RCL series circuit, impedance angle 60 °, resistance R = 10, then | Z | = (). Power factor = () How to calculate

RCL series circuit, impedance angle 60 °, resistance R = 10, then | Z | = (). Power factor = () How to calculate

Since the impedance angle is 60 degrees, if the impedance is Z, then the impedance vector is written as: Z ∠ 60 °. Obviously, this is an inductive impedance, which is written in complex form: z = R + J ω L and tg60 ° = ω L / R. therefore, ω L = R * tg60 ° = 10 * √ 3Z = √ [R ^ 2 + (ω L) ^ 2] ^ 2 = √ (100 + 300) = 20 Ω in the current, if the current = I, the total voltage

The simplest way to improve the power factor is parallel compensation capacitor. Its principle is to reduce the angle between u and I of RCL parallel circuit
The simplest way to improve power factor is parallel compensation capacitor. Its principle is to reduce the angle between u and I in RCL parallel circuit, that is, the power factor angle. The problem is that in RCL series circuit, series capacitor can also reduce the power factor angle. Why can't series compensation capacitor be used to improve power factor, while series capacitor compensation is only used for voltage regulation?

1. In terms of power supply quality: theoretically, for alternating current, capacitance is a conductor, but in fact, impedance is inevitable. Therefore, in the circuit diagram, it is expressed as capacitor series resistance. Due to the existence of resistance, voltage drop is inevitable, which will affect power supply quality. 2. In terms of power supply reliability, capacitance is relatively capacitive

When the RCL series circuit is under damped, what is the effect of reducing the loop resistance R on its transition process?

First of all, we should know whether RCLS are in series or in parallel; Q in series is WL / R, and R / WL in parallel; for series, increasing R will make them enter critical damping and under damping state; for parallel, only Q will be larger and more over damping

Given two points m (- 2,0), n (2,0), point P satisfies PM · PN = 12, then the trajectory equation of point P is______ ．

Let P (x, y), then PM = (− 2 − x, − y), PN = (2 − x, − y), ∵ PM · PN = 12, ∵ (- 2-x, - y) · (2-x, - y) = 12. After sorting, we get x2 + y2 = 16. So the answer is: x2 + y2 = 16

If the circumference of an isosceles triangle is 24, then the range of waist length x is and the range of bottom edge y is
If the circumference of an isosceles triangle is 24, then the range of waist length x is__ The range of the bottom edge y is__

2X + y = 24; 2x > y (the sum of two sides is greater than the third side); so y = 24-2x
The two formulas are: 2x > 24-2x; x > 6; y = x > 6; 8=

Point P is an arbitrary point on the chord ab of circle O, connecting PO.PC ⊥ OP, PC intersection and C. verification: PA * Pb = PC

So PC = PN connects an CB and proves that triangles are similar. CPB = ∠ APN vertex angle is equal. BCP = ∠ pan is equal to the same chord BN angle in the circle

As shown in the figure, in △ ABC, AB + AC = 15, point P is the intersection of bisectors of ∠ ABC ∠ ACB. Through P, make Mn ‖ BC intersection, AB AC, and calculate the perimeter of △ amn at point Mn

∵ BP bisection ∠ ABC
∴∠ABP=∠PBC
∵MN∥BC
∴∠MPB=∠PBC
∴∠ABP=∠MPB
∴BM=MP
Similarly, PN = NC
Δ amn perimeter = am + Mn + an = am + MP + PN + an
=AM+BM+CN+AN=AB+AC=15

As shown in the figure, in ladder ABCD, ad ∥ BC and BD are diagonals, and the median EF intersects BD at point O. if fo-eo = 3, BC-AD equals ()
A. 4B. 6C. 8D. 10

∵ EF is trapezoid, ABCD is median line, ∥ EF ∥ BC ∥ ad. ∥ ob = OD. ∥ BC = 2of, ad = 2oe. ∥ BC-AD = 2 (fo-eo) = 2 × 3 = 6

It is known that two of the quadratic equations x ^ 2 + PX + 2 = 0 are x1, X2, and x2-x1 = 2, then P=_____
As soon as possible!

x1+x2=-p
x1*x2=2
(x1-x2)^2=(x1+x2)^2-4*x1*x2=p^2-8=4
P = 2 radical (3) or - 2 radical (3)

Through the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, the line between a point P on the left quasilinear and the left focus f intersects the ellipse at two points a and B. if the vector PA = NaF, Pb = MBF, find n + M

The perpendiculars passing through two points a and B intersect the alignment lines at two points c and D. the intersection of the alignment line and the X axis is e,
AF / AC = e, BF / BD = e, vector PA = NaF, Pb = MBF,
Because triangle PAC is similar to triangle PBD, PA / AC = Pb / BD
M+N=PA/AF+PB/BF=PA/(e*AC)-PB/(e*BD)=1/e(PA/AC-PB/BD)=0