Why can't fuses be replaced by copper wire, aluminum wire or iron wire? extra urgent

The function of the fuse is to let the over-current fuse when it exceeds a certain current, so as to protect the circuit and electrical appliances. Its material must be low melting point and high resistance in order to fuse when it is overloaded. You say those materials are used as fuses. Once they are overloaded, they can't fuse by themselves, of course they can't play the role of insurance

Bend the wire repeatedly by hand, and the temperature rises at the bend of the wire. Some people say that the main reason is that the temperature of the hand is transmitted to the wire
Is that right? Why

Of course it's not right. If it's passed by hand, why isn't the whole wire hot? Obviously it's not right! Bend the wire back and forth to make the iron molecules move. There is friction between the iron molecules. Both the movement and friction generate heat energy, so it feels hot and only at the bending place

Resistance R1 is 50 ohm / (1W), R2 is 100 ohm / (0.5W) in series to the circuit of 15V power supply. When switch S is closed, what will happen to r1r2?

The closing current of the switch after series connection = V / r = 15V / (50 + 100) = 0.1A
Power on R2 = I square, XR = 0.1ax0.1ax100 Ω = 1 Watt
Power on R1 = I square, XR = 0.1ax0.1ax50 Ω = 0.5W
The actual power consumption of R 2 is 1 W, but the resistance is only 0.5 W
The actual power consumption of R1 is 0.5W, but the resistance is only 1W, no problem

I would like to ask three 1W LED patch lamp beads with 12V to 14V voltage, lamp beads are 300mA, with 5w20 ohm resistance and 2w15 ohm resistance what is the difference?

The voltage of three 1W LEDs is about 9v-10v when they work normally. It's just right to use 2w15 to reduce the voltage of 15 * 0.3 = 4.5V. The power consumption is 0.3 * 4.5 = 1.35W, which is within the 2W rated power consumption range of the resistor. With 5w20 to reduce the voltage of 20 * 0.3 = 6V, the power consumption is 0.3 * 6 = 1.8W, which is within the 5W rated power consumption range of the resistor, The remaining 6V -- 8V is not enough for the three LEDs to reach the rated working state

Can 1W 0.4 ohm wire wound resistor be replaced by other resistors

It depends on where you use it. As long as the resistance power is satisfied first, as long as it is not special, it can be replaced by other resistors with similar resistance value. But in some special places, you should be careful, because there is an inductance component in the wire wound resistance. As long as it is not very strict, you do not need or use a thin wire to measure the resistance value, as long as the installation place is large enough, It's OK to go up in a circle, ha! You can also find a car light bulb with similar resistance

Can the big white 7W, 2.7J resistor use 5W, 2.7K resistor?

Yes

In the circuit, a resistor R 1 is connected, the resistance is 100 ohm, the power is 5W, and the resistors R 2, R 3 are connected in parallel,
In the circuit, a resistor R 1 is connected, the resistance is 100 ohm, the power is 5W, in addition, R 2 and R 3 are connected in parallel, R 2 is 30 ohm, 0.5W, R 3 is 60 ohm, 0.5W. Can the three resistors work normally? If not, which one can be damaged?

R2 and R3 will burn out due to excessive dissipated power. Analysis: Taking the normal operation of the component with the maximum dissipated power in the circuit as the standard, R1 = 5W is taken as the standard, According to the formula u = IR and P = UI, it can be deduced that: P = IIR = IR, then u = [√ (P / R)] R (Note: power divided by resistance equals the square value of current, and the square value of current multiplied by resistance equals the voltage at both ends of resistance) u = [√ (5W / 100 Ω)] 100 Ω≈ 22.36v, because R2 and R3 are in parallel with R1, Therefore, the voltage at both ends of R2 and R3 is equal to the voltage at both ends of R1, that is, ur1 = UR2 = UR3. According to this voltage, the dissipated power of R2 and R3 can be calculated respectively. According to the previous formula, PR2 = (22.36v) / 30 Ω = 16.66w, PR3 = (22.36v) / 60 Ω = 8.33w, and R2 and R3 both exceed the maximum dissipated power. Therefore, R2 and R3 will be broken, and PR1 = (22.36v) / 100 Ω = 499.9696 / 100 Ω = 4.99966w ≈ 5W will be checked

We plan to use 0.05mm copper wire to wind a 5 ohm resistor. How long copper wire do we need?

For example, the resistivity of constantan is 0.49
The PI is 3.1416
Then according to: length = cross-sectional area * resistance / resistivity
And (0.05 / 2) * 3.1416 * 5 / 0.49 is about 20 mm

The diameter of copper wire is 0.1, the resistance is 150 Ω, and the number of turns is 1700; if the diameter is also 0.1, and the resistance is 120 Ω, what is the number of turns?

If the diameter of copper wire is 0.1 and the resistance is 150 ohm, the number of turns is 1700; if the diameter is also 0.1 and the resistance is 120 ohm, the number of turns is 1700
1360 laps

Cross sectional area of 0.1 ohm resistance of 1 meter copper wire
Find the formula
The resistivity of copper is calculated as 0.0175 ohm · mm...

The unit of resistivity is Ω. M
The formula of material resistance is r = PL / s, where p is the resistivity of material, l is the length of material, unit meter, and S is the cross-sectional area of uniform material, unit mm & #;
From R = PL / s, s = PL / r = 0.0175 * 1 / 0.1 = 0.175mm & # 178;

Why can't fuses be replaced by copper wire, aluminum wire or iron wire? extra urgent