The maximum current that a resistance wire can pass is 2A. When a voltage of 36V is applied to it, the current passing through is 0.5A______ (fill in "can" or "can't") used in 220 V circuit

The maximum current that a resistance wire can pass is 2A. When a voltage of 36V is applied to it, the current passing through is 0.5A______ (fill in "can" or "can't") used in 220 V circuit

The resistance value of resistor wire is r = u1i1 = 36v0.5a = 72 Ω. In the circuit of 220 V, the circuit current is I2 = u2r = 220 V 72 Ω≈ 3a, ∫ 3a ＞ 2A. It can not be used in 220 V circuit

With an LED lamp, the voltage is 220 V, the LED current is generally 20 mA, according to r = u / I, the current limiting resistance is 11 K, why is my resistance burnt out

R=U/I
U=220V I=20mA=0.02A
220 / 0.02 = 11000 Ω, power W = 220 * 0.02 = 4.4 w
Your resistance is not too small to withstand before burning

For a 60W bulb with a voltage of 220 V, the calculated resistance should be 807 ohm. Why do I use a multimeter to measure only 60 ohm?
When the pointer multimeter is not in use, what position should be set?

You use the multimeter to test the normal resistance of the bulb!
The filament of a lamp is a tungsten filament. When the current is applied to the filament, the tungsten filament will heat up. When the temperature reaches a very high level, the filament will become white hot. Most of the electric energy is consumed by heat, and a small part of the light energy is used by us!
Generally, metals are positive temperature materials, that is to say, when the temperature rises, the resistance of the metal increases, so if you raise the temperature of the filament to the white hot state, the resistance should be the same as what you calculate!

Why should the fuse have high resistance? The total voltage is 220 v. don't tell me that the current must be constant and the voltage must be constant. Isn't it because the resistance is small and the heat is much?

The principle of design and installation of "fuse" is: using the thermal effect of current q = I & sup2; RT, when the current in the circuit reaches (or exceeds) the designed maximum current, the fuse will produce "more" heat and make the temperature rise rapidly. When it reaches the melting point of the fuse, it will automatically fuse and disconnect the circuit, thus playing the role of insurance
If the resistance of the fuse is too small, it will not generate more heat, and the fuse will not blow automatically