In a RLC series circuit, r = 16, l = 102mh, C = 199uf, current, find: (1) impedance | Z |; (2) U; (3) phasor of u and I

In a RLC series circuit, r = 16, l = 102mh, C = 199uf, current, find: (1) impedance | Z |; (2) U; (3) phasor of u and I

Do not know the frequency, there is no way. 50 Hz power frequency simple series circuit, very good calculation

In RL Series circuit, r = 3 Ω, l = 12.74mh, total voltage U = 220 √ 2sin314tv

The inductive reactance of L is as follows
XL＝2×π×f×L＝2×3.14×50×0.01274≈4(Ω)
Total impedance:
Z = radical (R × R + XL × XL) = radical (3 × 3 + 4 × 4) = 5 (Ω)
1》 Current in the circuit:
I＝U/Z＝220/5＝44(A)
2》 Voltage at both ends of R and l:
Ur＝IR＝44×3＝132(V)
UL＝IXL＝44×4＝176(V)

How to calculate 5.48 - (1.4-0.52)

5.48-(1.4-0.52)
=5.48-1.4+0.52
=5.48+0.52-1.4
=6-1.4
=4.6

As shown in the figure, in △ ABC, ab = AC, CD bisects ∠ ACB intersects AB at point D, AE ‖ DC intersects BC at point e. it is known that ∠ e = 36 ° (1) prove that AC bisects ∠ BAE; (2) directly write all isosceles triangles except △ ABC in the figure

It is proved that: (1) CD bisects ACB, BCD = ∠ ACD, ∫ AE ∥ DC, EAC = ∠ ACD = ∠ BCD = ∠ e = 36 °, ∫ AB = AC, B = ∠ ACB = 2 ∠ ACD = 72 °, BAC = 180 ° - B - ∠ ACB = 36 °, BAC = ∠ EAC, i.e. AC bisects BAE. (2) △ CDB, △ DCA, △ CAE, △ EAB

Given that AB is not equal to 0, and (A2 + B2) ^ 3 = (A3 + B3) ^ 2 + 8A ^ 3B ^ 3, find B / A + A / b
Are there two situations?

a^6+3a^4b^2+3a^2b^4+b^6=a^6+2a^3b^3+b^6+8a^3b^3
3a^4b^2+3a^2b^4=10a^3b^3
ab≠0
Divide both sides by a ^ 3B ^ 3
3a/b+3b/a=10
So a / B + B / a = 10 / 3
Nothing else

Among the rectangles, the length of the rectangle is 15 Li meters, the width is 7 cm, the area of the triangle is 35 square cm smaller than that of the trapezoid, and the bottom of the trapezoid is 15 cm
Come on, don't talk nonsense

7

Let a > 0. A ≠ 1, the function f (x) = a ^ LG (x ^ 2-2x + 3) has the maximum, and find the monotone interval of function f (x) = ㏒ a (3-2x-x ^ 2)

From 3-2x-x ^ 2 = - x ^ 2-2x + 3 > 0, the domain is (- 3,1), not (1,3). Let t = LG (X & # 178; - 2x + 3), then f (x) = a ^ t, from t = LG (X & # 178; - 2X + 3) = LG [(x-1) & # 178; + 2] ≥ LG2, that is, t = LG (X & # 178; - 2x + 3) has the smallest

A snail, climbing a tree of 9m high, rises LM in the daytime and declines 13m at night. It starts to climb up one morning. How many days later can it reach the top of the tree?

(9-1) △ 1-13 + 1, = 8 △ 23 + 1, = 13 (days). A: you can reach the top of the tree after 13 days

Find the trajectory equation of the left vertex of the ellipse with eccentricity 12, passing through the fixed point m (1,2)

Because the ellipse passes through the point m (1,2) and takes the y-axis as the guide line, the ellipse is on the right side of the y-axis, and the major axis is parallel to the x-axis. Let the left vertex of the ellipse be a (x, y). Because the eccentricity of the ellipse is 12, the distance from the left vertex a to the left focus f is 12 times of the distance from a to the y-axis, so the coordinate of the left focus f is (3x2, y). Let d be the distance from the point m to the y-axis, then d = 1. According to | MF | D = 12 and the distance between two points From the formula, we can get (3x2 − 1) 2 + (Y − 2) 2 = (12) 2, that is, 9 (x − 23) 2 + 4 (Y − 2) 2 = 1, which is the trajectory equation

A parallelogram has a base of 1.6 meters and a height of 0.5 meters. The area of a triangle with the same base and height as it is is______ Square meters

1.6 × 0.5 ﹣ 2, = 0.8 ﹣ 2, = 0.4 (square meter), a: the area of the triangle with the same base and height is 0.4 square meter, so the answer is: 0.4