Why is the phase difference between current and voltage less than 90 degrees in the circuit with resistor and capacitor in parallel

Why is the phase difference between current and voltage less than 90 degrees in the circuit with resistor and capacitor in parallel

What is the phase relationship between current and voltage in pure resistance AC/

Pure resistance circuit is a circuit without inductive and capacitive load. In such a circuit, the voltage and current are in the same phase, and the phase between them is zero. If it is inductive circuit, the voltage is 90 ° ahead of the current. If it is capacitive circuit, the current is 90 ° ahead of the voltage

1/2+3/4+7/8+15/16+31/32+63/64+127/128+255/265
1 / 2 + 3 / 4 + 7 / 8 + 15 / 16 + 31 / 32 + 63 / 64 + 127 / 128 + 255 / 265 rule

1-1/2+1-1/4+1-1/8+1-1/16+1-1/32+1-1/64+1-1/128+1-1/256=8-(1-1/256

Known x1x2 Xn = 1, and x1, X2 Xn are all positive numbers, proving: (2 + x1) (2 + x2)... (2 + xn) > = 3 ^ n
Such as the title

Certification:
Let xi = 2yi, where I = 1,2 ,n
So x1x2 Xn=（2y1)(2y2)… (2yn)=2^n(y1y2… yn)=1
So y1y2 yn=1/2^n=（1/2）^n=0.5^n
And because
(2+X1)(2+X2)...(2+Xn)
=(2+2y1)(2+2y2)...(2+2yn)
=2^n(1+y1)(1+y2)...(1+yn)
So we have to prove that (2 + x1) (2 + x2)... (2 + xn) > = 3 ^ n
Just prove that 2 ^ n (1 + Y1) (1 + Y2)... (1 + yn) > = 3 ^ n
That is, (1 + Y1) (1 + Y2)... (1 + yn) ≥ 3 ^ n / 2 ^ n = (3 / 2) ^ n
From the binomial theorem, we can get
（3/2）^n=（1+0.5)^n
=1+n×0.5
+0.5²n(n-1)/2
+…
+C(n,t)0.5^n
+…
+0.5^n ①
In the above formula, C (n, t) is the combination number of t elements from n elements
Also known
(1+y1)(1+y2)...(1+yn)
=1+（y1+y2+...yn)
+(y1y2+y1y3+...+y(n-1)yn)
+…
+(y1y2… yt+…）
+…
+y1y2… yn ②
Where (y1y2 yt+…） It means to take t from the N numbers Y1 to YN, multiply them, and then add all such products
Let's prove that every term in formula 2 is greater than or equal to the corresponding term in Formula 1
The first, all 1, is tenable
The second term is defined by the mean inequality,
（y1+y2+...yn）/n≥(y1y2… yn)^(1/n)= (0.5^n)^(1/n)=0.5
That is, Y1 + Y2 +... Yn ≥ n × 0.5
It is established
...
The term T + 1 (i.e. the general term) is also defined by the mean inequality,
(y1y2… yt+…） /C(n,t)≥[（y1y2… yt）×… ]^（1/C(n,t)）
=[（y1y2… yn）^C(n-1,t-1)]^（1/C(n,t)）
=（y1y2… yn）^（t/n)
=（0.5^n）^（t/n)
=0.5^t ③
That is (y1y2 yt+…） ≥C(n,t)0.5^n
The conclusion is valid
③ Where C (n-1, t-1) means that all C (n, t) such y1y2 After multiplication of YT type formulas, from Y1 to YN, they will be multiplied by C (n-1, t-1) times
...
The last item (item n + 1) is y1y2 The conclusion is valid
The proof is over
Note 1: in fact, we only need to prove the general term. Write the first, second and N + 1 terms to understand more intuitively
Note 2: it is easy to see that a more general conclusion can be proved by using the above proof idea
Known x1x2 Xn = 1, and x1, X2 Xn is a positive number. For any positive number m, there is a
(m+X1)(m+X2)...(m+Xn)≥(m+1)^n
Of course, it can also be extended to a more general conclusion
Known x1x2 Xn = P ^ n, and x1, X2 Xn is a positive number. For any positive number m, there is a
(m+X1)(m+X2)...(m+Xn)≥(m+p)^n
The equal sign holds when X1 = x2 =... = xn

Area is 120 square meters, with the side length of 0.8 meters square tiles, how many tiles?

The floor area divided by the area of each tile, 600 * 600 is 0.36 square, 800 * 800 is 0.64 square, the number of kickboard is perimeter divided by the length of the brick (600 brick with 600 long board, 800 brick with 800 long board) plus 4 or 5, the wall brick is perimeter multiplied by (floor height - ceiling height) generally 2.5 meters, and then subtract the area of the door, the window does not need to reduce, because there is loss, 300 * 450 is 0.135 square meters
Floor tile auxiliary material consumption area * paving height * 0.25/0.04 = cement (bag)
Area * paving height * 0.75 = amount of sand (M3)
Wall brick auxiliary material consumption area * paving height * 0.33 / 0.04 = cement (bag)
Area * paving height * 0.66 = amount of sand (M3)
The floor tiles should be laid dry, the wall tiles should be laid wet, and the bathroom doorstone should be laid wet (for waterproofing)
The cement pasted on the floor tiles is better, and the cement pasted on the wall tiles is inferior (because the good cement solidifies too fast and easily cracks the wall tiles)

To find the minimum value of the function y = x ^ 2-2x + 6 / (x + 1) (x is greater than 0), do we use the mean inequality?
Bracketed

y=x^2+2x+1-4x-4+9
=[(x+1)^2-4(x+1)+9]/(x+1)
=x+1-4+9/(x+1)
>=2√9-4=2
Take the equal sign if and only if x + 1 = 9 / (x + 1), x = 2
:.y>=2

What is the original function of arcsinx?

1 / (radical 1-x)

Use 100 small squares with side length of 2 cm to make a big square. How many square centimeters is the area of the new big square
You must be right

The side length of a large square is 10 * 10 * 4 = 400 square centimeters, that is, 10 rows. Each row has 10 small squares with a side length of 2 centimeters. The area of a small square is 4, that is, 4 * 100 = 400 square centimeters

When designing a rectangular clock face, an extra-curricular learning group wants to make the width of the rectangle 20 cm, the center of the clock is at the intersection of the diagonal lines of the rectangle, the number 2 is at the top of the rectangle, and the numbers 3, 6, 9 and 12 are marked at the midpoint of the side, as shown in the figure. (1) how long should the rectangle be? (2) Please point out the position of the number 1 on the rectangular box, and explain the method to determine the position; (3) please point out the position of the remaining numbers on the clock face on the rectangular box, and write the corresponding numbers (Note: draw necessary auxiliary lines to reflect the solution idea)

(1) According to the meaning of the title, we know that ∠ AOC = 2 ∠ BOC, ∵ AOC + ∠ BOC = 90 °∠ BOC = 30 °, AOC = 60 °, ∵ tanb = bcob = 33, that is ob = 3bC, ∵ rectangle ABCD is three times the width, and ∵ rectangle is 203 cm long

Let x = 1 + T ^ 2, y = cost find which pair of dy / DX and d ^ 2Y / DX ^ 2 sin tcost / 4T ^ 3 and sin tcost / 4T ^ 2?
Let x = 1 + T ^ 2, y = cost find dy / DX and d ^ 2Y / DX ^ 2
Which is right between sint tcost / 4T ^ 3 and sint tcost / 4T ^ 2?

∵x=1+t²,y=cost
==>dx/dt=2t,dy/dt=-sint
∴d²y/dx²=d(dy/dx)/dx
=(d((dy/dt)/(dx/dt))/dt)/(dx/dt)
=(d((-sint)/(2t))/dt)/(2t)
=((sint-tcost)/(2t²))/(2t)
=(sint-tcost)/(4t³)
So (sint - tcost) / (4T & # 179;) is right