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Does the 220 V voltage of the utility power mean that the effective value is 220 V or the peak value is 220 V?
The 220 V voltage of the utility refers to that the RMS is 220 V, and its single peak value is twice the root of the RMS. That is 220 * 1.414, which is about 310 v. the peak to peak value is the difference between the positive peak value and the negative peak value = 310 V - (- 310 V) = 620 v
The relationship between current and voltage in series and parallel circuits?
In series: I = I1 = I2 =. = in
U=U1+U2+.+Un
Parallel: I = I1 + I2 +. + in
U=U1=U2=.=Un
How to calculate the current and voltage in series and parallel circuits
In parallel connection, the total voltage is equal, the total current is equal to the current of each parallel branch, and the total current is equal to the current of each parallel branch
Series connection means that the total current is equal and the total voltage is equal to the sum of the voltages on each component
To put it simply, it is shunt in parallel and partial pressure in series
Simplified fraction: (2A2 + 3A + 2) / A + 1 - (a2-a-5) / A + 2 - (3a2-4a-5) / A-2 + (2a2-8a + 5) / A-3
-12- 3/a
Put the nine numbers - 2, - 1,0,1,2,3,4,5,6 into the nine palace lattice respectively, so that each row, each column and the diagonal
1 6 -1
0 2 4
5 -2 3
Using substitution elimination method to solve equations 3 (Y-2) = x-17,2 (x-1) = 5y-8
3(y-2）=x-17,
3y-6=x-17
So x = 3Y + 11
Substituting 2 (x-1) = 5y-8
2(3y+11-1)=5y-8
6y+20=5y-8
y=-28
x=3y+11=-73
How to write question 2.3.4 on page 6 of supplementary exercises in Volume I of sixth grade mathematics of Jiangsu Education Press
2. 120x-85x=52.5
3. 324x-319x=120
4. 1.8x-1.2x=54
If the value of algebraic formula 2 times the square of X + 3 times x + 7 is 12, find the value of algebraic formula 4 times the square of X + 6 times X-10
2x^2+3x+7=12
2x^2+3x=5
4x^2+6x-10
=2(2x^2+3x)-10
=2*5-10
=0
How to calculate 1.3 divided by 0.7 + 0.8 divided by 0.7
=(1.3+0.8)x10÷7
=3
Given that the perimeter of triangle is 8, how to find the area
If the point E (x, 0) moves on the line AB and the point F moves on the broken line ACB, when the line EF bisects the perimeter of the triangle ABC, let the area of the triangle AEF be y, and find the relationship between Y and X. (where AB is on the X axis, the perimeter of the triangle ABC is 16, the triangle AFE is 8, a (- 3,0), C (4,0), B (3,0))
The number of C is (0,4) wrong. It should be AE + AF = 8. I have the wrong number
Is the perimeter of the triangle AFE 8? If EF bisects the perimeter of the triangle ABC, shouldn't AE + AF = 8?
RELATED INFORMATIONS
- 1. It is known that the resistance of the electric iron r = 1930 Ω, which is connected to the sinusoidal AC circuit of u = 220 V to calculate the current I flowing and the power P consumed
- 2. When analyzing the amplifying circuit, draw the AC path diagram. Why can the DC power supply be regarded as a short circuit?
- 3. The principle of drawing small signal equivalent circuit in analog circuit
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- 6. How to calculate the polar radius What does r = 2asin θ mean?
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- 8. When polar coordinates are transformed into rectangular coordinates, such as: x ^ 2 + y ^ 2 = x r^2=r*cosQ r=cosQ Why can we make an appointment with R? R > = 0 on both sides? If it is equal to 0, then we can make an appointment? r^2-r*cosQ=0 r(r-cosQ)=0 R = 0 or r-cosq = 0 Because r-cosq = 0 contains r = 0, it can be directly written as r-cosq = 0 Is that right? Is it wrong? R ^ 2 = R, can r be reduced in this case? R ^ 2 = R -- > R = 1 or r = 0; if R is reduced to r = 1, won't r = 0 be missing?
- 9. Do polar coordinates have to be converted to rectangular coordinates For example, there is a problem, r = a (1-cos θ), to find the tangent equation at (R, θ) = a certain point I know it's a polar equation, x = RCOs θ y = rsin θ Why can't r = a (1-cos θ) be directly regarded as y = a (1-cos x) and find the tangent equation at a certain point (y, x)? Why is this wrong?
- 10. Converting polar coordinates into rectangular coordinates ρ（2COSθ +5sinθ )-4=0 ρ=2cosθ -4sinθ
- 11. How to change 220 V AC into 380 V AC
- 12. The voltage of sinusoidal current for lighting is 220 V, what are its effective value and peak value?
- 13. It is known that the effective value of sinusoidal current I = 1, f = 50 Hz, and the initial position π / 3. The instantaneous value expression and waveform are obtained
- 14. The effective value of a sinusoidal voltage is 380V, the frequency is 50 Hz, the starting value of timing is equal to 380V, and the expression of its instantaneous value is
- 15. 3.3.7.7 how to calculate equal to 24
- 16. The quadratic function y = ax ^ 2 + BX + C (a) is known
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- 19. In the triangle ABC, the length of edges a and B are two of the equation x2-5x + 2 = 0, and the angle c = 120, then the edge C =?
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