High School Physics Transformer Problems Guys, help me

The power of both ends of the transformer is equal, and if the current of one section is zero, then his power is zero, resulting in the power of the other end is also zero, so there is no current; The original coil is connected to the generator, which is the conversion of energy, there will be no voltage without current, and the potential energy of water will not be converted into electrical energy; When transmitting power over long distances, the user's load is uncertain, that is, the resistance is uncertain, and cannot be considered from the aspect of resistance. It should be thought that when transmitting electricity over long distances, it is to minimize the loss of electric energy, and W=PT means that the loss of power is as small as possible. When increasing the transmission voltage, it can be seen from p=ui that when the same amount of power is transmitted, the voltage is high, the current is small, and the q i rt is small during transportation.

Hope it helps!

From the perspective of high school physics: when the output current of the transformer is 0, the transformer is equivalent to a coil with a large amount of inductance, although there is a voltage on the top, the current is not (the inductor is DC and AC). In reality, when the transformer is open in output, there is a small current in the input, which is called the excitation current, and its function is to maintain a certain magnetic field in the transformer core.

The damage in power transmission is mainly on the transmission line, you only consider the loss of the transmission line, the resistance of the transmission line is certain (the same area and length of the wire), the transmission power is certain (p=u*i), the greater the current, the greater the damage to the line (p=i*i*r), so the choice of high voltage, small current mode can reduce the damage of the line.

If the power plant generates electricity, the user does not use electricity, and the power plant does not send electricity to the outside world, and how many people use it to send as much electricity.

High-voltage transmission is the improvement of transmission efficiency, for example, at the beginning of the investment of 10 yuan loss of 1 block, now the investment of 1000 loss of 10 yuan, or the cost.

Your high school is about the ideal transformer, and if you go to college, the power consumption of the primary coil is really zero.

1. If the power of the secondary coil is zero, it means that there is no load on the secondary side of the transformer, there is no current, according to the principle of power conservation P1 = P2 (this is an important formula, of course, it is only applicable to the ideal transformer stage in high school, in fact, the transformer has its own loss after passing the current in college), that is, the input power is equal to the output power, then the secondary coil power = 0 = primary coil power.

The landlord's confusion is that there is alternating current at the primary level, this is certain, the transformer is dependent on this alternating current to generate magnetic flux, and the primary secondary coil of the transformer induces electromotive force, and the primary induced electromotive force = the connected power supply voltage.

2. The secondary coil is not connected to the load, the circuit is not connected, there is naturally no current, the load circuit is connected, and the current flows through, the landlord said that the load changes the magnetic flux, and the load does change the main magnetic flux after passing the current, but it is offset by the magnetic flux generated by the increased current of the primary coil, so it is still the main magnetic flux generated by the original (this is the reason why the primary current increases with the secondary side).

3. According to Ohm's law r=u i, if the voltage of the secondary coil is unchanged, the resistance becomes larger and the secondary coil current becomes smaller.

The landlord's ambiguity can be discussed.

1.The transformer itself does not consume power.

2.The transformer itself does not produce or consume power, it only transmits power. The secondary load is connected to the load, and the load consumption power is transmitted from the primary.

3.Voltage is the potential energy and current is kinetic energy. In the circuit, no matter how high the voltage is, there is no current, and it will not work. If there is current, there must be voltage, and there is also power consumption.

1. To be precise, the power of the secondary coil is zero, and the primary will not be zero, considering the loss of the excitation part. The primary coil is connected to the power supply, will there be alternating current? There is also a corresponding voltage that will have a magnetic flux.

2. There is only the excitation current without load, the current is very small, and after the secondary load is added, the primary current is composed of the excitation current and the secondary current, and the current will become larger.

3. I didn't understand the third question.

The diode in the secondary coil causes the circuit to generate a current, and the direction does not change, but the magnitude changes, then the effective value of the secondary coil is calculated as follows."

44) 2 r*t 2=u 2 r*t u=22 2v , if there is p1=p2 then i1:i2=1:5 2 then ab is wrong, c is correct.

The resistance of the d-slide rheostat becomes smaller, the voltmeter remains unchanged, and the current becomes larger, the meter indicates that the number becomes larger d error.

Hope it helps!

Solution: a. The current in the primary and secondary coils is inversely proportional to the number of turns, so the ratio of the current is 1:5, and the term A is correct;

The voltage of b, c, primary and secondary coils is proportional to the number of turns, so the voltage at both ends of the secondary coil is 44 V, but the diode has unidirectional conductivity, according to the definition of the effective value, there are 442 R t 2 = U 2 effective r t, so that the effective voltage at both ends of the voltmeter is U effective = 22 2 V, then the heat generated in 1 min is Q = u 2 effective r t = u 2 effective r 60 = 2 904 J, C is correct, b is wrong;

d. Slide the slider of the sliding rheostat downward, and the resistance value in the access circuit becomes larger, but it has no effect on the voltage at both ends of the primary and secondary coils, that is, the reading of the voltmeter remains unchanged, and the reading of the ammeter becomes smaller, so D is wrong

So choose AC

aThe ratio of current is inversely proportional to the ratio of voltage.

B because the diode is one-way, the equivalent voltage is only half of the original, 22V, do not choose C access 20 resistance current is 22 20 I2RT=1452 is not equal to 2904, do not choose.

The d-resistance value decreases, and the voltmeter value decreases. The power increases, the current increases, so the answer is A

Hope it helps.

The correct answer is AC. The above answers, the analysis of a, b, d is correct, but c is also correct, after the diode rectification is the half-wave pulsating DC, within 1 minute, when the half-wave exists, the effective value is still 44 volts, and the heat generated in 30 seconds is 30x44x44 20=2904j, and the other 30 seconds without voltage do not produce heat, so c is also right.

Because the ratio of turns n1:n2=2:1, then the ratio of current is 1:2, the voltage of lamp b is ub, then the voltage of lamp a is ub 2

u-ub 2=2ub ub=option d

is answer d. But I believe that the questioner will choose a for this question, which should be confused by n1 n2=u1 u2, right? Please pay attention to the position of the bulb.

The resistance r of lamp b is equivalent to 4r on the n1 side, the voltage division is , and the voltage to the n2 side is .

A: First of all, we need to know that the voltage and current in the question are both effective values, and the schematic diagram of the whole process is as follows.

The general topic does not specifically refer to the power loss of the transformer, we can think that the loss is mainly the loss on the transmission wire, and the electrical energy lost on the wire is the heat energy generated on the wire, which can be ...... with p=i2 ror p=u2 r......From the meaning of the title, we know that the maximum P loss = 600W, and the current of the wire can be found to be 6A, that is, the current you said, and the voltage at both ends of the wire can be found to be 100V. So the process is to first drop the 240V voltage to 100V, and then raise the 100V voltage to 220V. Therefore, the voltage at both ends of the step-up and step-down transformer is known, and now you only need to find the ratio of turns required by the problem according to the voltage value at both ends of the primary and auxiliary coils proportional to their turns.

For bucking: n original n sub = 240v 100v = 12 5

Boost: n original n sub = 100v 220v = 5 11

The current of the middle line is the current of the high-voltage line in life.

B does not need to calculate, the voltage conversion of the transformer only looks at the ratio of the number of turns of the coil, and the energy loss can be ignored, and the figure on the right, the energy is lost in the resistance, GH at both ends plus 110V is equivalent to E series a resistance connected to the 110V power supply, because E and F open circuit there is no current, so the nano EF voltage is 110V, if the EF is a series of resistors, then the voltage of the EF will be how many holes?

Answer: (b) A is the transformer mutual inductance principle, B resistor divides the voltage, due to the no-load between EF, then UEF=UGH

The ratio of turns of the primary secondary coil is n1:n2=2:1

The voltage ratio of the primary and secondary sides is u1:u2=2:1

The pay-as-you-go voltage is half of the primary-side voltage. If the rated voltage of the three bulbs is the same, and the voltage applied on L1 is the rated voltage, then the voltage obtained on L2 and L3 is only half of the rated voltage, the current through L2 and L3 is only half of the rated current, and the actual power of L2 L3 is only 1 4The brightness is greatly reduced.

If the trunk current is i, it will pass through the transformer, and it will become 2i, l1, l2, l3 to the right, which is the same, so they all emit light normally.

U1 = 4V, L1 emits light normally, the partial voltage is 2V, and the power supply voltage is 6V