A high school physics question about inductance

L1 and L2 are inductors, the moment you close them to the moment they turn on, they are equivalent to two batteries, and the direction of this current is opposite to the direction of the original current, so AB becomes bright first and then dark.

When the electric bond is closed, the three bulbs are connected in parallel, with the same voltage, the same bulb, the same resistance, and the same current passing through.

Parallel shunt, let the resistance of each bulb be R, the parallel resistance of B and C is R2, and then parallel with A, then the current through L1 is 2i, the current of A is in parallel with C, the current through B is 1i, and the current through L2 is 1i, that is, the current of C is 1i.

When the electric key is disconnected, L1 and L2 generate current in the same direction as the original current.

L1 generates a current of 2i, which is shunted by b and c, i.e., 1i passes through b, 1i passes through c, and the current through the trunk a is still 2i.

L2 generates a current of 1i through the trunk circuit C1i, which is divided into A and B, respectively. where the current divided on b is counterclockwise and the current of L1 on B 1i is clockwise. i.e. the current on b is. Whereas, the current of a is 2i.

The above is my personal opinion, and I will discuss it with my classmates.

When the power supply is closed, there is no problem if the current of the three lamps is i, and according to Kirchhoff's law, L1 is 2i and L2 is I.

When the power supply is disconnected, note that the inductance remains in its original state, so that initially L1 is 2i and L2 is I. At this time, according to Kirchhoff's law, a is 2i, and both b and c are i, does the landlord understand here? The nodes here are analyzed at the upper end of A and the upper end of B, and the current flowing into the node is equal to the current flowing out of the node.

Induced current. The essence of the production is:

Magnetic flux. changes, including Lou Oak Bridge said under the bridge.

Cutting is like cutting magnetic inductance lines.

The essence of this is also that the magnetic flux increases due to a change in area s (=bs). The change in magnetic flux results in e = t where is not 0 (i.e. the amount of change in , 0 means that is a fixed value, not changing).

Induce electromotive force.

e, with Kaizhi electromotive force, the induced current i is naturally generated

As shown in the figure below, the third section is the coil in the stage of leaving the magnetic field, at this time the AB edge is cutting the magnetic inductance line, if the right hand is used, the right hand should be placed on the AB edge, the magnetic field is perpendicular to the paper side outward, the palm should be facing down (let the magnetic inductance line penetrate the palm), the thumb points to the direction of speed (speed to the right), at this time the four fingers point to the direction of the current in the AB, the current direction in the AB is downward, and the current direction in the coil is counterclockwise, so it is correct.

According to Lenz's law, the direction of the magnetic field of the induced current in the coil should be the same as the direction of the original magnetic field, the direction of the original magnetic field is perpendicular to the paper side, and the direction of the magnetic field of the induced current should also be perpendicular to the paper side, and then use the spiral rule of the right hand, the thumb is perpendicular to the paper side outward (pointing to the direction of the magnetic field of the induced current), the four fingers refer to the direction of the induced current, and the current direction in the coil is counterclockwise. As shown in the figure below:

This one is very simple.,And isn't there an analysis next to it.。

Choose C, both methods are correct. **If something goes wrong, you can mention it.

When the switch is closed, the self-inductive obstruction effect of the coil can be regarded as resistance, and the coil resistance gradually decreases, and the resistance of the parallel circuit gradually decreases. The voltage gradually decreases; When the switch is closed and then disconnected, the induced current of the coil is in the same direction as the original current, forming a loop, and the current of the bulb is opposite to the original and gradually decreases to 0, so B is selected for this question.

Answer: Because the bulb is connected in series with a capacitor or a coil, they both have impedance and need to divide voltage, so AB is wrong;

Because "the voltage of the power supply is the same as the effective value of the AC voltage", it seems that it is connected to the DC power supply, and the self-inductance coil is connected in parallel with the bulb, so that the lamp is short-circuited and the bulb does not light up. d correct.

On the second floor, the title clearly says that it is connected to the AC power supply!

If this question is not wrong, it is to ensure that the power supply is not written wrong, if it is AC power, the light emitting of the bulbs C and D (if it is the same bulb) is the same. They are all connected to the same power supply, and whether it is a parallel inductor or a parallel capacitor has no effect on the bulb.

The inductor is on direct resistance AC, and the capacitor is on AC resistance, so for this problem, first of all, the inductor must not be connected in parallel with the bulb, c is not right; The capacitor must not be connected in series with the bulb, a is not true; The answer is still D

From the conservation of energy, fx=1 2mv +q, that is, the work done by the external constant force f is converted into heat energy and the kinetic energy of the rod;

From bvl=u, u r=i to obtain v=ri bl, and the above equation and data can be used to obtain the answer. We must know how to flexibly apply the law of conservation of energy.

BLV R=I0, FX-W Magnetism = 1 2*M*VV, Energy Consumed by Resistance = W Magnetism.

Hello, that's how it came.

The rod starts the source. If the velocity is V, then the instantaneous EMF is BVL. For a dry circuit, the resistance is r 2 (understand?) ), then the current in the trunk circuit is BVL (R 2), then the ampere force it receives is BIL, and the substitution is option A.

You know that the left and right resistors are the same, then q heat is generated on the AC resistor, and the resistance at the other end is also q, so the total heat is 2q, looking at the whole process, it is known by the conservation of energy, kinetic energy-heat = elastic potential energy, so it is option C.

I don't know how to ask questions.

Big brother didn't try to guess it?

The fast edge of the A revolution can be seen as an increase in the current! The original magnetic field increases!

So according to Lenz's law, the induced magnetic field will hinder the change of the original magnetic flux!

Magnetic flux = BSSIN

It can be seen that the original magnetic field increases, so B has a tendency to decrease the area S.

The original magnetic field and the induced magnetic field are still mutually repellent!

The thread tension is reduced.

b Although there is no figure, according to the title formulation and Lenz's law, the magnetic flux in the metal ring tends to remain constant, so the area tends to decrease, and the distance between the two tends to increase.