is low, to be able to supply rated power, the system draws increased current (I). With increased current IR and copper losses increase resulting over heating of the motor.
Yes.If the angle between them is 90 degrees. As we know that A.B=|A| |B| cos (phi). When phi=90 degree,cos 90=0. Hence A.B= |A| |B| *0 =0.
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.
3 phase kVA = V*I*sqrt(3) Where voltage is line to line, and current is the actual RMS current flowing in the a wire. kW = V*I*sqrt(3)*Cos (phi), where phi is the angle between the voltage and current; Cos (phi) is also known as the power factor. kVA is the vector sum of kW (real power) and kVAR (reactive power). As the equations above suggest, you must know the voltage to correctly calculate the current.
(a+b)/a=a/b=phi (the golden ratio, as defined) (a+b)/a=phi (we'll solve this equation) 1+b/a=phi (just changing the form of the left side a little) 1+1/phi=phi (a/b=phi so b/a=1/phi) phi+1=phi2 (multiply both sides by phi) phi2-phi-1=0 (rearrange) From here, we can use the quadratic equation to find the positive solution: phi=(-b+√(b2-4ac))/(2a) phi=(1+√(1+4))/2 phi=(1+√5)/2≈1.618
because 3Phase power - V X I X Cos Phi X Square root of 3 and square root of 3 - 1.73
To convert kVA to Amps, you need the voltage at which the calculation is being done. Without the voltage, the conversion cannot be accurately calculated.
The formula to calculate split times is given by percent of time= (2-4*phi)* (percent of distance) ^2 + ((4*phi)-1) *percent of distance). Positive splits and negative splits are the two types of splits.