A physics and electricity problem, find answers and explanations, focus on explanation!

Let the voltage at both ends of b be u, that is, u2=u, and let both resistors be r, so u1=2u, i1=(u-2u), i2=u, because i1:i2=1:2=(u-2u): u

Get u=2 5u for d

Let the voltage at both ends of the primary coil be U1, the voltage at both ends of the secondary coil is U2, the voltage at both ends of A is ul, and the voltage at both ends of B is U.

By the principle of partial voltage, the relationship between the primary and secondary coils can be executed by equations:

u1=u-ul

u1=2u2

2ul/r=u2/r

Solution: u2=2u 5

Since the B lamp is directly connected to the secondary coil, the voltage at both ends of B is equal to the voltage at both ends of the secondary coil, that is, U=U2=2 5U.

Therefore, the answer is D.

Assuming that the B voltage is U current when i, then the voltage supplied by the left end of the transformer is 2U and the current i 2 is the same as AB.

So a on voltage U2

So u=2u+u2=5u2

u=2u/5

This only considers voltage, does not consider current and energy, voltage ratio = inverse ratio of the number of turns, I didn't consider the voltage drop of A, downstairs is right, choose D

Let the resistors of A and B be R, then the equivalent current on the left U R, the equivalent current on the left side I2 1 2, the current on the right side 2U R, then multiply R to obtain the voltage of 2U

After reading the answers of others, I knew that I had ignored the problem of partial pressure and should not use ohms.

Originally, it was necessary to consider the resistance of electric lights.

Let the resistance value of the small bulb remain unchanged.

According to the words on the bulb label (4V, 4W) and the formula p = u r, the bulb resistance r = 4 can be obtained

The slide vane of the sliding rheostat is at the A-side, where only the bulb is connected to the circuit and emits light normally, i.e. U = 4 V

When the sliding vane is at the midpoint of AB, the sliding rheostat is connected to the resistor r, =4 r, that is, the lamp bidivides the voltage U at the sliding rheostat

So according to the formula, the power of the bulb at this time p = 2 4 = 1 w

The bulb 4W is the standard power, so the standard voltage should be used when finding the resistance: r=4 4=4 The resistance does not change, it is always 4

Normal luminescence at point A indicates that U=4V.

At the midpoint, the AB resistance is 4, and the actual power of the bulb can be found by the current and resistance, and I r=1W

Therefore, the standard voltage should be used when finding the resistance: r=4 4=4 The resistance does not change, it is always 4

Normal luminescence at point A: U=4V.

At the midpoint, the AB resistance is 4, and the actual power of the bulb can be found by the current and resistance, and I r=1W

The resistance of the bulb can be found by deriving r=u*u p using the formula p=u*u r. i.e. r=4v4v4w=4. The power supply voltage is 4V.

When the sliding rheostat slide is moved to the midpoint, the resistance connected to the circuit is 8 2 = 4The total resistance in the circuit is 4 +4 = 8. The current in the circuit is 4V 8=.

So the actual power of the bulb =.

Fill into.

If the charges on the two spheres are evenly distributed, then the electric field it generates externally, equivalent to all the charges concentrated in the center of the sphere, is equivalent to the point charge at the corresponding position.

In fact, due to the interaction of the charges of the two charged spheres, the charges will move in opposite directions, and most of the charges are close to each other, which means that the distance between the two charges +q and -q is less than the distance d, so the mutual attraction is greater than that of two point charges that are separated by d.

When the switch S1 and S2 are closed, and the sliding blade P of the sliding rheostat is placed in a certain position, the indication of the voltmeter with a range of "0 6V" is 4V, and the small bulb L marked with the words "6V 3W" emits light normally. It can be seen that R is short-circuited, RL is connected in series with RP, and the total voltage of the circuit is equal to the sum of the voltages of the parts.

Voltmeter UV=UP=4V, the small lamp emits normal light UL=6V, then U=UV+UL=......=10v

Hence a mistake!

Because p=ui i=u r

then rl=ue 2 pe=......=12ω

ip=ile=pe/ue=……=

Hence rp=up ip=......= 8 ohms.

When the switch S1 is closed and S2 is disconnected, and the slider P of the sliding rheostat is placed at the right end of the rheostat, the small bulb L still emits light normally. 】

Knowing that RP is short-circuited, RL is connected in series with R, and the small light is still shining normally, then R=RP=8 Therefore, B is wrong!

There is no condition in the problem to say what is the maximum value of RP, so it cannot be solved! But the RP max is 8 so C is also incorrect!

When the total resistance of the circuit is maximum, r total rl+r+rp=......=28ω

Then i min u r total =

then pl'Min i min 2 rl=......= - only d is possible - it depends on the maximum value of rp!

This question should be a multiple-choice question that is not rigorous enough! - There is no definite answer!

The process of option C: After the sliding rheostat is fully connected to the circuit, the larger its resistance value, the greater the distributed voltage, but its maximum value cannot exceed the range of 6V of the voltmeter connected in parallel.

When switches S1 and S2 are closed, and the sliding blade P of the sliding rheostat is placed at the leftmost end, the maximum voltage distributed by the sliding rheostat is 6V. From this, the maximum resistance value is: 6V (U lamp R lamp) = 6V (10-6) V 12 ohms = 18 ohms.

S2 is closed, R is short-circuited, the voltage is 4V, the small bulb emits light normally, so the bulb voltage is 6V, so the power supply voltage is 4+6=10V

S2 is disconnected, P is at the right end, the sliding rheostat is equivalent to the wire, and the small bulb is still glowing normally, then the voltage on R is 4V, and according to "6V, 3W", you can get the current of the normal light-emitting small bulb is, the bulb resistance is 6x6 3=12 ohms, the resistance of r is 4 ohms, and the maximum resistance of the sliding rheostat must be greater than the resistance of r 8 ohms.

3,4 is grounded, then the positive charge of qa is neutralized by the negative charge given by the ground, and it is not charged, so 3,4 must not change. Exclusion option A

First of all, it is necessary to understand that the resistors R1 and R2 of the two parts of the sliding rheostat are connected in parallel in the circuit.

When P is at the top, R1=0, after P begins to decline, R1 gradually increases from zero, the total resistance gradually increases in the previous stage, and the total current gradually decreases, it can be seen that the internal voltage decreases, the external voltage increases, the number of voltage representations increases, and the total current decreases, then the voltage between R and R decreases, the parallel voltage increases, the R2 voltage increases, and the number of current representations increases.

When R1=R2, P continues to slide, the parallel resistance decreases, the total resistance decreases, and the total current increases, it can be seen that: the internal voltage increases, the external voltage decreases, the voltage representation decreases, the total current increases, then the voltage between R and R increases, the parallel voltage decreases, the R1 voltage decreases, and the R1 blocking increases, so the R1 current decreases, the R2 current increases, and the current representation increases;

It can be seen that the number of voltage representations increases first and then decreases, and the number of current representations increases all the time.

Qualitative derivation is a bit too much.,I don't know if I can understand it.。。。

The voltage is present because the potential is different;

The existence of electric current is driven by external forces such as voltage and magnetic field;

There is a potential" and "can this potential drive the directional flow of charge" are two different things;

Magnetic inductance lines can drive electrons to move, but in the absence of circuits, electrons (charges) are crowded into a dead end, forming an electric current, but the potential has already formed.

- The explanation in the book is:

The formation of the current requires a power supply and a closed loop, both of which are indispensable.

Voltage is an expression of the power supply capability.

If there is current, there must be voltage, because voltage is the cause of the current. There should be no doubt about this? But why can there be voltage but no current?

There is a very classic counter-example: a conductor rod moves in a magnetic field, due to the action of the ampere force, it moves the charge, but because of the action of the ampere force, although there is a voltage, the voltage cannot guide the electrons to move freely, they are balanced. Another example is that the two plates of a capacitor have voltage, but the circuit is disconnected, so there is no current.

So when there are certain external conditions, there is a voltage and not necessarily a current Dare to ask what is not an example? Do you believe me just talking about theory? Is that clear?

Where do you want me to talk?

There are two conditions for the formation of current in the circuit: one is that there is voltage at both ends of the circuit, and the other is that the circuit is a path.

Therefore, if there is voltage at both ends of the conductor, there is not necessarily a current in the circuit.

But there is an electric current in the conductor, and there must be a voltage at both ends of the conductor.

Needless to say, there are three conditions for generating current, 1 power supply (electromotive force) 2 conductor There is also a closed loop between the two, so there must be voltage if there is current, and there is no current if there is voltage and does not constitute a loop.

7.Correct answer a

The key to this problem is that when the output power of the power supply is maximum, the efficiency of the power supply outside R=R is 50%, and R1=R2+R

r=10 r=(r2+r)r1 (r1+r2+r)=10 a correct.

During the sliding process, the external resistance decreases, the total current increases, PE=IE increases, and B is wrong.

pm=e^2/4r=3600/40=90w