What role does a 0 ohm resistor play in a circuit board 0 ohm resistor is generally placed between the G pole and S pole of the switch tube, and there is a pin into the IC, but what is its role?

Some of them are only used to isolate analog ground and digital ground, and connect them with 0 ohm resistor. Sometimes, because the wiring on PCB is not good enough, they can only be short circuited with wires. However, during production, the placement machine can not install wires, so it can replace wires with resistors. In this way, it is easy to assemble. There are other functions, such as fuse, which will burn when the current is high, so as to protect the circuit, and the EMC Effect of high frequency,

What are the functions of diodes, triodes, resistors, electrolytic capacitors and inductors
Don't copy a bunch of things, just say what they are for
Also, I see that some people connect a small capacitance (as big as chopsticks) to the output end of the pneumatic fixture. I can't understand whether that connection can stabilize the voltage. Under what circumstances should it be connected like that? I'm also a new person to connect the starting fixture,

The function is:
Diodes are used for rectifying, detecting and switching
Triode is used for amplifying, switching and stabilizing voltage
Resistance has the function of voltage division and current limitation
The function of electrolytic capacitor is filtering and coupling
Inductor is also filtering and coupling, but inductor is filtering high frequency current signal. These components can form electronic and control circuits with different functions, and each has its own specific functional circuit

In a series circuit, if the ratio of two resistors R1: R2 = 3:2, the ratio of current through resistor r1r2 I1: I2 = the ratio of voltage at both ends of r1r2 U1: U2=

∵ two resistors are connected in series, and the ratio of current through resistor r1r2 is I1: I2 = 1:1
According to Ohm's law, u = IR, R1: R2 = 3:2, I1: I2 = 1:1
R1r2 voltage ratio U1: U2 = 3:2

There is only one rated power and one actual power for each consumer

Wrong, many electrical appliances have more than one rated power and actual power, such as induction cooker, baking stove, etc. different uses, gears, rated power and actual power are different

Write down the measurement steps for measuring the rated power of small bulb

1. Connect the small bulb with sliding rheostat, ammeter, switch and power wire to form a series circuit, and connect the voltmeter in parallel at both ends of the small bulb (switch on, sliding rheostat resistance to the maximum)
2. Close the switch, move the slide of the sliding rheostat, make the indication of the voltmeter equal to the rated voltage UL of the small bulb, read out the indication IL of the corresponding ammeter, observe the brightness of the small bulb, then the rated power of the small bulb is: PL = ulil
3. Move the slide of the sliding rheostat to make the indication of the voltmeter equal to 1.2 times of the rated voltage of the small bulb, observe the brightness of the small bulb, read out the indication of the ammeter, and then the actual power of the small bulb can be calculated
4. Move the slide of the sliding rheostat to make the indication of the voltmeter lower than the rated voltage of the small bulb, observe the brightness of the small bulb, read out the indication of the ammeter, and then the actual power of the small bulb can be calculated
5. Compare the relationship between the brightness and power of the third bulb
6. Arrange the equipment

Ji Xiaoming found that the voltmeter was damaged when measuring the rated power of a small bulb with a rated voltage of 2.5V
Xiaoming found that the voltmeter was damaged when he measured the rated power of the small bulb with the rated voltage of 2.5V, but the ammeter, power supply, switch, lamp, sliding rheostat and wire were in good condition. In order to let him measure the rated power, the teacher added a constant resistance R0 with known resistance value. Please help Xiaoming complete the experiment
(1) Draw the circuit diagram in the dotted box
(2) The experimental steps were as follows
A connect the real object according to the circuit diagram
B connect the ammeter first___ Adjust the slide P so that the ammeter reading is
C. connect the ammeter_ , slide p_ (fill in "not moving" and "sliding to the maximum resistance"), read out the number of ammeter and record it_ .
D calculate the amount of P=_____
Mainly tell me the circuit diagram

Zixue Lingyun, Hello: (1) voltmeter damage, the most important thing is to make a voltmeter, you need to understand the essence of a voltmeter is an ammeter and infinite resistance in series, understand this, you can easily solve the problem. Therefore, making a voltmeter will first make the ammeter and sliding rheostat in series. This is a voltmeter

Power supply voltage is 4.5V, rated voltage of bulb is 3.8V, filament resistance is about 10 ohm, sliding rheostat is marked with "10 ohm 1A", ammeter (0 ~ 0.6A, 0 ~ 3a), voltmeter (0 ~ 3V, 0 ~ 15V)
When measuring the rated power of a small bulb, if the circuit is properly connected and operated in the correct order, the lamp will not light up when the switch is closed,
If the filament is broken, what is the indication of voltmeter and ammeter?
If the small bulb is short circuited, what are the readings of voltmeter and ammeter?

If the filament is broken, the voltage indication is about 4.5V, and the current indication is about 0.5V
If the small bulb is short circuited, the voltage indication is 0 and the current indication is about 0.45 a

A student did the experiment of "measuring the rated power of small light bulb". The power supply used was a new dry battery. The rated voltage of small light bulb was ambiguous, which could be 3.5V or 2.5V. It was estimated that the rated power of small light bulb was 0.5V; The student chose two dry cells in series and correctly connected the circuit. The experimental steps were correct. He closed the key and observed that the indication of voltmeter and ammeter were 1 V and 0.1 A respectively. When he moved the slide of sliding rheostat, the brightness of the small lamp changed, but it was always dark and couldn't light normally. After thinking, the student found the problem and made appropriate adjustment to the experimental equipment Adjust, and then re experiment, close the key, move the slide of the sliding rheostat, and find that when the slide is in the midpoint position, the small lamp will light normally. The reason why the small lamp can't light normally before adjusting the experimental equipment The rated voltage of small lamp is When the lamp is normally on, the resistance of the rheostat connected to the circuit is Oh, the rated power of the small lamp Tile

According to "moving the slide of the sliding rheostat, the brightness of the small lamp changes, but it is always dark", it can be seen that at this time, the bulb can not light normally because the power supply voltage is too small. At the same time, it can be seen that two dry batteries (3V) can not make the bulb light normally, so the rated voltage of the small bulb must be 3.5V, and at this time, the power supply must use three dry batteries as the power supply, so the power supply voltage is too low It is 4.5V; when two dry batteries are connected, close the key and observe that the indication of voltmeter and ammeter are 1V and 0.1 A respectively. It can be seen that the voltmeter is connected in parallel at both ends of the bulb, so the resistance of the bulb is: I = ur: R lamp = UI = 1v0.1a = 10 Ω. And the rheostat is connected in series with the bulb, so the voltage of the rheostat is u = 3v-1v = 2V; the current through the bulb is 0.1A, so the rheostat The total resistance of the rheostat is: I = ur: r = UI = 2v0.1a = 20 Ω; the experimental equipment is properly adjusted, and then re experiment, close the key, move the slide of the sliding rheostat, and it is found that when the slide is in the midpoint position, the small lamp lights normally, that is, the resistance of the rheostat connected to the circuit is half of the total resistance, that is, R rheostat = 10 Ω; at this time, the total resistance of the circuit is: R total = R Lamp + R rheostat =10 Ω + 10 Ω = 20 Ω; therefore, the total current of the circuit is: I total = u total R total = 4.5v20 Ω = 0.225a; because the bulb is normally emitting at this time, the rated current of the bulb is 0.225a, so the rated power of the bulb is: P = UI = 3.5V × 0.225a = 0.7875w; therefore, the answer is: the power supply voltage is too low; 3.5; 10; 0.7875

The rated power of electrical appliances refers to the rated power of electrical appliances A light bulb marked with "220v60w" is connected to a 110V circuit. The actual power of the light bulb is ．

(1) Rated power refers to the electrical power of the appliance working under rated voltage; (2) from P = u2r, the resistance of the bulb: r = u2p = (220V) 260W = 24203 Ω, when connected to the 110V circuit, the actual power of the bulb: P actual = U2, actual r = (110V) 224203 Ω = 15W

The electric power marked on the electric appliance refers to ()
A. How much electricity consumption B. rated power c. consumption D. actual power

The electric power marked on the electric appliance refers to the rated power of the electric appliance

What role does a 0 ohm resistor play in a circuit board 0 ohm resistor is generally placed between the G pole and S pole of the switch tube, and there is a pin into the IC, but what is its role?