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
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.
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.
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.
because 3Phase power - V X I X Cos Phi X Square root of 3 and square root of 3 - 1.73
(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
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.
Cos phi is a European term used on a motor to differentiate between apparent power (kVA) of an inductive load as compared to the active power (kW) that is really used by the load. Cos phi= active power/apparent power. In North America a more familiar term would be power factor. Power factor = kW/kVA
The power factor for a three phase generator is 80 percent. The generator consumes 36 kilowatts and a line to line voltage of 400 volts.