An Electrophysics Question 200

Amplitude = 2 RMS (here is the maximum value im = 5 2); The relationship between the period t, frequency f and angular frequency is f=1 t, =2 t=2 f. The initial phase is between -.

In this problem, the expression of the instantaneous value of the current is obtained by the expression i=imsin( t+ ).

i=5√2sin(2πft+π/3)=5√2sin(314t+π/3)。

Waveform diagram: is a function curve of y=5 2sin(314t+3), when t=0, y=5 2sin(3)=5 6 2, the curve intersects on the positive half axis of y. When y=0, sin(314t+3)=0, that is, 314t=- 3, the curve intersects on the negative half axis of t (if 314t is used as the x-axis, the curve is a sinusoidal waveform sin(314t) shifts to the left 3), i=5 2sin(314t+3) lags behind i=5 2sin(314t)3.

1. The time it takes for the alternating current to change for a week is called the cycle.

2. Sinusoidal alternating current has to go through a 360-degree angle to complete a one-week change, and the specified angle is called an electric angle.

3. The three elements of a sinusoidal quantity are the maximum, angular frequency, and initial phase.

4. The electrical angle at which the alternating current changes in one second is called the angular frequency.

5. Sinusoidal alternating current can be represented by the stationary vector of the initial position, which is called the phasor diagram of the sinusoidal quantity, and the length of which is equal to the effective value is called the phasor diagram of the significant value.

6. The instantaneous power of the capacitive element in the sinusoidal AC circuit is positive and negative, which indicates that it is constantly exchanging power with the outside, and the maximum value of its power is used to represent the scale of the exchange power, which is called reactive power.

7. When the capacitor is connected in parallel with the inductive load, its capacitive current can compensate for the reactive power component of the load current, so that the power factor of the line is increased and the total current is reduced.

8. Sinusoidal AC electromotive force is generated by the alternator, and there is a coil that can rotate between the n and s poles. When the coil rotates counterclockwise at a constant angular velocity, the two conductor sides of the coil cut the magnetic field lines to generate an induced electromotive force.

Apply the kinetic energy theorem.

Let the distance between point A and the bottom end of the inclined plane be l, the inclination angle of the inclined plane is , and the friction factors of the inclined plane and the horizontal plane are 1 and 2 respectively.

No electric field: mglsin - 1mglcos - 2mgs1=0-0 to get: 2s1=lsin - 1lcos has an electric field:

mg+qe)lsinθ-μ1(mg+qe)lcosθ-μ2(mg+qe)s2=0-0

Get: 2s2=lsin - 1lcos Comparison of the two formulas: get s1=s2

The first time: mgh = umghcosa sina + umgs1, the work done by gravity is all used to overcome the friction of the inclined plane and the plane.

The second time: (qe+mg)h=u(mg+qe)hcosa sina +u(mg+qe)s2, the work done by gravity and electric field force is all used to overcome the friction force of inclined planes and planes.

Note: h is the height of the inclined plane, and a is the angle between the inclined plane and the horizontal plane. Pick B

If I want to choose B, the electric field force does the positive work, but at the same time the resistance also increases. The best way is to add the point field, which is directly equivalent to the increase of the gravity of the ball, so that after using the kinetic energy theorem, the mass of the ball will be reduced, regardless of the mass. My phone is not convenient for formulas.

Answer: choose B, apply the conservation of mechanical energy, let the distance between point A and the bottom end of the inclined plane be L, the inclination angle of the inclined plane is , and the friction factors of the inclined plane and the horizontal plane are 1 and 2 respectively. For the first time, the potential energy is equal to the work done by the friction force on the inclined plane plus the work done by the friction force on the horizontal plane, so there is the first time:

mglsin = 1mgcos *l + 2mgs1(1), the second electric potential energy plus the mechanical potential energy is equal to the work done by the friction force on the inclined plane plus the work done by the friction force on the horizontal plane, so there is a second time: (qe+mg)lsin = 1(mg+qe)cos *l+ 2(mg+qe)s2 simplification: mglsin = 1mglcos + 2mgs1(2) ; 1) (2), so choose b

Select C, under the action of uniform and strong electric field, the sliding friction becomes larger.

When the switch is disconnected, the capacitor is not connected to the circuit, so the potential is 0

After closing, it is connected in parallel with R3.

In this case, the voltage is the voltage of R3.

q=cu3 ①

Regardless of the resistance of the capacitor in series with it, the distribution of the voltage satisfies.

u1=u4+u3 ②

r3+r4)r1/(r1+r3+r4)+r]i=e ③u4/u3=r4/r3 ④

The charge flowing through is.

q=q is solved by the five formulas on the synthesis.

q=36 29*10 -4c (I can't see which r4 is.) The answer is that r4 is the topper-most resistor).

q=2*10 -4c (this is the answer when r4 is connected in parallel with the capacitor).

Set the voltage to up=u r

u²／r1=6①

u r1+u r2=9 so u r2=3 gets; r1:r2=1:2 2r1=r2

After series p=u (r1+r2)=u 3r1=6 3=2, so the total power consumed by the two resistors after series connection is 2 watts.

The total power is 2W

Analysis: Oops, it's too hard to write a formula on this, so I won't write it. I hope you forgive me It's a column equation, and when you connect to r1, an equation contains unknown quantities r1 and u squared.

Another equation when parallel r2 contains the unknown quantities r1, r2, u squared.

Calculate R1 and R2 with an unknown amount of U squared

In the end, the total power is an equation that can be summed up to cancel out the u-squared. All right!

Possibility 1: Before the measurement, there is no [zeroing] of the two ammeters, and at least one fast ammeter pointer is not at the zero scale mark (the pointer of A1 is more right-hand than the pointer of A2);

Possibility 2: The actual range used by the actual access circuit of the two ammeters is different (A1 uses the range, A2 uses the 0---3A range), and the current value is read according to the same range when reading.

If all of them are connected in series, and there is no parallel connection at all, then there must be an error in the table, and it is gone.

First of all, look at the connection mode of the circuit, because the capacitor is disconnected, that is, there is no current in the branch, so only R1 and R2 are connected in series on the external circuit, and there is no current on R3, so the voltage at both ends is zero. i.e. R3 is equivalent to a wire.

The current in the circuit i=e (r1+r2+r)=2ac1 is equal to the voltage at both ends of r2.

u2=ir2=2a*2ω=4v

The charged amount of C1 is Q1=C1U2=

The C2 terminal voltage is equal to the circuit terminal voltage.

u=i（r1+r2）=10v

The charge of C2 is Q2=C2*U=

Solution: From the meaning of the title, it can be seen that the voltage at both ends of C1 is equal to the voltage at both ends of R2, and the voltage at both ends of C2 is equal to R1 and R2.

Voltage at the end. The current in the circuit is: i 12 (3+2+1)=2(a)r2 The voltage at both ends is: 2 2=4(v).

Then the charge of C1 is: 4 4 = 16 (coulombs) R1, the voltage at both ends of R2 is: 5 2 = 10 (V) then the charge of C2 is: 10 1 = 10 (coulombs).

R3 has no current passing through, and the effect is that there is no voltage drop in the wire. Make an equivalent circuit diagram as shown in the figure (the left end of C1 is connected to the negative end of the power supply, the right end is connected to the left end of R1, the left end of C2 is connected to the negative end of the power supply, and the right end is connected to the positive end of the power supply

i=e/(r+r1+r2)=2a,∴u2=ir2=4v;u1=ir1=6v;U-end = u1+u2=10v

q1=c1*u-end=4 10 -6 10=4 10 -5cq2=c2 u2=1 10 -6 4=4 10 -6c( b sweat! It's not good to realize that the network speed is slow and can't be transmitted)

The answer is that the fault in BA is that the voltage cannot say the voltage that passes through, only the voltage at both ends of the conductor.

c has a free charge, and without voltage, it cannot move directionally and cannot form an electric current.

d, there is voltage in the circuit and the circuit is closed before there is current.

Choose, the voltage can not be said to pass through the conductor; c. It should also be able to move in a directional manner; d, it also needs to form a pathway.

The conductor of the selected BA should be a closed loop.

The mistake is that the voltage is not said.

d does not say closed loop.

It is enough to grasp the necessary conditions for the formation of current: voltage, conductor, closed loop, all are indispensable, voltage provides electromotive force, conductor provides free electrons, and closed loop is the path of continuous current.