A circuit physics problem, experts please enter

Explain two questions you don't understand:

1. After K is combined, if the motor does not rotate, it is a pure resistance, and the current is expressed as the number;

If the motor rotates under the action of ampere torque, the coil will move to cut the magnetic field lines, resulting in an induced electromotive force, which is in the opposite direction to the applied voltage (a reverse power supply is strung in the motor circuit).

Therefore, the motor current is less than 1A, and the current representation is less than that.

2. Why does the voltage remain the same?

This is a known condition given by the question.

Moreover, the general good battery (such as battery), the internal resistance is only a few percent of the ohm or less, if the internal resistance is European, the current increases by 1 ampere, and the terminal voltage will only decrease volts, compared to the power supply voltage in the topic of 20 volts, yes, you see, is the voltage unchanged? !

In high school, in general, less than 1% is negligible.

1. The voltage is the power supply voltage, which generally does not change due to the access of electrical appliances.

2.The motor is not a pure resistive electrical appliance. When he was working, he wanted to be a resistor, an ideal motor.

Therefore, the voltage of its resistance is much smaller than the voltage of R1, so B is right; By the resistance R1=20, the motor winding resistance R2=10, when the electric key S is disconnected, the indication of the ammeter is I1 = the total voltage can be known to be 10 volts, the total current is <, so P<15W

Before the closure of the power building i* r+i * r1=u r is the internal resistance of the power supply, the voltage on the R1 obtained at this time is already the voltage of the power supply minus the voltage of the internal resistance of the power supply, after closing, due to the parallel, the voltage on the motor is equal to the voltage of R1, and the voltage at both ends of the circuit remains unchanged.

1. After the S is closed, the motor rotates, and the power supply voltage remains unchanged; This condition gives that the power supply is a constant voltage source.

2. Before S is closed, the ammeter reading is the current flowing through R1, the power supply voltage = I1 * R1 = 3, after S is closed, the motor rotates, the power supply voltage remains unchanged, and the current flowing through R1 remains unchanged;

4. After S is closed, if the motor is locked so that it does not rotate, the motor is equivalent to a pure resistance, and the current flowing through the motor is:

10v 10 ohms = 1a

5. After S is closed, the motor rotates to produce a back electromotive force E, which is the current flowing through the motor

10-e）/10<1a

Therefore, the total current is less than and the total power is less than 15W

When S is closed, R1 and R2 are connected in parallel, and the voltage at both ends of the circuit is provided by the controllable circuit, and the problem assumes that the voltage at both ends of the whole circuit is constant.

The first S is opened, the voltage at the end of the circuit can be found to be U=I1R1=10Vs closed, the voltage at both ends of the parallel circuit is equal, and the total voltage is adjusted by the theme to keep the total voltage unchanged, the internal coil of the motor is a non-pure resistance circuit, so Ohm's law does not apply, the actual current through the motor is much smaller than Ohm's law, if it is calculated by Ohm's law: i2=U R2=10 10=1A

Therefore, it can be seen that the total current is less than, and the total power p=u*i is less than 15W

In this question, it should be considered that the motor is an inductive load, that is, in addition to the resistance, there is also an inductive impedance, otherwise the motor will not rotate. Or you can understand that the power consumed by the motor is divided into active warp rate and reactive power, the active part is the work of the motor), and the reactive part is the heat generated by the motor, the magnetic field loss, etc.

The specific algorithm, when there is no closing s, is regarded as a pure resistance circuit, and the voltage can be calculated as 10 volts. After closing S, if the motor is a pure resistance element, regardless of the line loss factors such as the internal resistance of the power supply, the ideal current A is just A, but it has to be considered that the electric rotation is not a pure resistance device, and the impedance of the whole circuit is definitely higher than that of a pure resistance device, so the loop current should be less than A.

By the same explanation, answer D can be derived.

As for why the voltage you are talking about does not change, this is a premise given in the title, that is, it is assumed that the supply voltage does not change. In fact, if the power supply is stable, it can be regarded as unchanged for a little time after closing the S, but it will fluctuate a little when it is turned on just now.

To say that the voltage does not change means that the power supply is the ideal power supply, so I can understand it.

s is not closed is, v=ri=10

When closed, the R2 circuit I 10 10=1 because the electric motor wants to convert electrical energy into mechanical energy, and it also generates resistance.

Because the total resistance of the motor circuit is greater than 10 ohms, the total power is less than 15 watts if the power is less than 10 watts.

Hey, the constant voltage across the circuit is an assumption that the device that supplies the electrical energy is a so-called constant voltage source.

The motor will convert electrical energy into other forms of energy, so the power of the motor cannot be calculated with p=ui, which is less than this value, and only the whole circuit is considered the internal resistance of the power supply, and some circuits are not used.

The title says, "When the motor turns, the voltage at both ends of the circuit does not change."

The visible power supply is a constant voltage source.

The voltage at both ends of each component of the series circuit is equal!!

Current i=q t=20c 30s=2 3a=current workw=uit=220v*2 3a*30s=4400j, what the current does is the electric energy consumed is 4400j

Resistance r=u i=220v ohms.

The voltage is 220V, the current through the electrical appliance is I=Q T=20C 30S=2 3A, the current workmanship is the same as the electric energy consumed, which is equal to W=UIT=220V*2 3A*30S=4400J, and the resistance is R=U I=220V=220V (2 3A)=330 ohms.

The current is equal to the amount of charge divided by the time i.e. 2 3 A

Resistance is equal to voltage divided by current i.e. 220 (2, 3).

The work is the square of the voltage divided by the current.

us1 = i1*r1+i3*r3

us2 = -i2*r2+i3*r3

i1 = i2 + i3

The above three equations can be obtained by solving the above equations.

Solution: (1) When the switch is disconnected

When R1 is connected in series with L. du

rl=ul 2 pl=(6v) 2 6w=6 ohm zhi i=u (r1+rl)=9v (9 ohm 6 ohm) = (2)pl

dao=il 2*rl=(oh

3) When the inner switch is closed, the sliding rheostat is connected in parallel with R1 and connected in series with the small bulb.

In order to make the small bulb shine normally, the voltage at both ends of the small bulb is 6V through the current of the small bulb IL=PL UL6W 6V=1A

Let R2 and R1 be connected in parallel, and the voltage at both ends is U1.

u1=u-ul=9v-6v=3v

The current passing through R1.

i1=u1 r1=3v 9 euro 1 3a

The current passing through the sliding rheostat.

r2=u1 (il-i1)=3v (1a-1 3)=ou hope this helps you.

(1) Lamp resistance.

Come rl=u squared.

Since p=6*6 6=6

i=u/(r1+rl)=

2) Actual work.

BAI rate p1 = i squared * rl =

3) If the lamp works normally, the lamp gets the voltage zhi6v, so daou*rl (rl+r1 r2)=6, so r2=

First, calculate the resistance of the bulb: r=u p=6

1. Total resistance = (6 + 9) = 15, total current = u 15 = 9v 15 = , the electrical circuit is in series.

2.The actual power response rate p = r* =

3.In order for the bulb to shine normally, the voltage of the bulb is 6V, the current is 1A, and the voltage of the resistor part is 3V. The current of the circuit is the current of the bulb 1A, the total resistance of the resistor part = 3V 1A = 3, and according to the calculation of the parallel resistance, R2 = is obtained

According to the bulb data, according to the formula p=u square divided by r, the bulb resistance can be calculated as 6 The first bai question: s is disconnected.

Du then the resistance is connected in series with the bulb and the total voltage divided by (daor+r lamp) to get the current amperes The second question: To calculate the actual power of the small bulb, as long as you know the current flowing through the small bulb, the current is found in the first question, so the power of the bulb is p=i, square r, and the lamp is solved to get p=

1.Find the number of current representations.

R3 R4 = 6*3 (6+3) = 2 and R1 with a voltage of = V

The current is expressed as + = + = a2Find the voltage representation.

r1//(r2+r3) = 10*(4+6)/(10+4+6) = 5ω

The partial pressure on R4 is = V

The upper partial pressure is , and the upper partial pressure of r3 is = v

So, the voltage is expressed as + = v

I can't see the picture on my mobile phone, and this kind of problem is very simple to solve with Kirchhoff's law, and the equation is listed.

Because the voltages at both ends of 3 and 4 are equal).

Only the outgoing current through R4 is required, and it is subtracted from the total.

The resistance after 3 and 4 in parallel is: 2 ohms.

The resistance of 1 is 10 and 2 is 4, so the total resistance is 3 ohms.

The total current is ampere.

2.After changing to voltmeter, it can be ignored, according to the electromotive force R2 The potential difference between the upper end and the lower end of the voltmeter is , and according to the calculation, the potential difference between the upper and lower ends of R2 is , so the potential difference between the upper and lower ends of the voltmeter is - = v

u3=12v

Suppose the current vector is i and the initial phase is 0 degrees. Then calculate the voltage vectors of capacitance, inductance, and resistance separately. According to KVL's law, it is easy to get ur+ul+uc=u. Reverse solution u3

It's easy to see by drawing a picture.

It is necessary to pay attention to the problem that the inductor voltage and the capacitor voltage are just reversed

The role of the power supply in the circuit is to provide continuous current, and when calculating such a problem, the current contributed by each power supply to the circuit can be calculated by considering it one by one (if one direction is positive, then the current flowing in the opposite direction is negative), and then superimposed to obtain the current flowing through a certain point.

Step (when the power supply does not count the internal resistance):

1. Calculate the current contributed by E1 in the circuit.

Reduce the voltage of E2 to 0, then R2 and R3 are short-circuited, through R2, the current of R3 i2=i3=0, the current through R1 i1=E1 R1, the current direction is from left to right, let this direction be positive;

2. Calculate the current contributed by E2 in the circuit.

Reduce the voltage of E1 to 0, then the circuit can be equivalent to R1, R2, and R3 in parallel at both ends of E2, and the currents through R1, R2, and R3 are respectively

i'1=e2 r1 and the current direction is still from left to right i'2=e2 r2, the direction of the current is from right to left, which is negative i'3=e2 r3, the current direction is from right to left, which is negative

i(r1)=i1+i'1=

i(r2)=i2+i'2=-7a

i(r3)=i3+i'3=

If the internal resistance of the power supply is considered, the above method can still be used.

R1 voltage at both ends 42-21=21V current 21 12 A

R2R3 voltage at both ends 21V R2 current 21 3A=7a R3 current 21 6A=7 2A

The positive electrode of E2 is equivalent, the positive electrode of E1 is 42V higher than the negative electrode, the negative electrode of E2 is 21V lower than the positive electrode, the voltage of R1 is 63V, and the current is 21 4A; R2R3 voltage at both ends 21V R2 current 21 3A=7a R3 current 21 6A=7 2AThat's it.

It's better to ask the teacher, it will be convenient.

The university method is very simple: the current of r2 r3 is directly removed with e2, because e2 is a constant voltage source.

And then the current of R1 is e1 + E2 removal, and if the current of R1 is in the positive direction, then the current of R2 and R3 are both in the negative direction.

i1=i2=-7vi3=