When the constant voltage U is applied to both ends of a metal wire with uniform thickness, the current intensity passing through the metal wire is I, and the average rate of directional movement of free electrons in the metal wire is v. if the metal wire is elongated to make its length twice the original length, throw it to both ends and apply the constant voltage u, then The rate of directional movement of free electrons is v / 2 I know that the current becomes a quarter. Why does the rate become a half?

When the constant voltage U is applied to both ends of a metal wire with uniform thickness, the current intensity passing through the metal wire is I, and the average rate of directional movement of free electrons in the metal wire is v. if the metal wire is elongated to make its length twice the original length, throw it to both ends and apply the constant voltage u, then The rate of directional movement of free electrons is v / 2 I know that the current becomes a quarter. Why does the rate become a half?

When the length is 2 times of the original, the cross-sectional area becomes half of the original. R = ρ L / s, l becomes 2 times, s becomes 1 / 2, so the resistance becomes 4 times of the original, and the current becomes 1 / 4 of the original
Is the current half?
I can't think like this!!! It's the internal electric field velocity!!!
Cable insulation resistance measurement yjy62-8.7/10kv 1 * 70 cross-linked cable use 2500 V megger to measure the insulation resistance. What is the resistance? What is the general temperature for testing? Is the resistance closely related to the site temperature?
Test at 5, 10 and 20 degrees respectively. What is the resistance?
The temperature is inversely proportional to the insulation resistance. The higher the temperature is, the lower the insulation resistance is and the greater the conductor resistance is. In this model, YJ crosslinked polyethylene has little effect on the temperature. The test results at 5, 10 and 15 degrees will not change in order of magnitude, so it can be measured at ambient temperature. This model should be identified according to GB / T 12706.2-2008, The insulation resistance test method is in accordance with the national standard GB / t3048.5-2007. If the 2500V megger is used for testing, it should be out of range normally. Because the insulation resistance of medium and high voltage cable is not specified in the national standard, but the dielectric loss is used to see the insulation performance. According to the volume resistivity of cross-linked polyethylene insulation at 20 ℃, the insulation resistance should be above 10000m Ω· km, Medium and high voltage cables are generally subject to strict partial discharge test and voltage test before leaving the factory. Therefore, in general, medium voltage cables can reach several hundred thousand M Ω. Km or more. Of course, the larger the better
A 220 V 100 W light bulb is connected to the 220 V circuit, and (1) the resistance of the filament, 2 the working current and 3 degrees of electricity are calculated to make it work
A 220 V 100 W bulb is connected to the 220 V circuit to find out (1) the resistance of the filament, (2) the working current, (3) how many hours a degree of electricity makes it work normally, (4) if no other electrical appliances are used, how many such lamps can be connected on the 220 V 2A watt hour meter. (5) the voltage of 150 V is more than ten times the power of the lamp
(1) ∵ I = P / u ∵ I = 100W / 220V ≈ 0.4545a ∵ r = u / I ∵ r = 220V / 0.4545a = 484 Ω (2) ∵ I = P / u ∵ I = 100W / 220V ≈ 0.4545a (3) ∵ w = Pt ∵ t = w / T = 1x10 ^ 3x1 / 100 = 10h (4) ∵ P = UI = 220V × 2A = 440W ∵ 440W / 100W = 4.4 ∵ up to 4 lamps (5) P = u ^ 2 / R
(1) R = u / I = u ^ 2 / P = 220 ^ 2 / 100 = 484 ohm
(2)I=P/U=100/220=0.45A
（3）t=W/P=1X10^3X1/100=10h
(4) 2 / (100 / 220) = 22 / 5 = 4.4, 4 at most
(5)P=U^2/R=150^2/484=46.5W
(1) The resistance of filament r = u ^ 2 / P = 220x220 / 100 = 484 Ω
(2) Working current I = u / r = 220 / 484 = 0.455a
(3) Normal working time of one degree electric energy: from a = Pt, a = 1 degree, P = 100W = 0.1kw, t = A / P = 1 / 0.1 = 10 hours
(4) If no other electrical appliances are used, the maximum number of lamps that can be connected to 100W on the 220 V 2A watt hour meter is as follows:
IU / P = 2x2... Expansion
(1) The resistance of filament r = u ^ 2 / P = 220x220 / 100 = 484 Ω
(2) Working current I = u / r = 220 / 484 = 0.455a
(3) Normal working time of one degree electric energy: from a = Pt, a = 1 degree, P = 100W = 0.1kw, t = A / P = 1 / 0.1 = 10 hours
(4) If no other electrical appliances are used, the maximum number of lamps that can be connected to 100W on the 220 V 2A watt hour meter is as follows:
IU / P = 2x220 / 100 = 4.4 = 4
(5) When the voltage is 150V, the actual power of the lamp is: P = u ^ 2 / r = 150x150 / 484 = 46.5w
According to w = UI = u square divided by R, r = 484 ohm, I = 0.45a.
One degree of electricity is kilowatt per hour. This is a 100W light bulb. One degree of electricity can work normally for 10 hours.
2 / 0.4545 = 4.4, up to 4 such lamps can be connected.
According to w = u square divided by R, = 150 * 150 / 484 = 46.49w
According to w = UI = u square divided by R, r = 484 ohm, I = 0.45a.
It's 100 kilowatts per kilowatt hour.
2 / 0.4545 = 4.4, up to 4 such lamps can be connected.
According to w = u square divided by R, = 150 * 150 / 484 = 46.49w
Or (1) ∵ I = P / u
∴I=100W/220V≈0.4545A
... unfold
According to w = UI = u square divided by R, r = 484 ohm, I = 0.45a.
One degree of electricity is kilowatt per hour. This is a 100W light bulb. One degree of electricity can work normally for 10 hours.
2 / 0.4545 = 4.4, up to 4 such lamps can be connected.
According to w = u square divided by R, = 150 * 150 / 484 = 46.49w
Or (1) ∵ I = P / u
∴I=100W/220V≈0.4545A
∵R=U/I
∴R=220V/0.4545A=484Ω
（2）∵I=P/U
∴I=100W/220V≈0.4545A
（3）∵W=PT
∴T=W/T=1X10^3X1/100=10h
(4) ∵P=UI=220V×2A=440W
Ψ 440W / 100W = 4.4
A maximum of 4 lamps can be installed
(5) P = u ^ 2 / r = 150 ^ 2 / 484 = 46.5w
When a voltage of 6V is applied at both ends of a constant value resistor, the current passing through it is 0.4A; if a voltage of 3V is applied at both ends of the resistor, the current passing through it is ()
A. 0.2AB. 0.4AC. 0.8AD. 1.2A
According to I = ur, the resistance of the constant resistance is r = UI = 6v0.4a = 15 Ω, ∵ resistance is a property of the conductor itself, which has nothing to do with the voltage at both ends and the current passing through. When 3V voltage is applied at both ends of the constant resistance, the resistance of the constant resistance is still 12 Ω, and the current passing through is I '= u' = 3v15 Ω = 0.2A
How much power load can 3 * 16 + 1 * 10 cable carry? How much power load can 3 * 6 + 1 * 4 cable carry? How to calculate the current carrying capacity of multi strand copper wire
What is the difference between 3-phase and 2-phase current carriers?
16 square = π R & sup2; so the diameter is 5mm, right?
What does it mean to multiply 9 below 2.5 and subtract one from the top, then go 35 * 3.5 in pairs and form a group of - 0.5
Is the current of 3 * 16 & sup2; cable the same as that of 50 & sup2; single strand? Why?
A 3 * 16 + 1 * 10 cable can carry a three-phase symmetrical load of 48 kilowatts. A 3 * 6 + 1 * 4 cable can carry a three-phase symmetrical load of 18 kilowatts. If the load is asymmetrical, the zero line should be considered
A lamp of "220 V 100 W" and B lamp of "110 V 220 W" are connected in series in the circuit. The ratio of the actual power consumed by the two lamps is pa: Pb = (): ()
The ratio of the actual power consumed by two lamps is pa: Pb = (44): (5) resistance of a lamp: RA = UA & # 178 / / PA = 220 & # 178 / / 100 = 484 (Ω) resistance of B lamp: RB = UB & # 178 / / Pb = 110 & # 178 / / 220 = 55 (Ω)
RA=(220V)²/100w=484Ω,RB=(110V)²/220w=55Ω
The series current is equal, according to P = UI = I & # 178; R. The power is proportional to the resistance, PA: Pb = 484:55 = 44:5
When the voltage at both ends of a fast conductor is 3V, the current through the conductor is 0.3A,
When the voltage at both ends of a conductor is 3V, the current passing through the conductor is 0.3A. If the voltage at both ends of the conductor is increased by 6V, then the resistance of this problem and the current passing through are
fast
When 3 V is applied, the resistance R = u / I = 3 / 0.3 = 10 Ω,
When 6 V is applied, the resistance will not change due to the change of voltage, so it is still 10 Ω,
At this time, the current I = u / r = 6 / 10 = 0.6A
Resistance 10 ohm
Current 0.6A
How to calculate the resistance of wire? For example, 0.2 copper stranded wire, 2.5 square wire. What is the resistance?
1、R=pL/S
2. The resistivity of copper conductor P ≈ 0.017
3. R = resistivity of copper conductor P ≈ 0.017 * 100 / 2.5 = 0.68 ohm (100m)
According to the cable material, its resistivity is also different, known wire section size, length into the resistance calculation formula. This problem does not explain the material and length of the wire. Ask: did not say is 0.2 copper stranded wire???? 2. Twisted into 2.5 square wires. The length is 100 meters.
Connect the bulb marked "PZ 220 V 100 W" into the 220 V and 110 V circuit respectively, and calculate the resistance of the bulb and the corresponding actual power
The filament resistance is assumed to remain constant
From the Title Meaning: P = UI, P = UU / R,
100W=48400/R R=484
Access to 220 V: P = UU / R P = 48400 / 484 = 100 W
Access 110V: P = UU / R, P = 12100 / 484 = 25W
Filament resistance R = U2 / p;
Actual power: P real = (1 / 2U) / R can be concluded: when the actual voltage becomes half of the rated voltage, the actual power becomes one fourth of the original. It's very useful. Remember it. It's easy to take exams.
There is a wire with 4A current and 2V voltage at both ends
If the conductor is evenly elongated to 3 times of the original length and the same current is applied, the voltage at both ends is calculated
The resistance of wire before stretching is U / I = 2 / 4 = 0.5 Ω
The resistance of the elongated wire is 0.5 * 3 * 3 = 4.5 Ω (when the cross-sectional area of the wire is reduced to 1 / 3 of the original, the resistance is increased by 3 times, and the length is increased by 3 times, and the resistance is increased by 3 times on the original basis)
The voltage of the elongated wire is IR = 4 * 4.5 = 18V