The voltage at both ends of a conductor is 12 volts, and the current intensity passing through it is 0.9 amperes, so that the current intensity passing through it is 0.6 amperes, and the voltage applied at both ends of it should be more than 10 amperes

The voltage at both ends of a conductor is 12 volts, and the current intensity passing through it is 0.9 amperes, so that the current intensity passing through it is 0.6 amperes, and the voltage applied at both ends of it should be more than 10 amperes

The resistance does not change
R=U1/I1=U2/I2=12V/0.9A=U2/0.6A
u2=8V
12/0.9=X/0.6
X=8 V
There are two bulbs in series in the circuit, the current in the circuit is 0.1 A, the voltage at both ends of bulb L1 is 3 V, and the voltage at both ends of bulb L2 is 9 v. find out: (1) the resistance of the two bulbs; (2) the total resistance in the circuit; (3) the power supply voltage
(1) According to: I = ur, bulb resistance: R1 = u1i = 3v0.1a = 30 Ω, R2 = u2i = 9v0.1a = 90 Ω; (2) total resistance of circuit: r = R1 + R2 = 30 Ω + 90 Ω = 120 Ω; (3) power supply voltage: u = U1 + U2 = 3V + 9V = 12V; answer: (1) resistance of two bulbs is 30 Ω and 90 Ω respectively. (2) total resistance of circuit is 120 Ω. (3) power supply voltage is 12V
An ammeter with internal resistance R = 1000 ohm and full bias current I = 0.1A should be converted into a 3V voltmeter in series with a resistor of R1. To convert it into an ammeter of 0.6A, a resistor of R2 should be connected in parallel
1 mi = 0.1 A or unreasonable
IR+IR1=3
R1 = (3-Ir) / I = [3-0.1 * 10 ^ (- 3) * 1000] / 10 ^ (- 4) = 29900 Ω
IR=I'R2
R2 = IR / I '= 0.1 * 10 ^ (- 3) 1000 / 0.6 = 1 / 6 = 1.67 Ω
Refit into a voltmeter to use partial voltage, so the current is equal
3/(R+1000)=0.1
Refit into ammeter to use shunt, so the voltage is equal
0.1*1000=0.5*R
R can be solved respectively
I=0.1A
IR+IR1=3
R1 = (3-Ir) / I = [3-0.1 * 10 ^ (- 3) * 1000] / 10 ^ (- 4) = 29900 Ω
IR=I'R2
R2 = IR / I '= 0.1 * 10 ^ (- 3) 1000 / 0.6 = 1 / 6 = 1.67 Ω
The power of three-phase 380V motor is 1.5KW. We want to change the star connection method to the triangle connection method, and the power supply is three-phase 220V
This motor is a motor with braking function. It can run through the three-phase 220 V power output by the frequency converter. After changing the triangle connection method, it can't be started. It can be started through the three-phase 380 V power supply. I don't know why. Please advise
1. Star connection method: I line = I phase = 380 / (1.732 * z), motor power: P star = 1.732 * 0.38 * I line = 1.732 * 0.38 * 0.38 / (1.732 * z) = 1.732 * 0.22 * 0.38/z; 2. Delta connection method: I line = 1.732 * I phase = 1.732 * (220 / z), motor power: P △ = 1.732 * 0.22 * I line = 1.732 * 0.22 * 0.38/z; 2
200 meters of 6 square aluminum wire, how much power can it bring?
If calculated according to the load capacity of the wire, a square millimeter of aluminum wire is 4 a
Under the condition of - 5% ～ 10%, the low-voltage apparatus can operate normally
That is to say, the voltage drop on the wire should not be greater than 5% of the supply voltage, and it is 11 V at 220 v
At 20 ℃
The resistivity of aluminum is 0.028, and the resistance of 200 meter conductor is 5.6 ohm,
11 / 5.6 = 1.96a, 6 pieces are equivalent to 1.96 * 6 = 11.786a in parallel
220*11.786=2.592KW
Test method for volume resistivity and surface resistivity of solid insulating materials (GB 1410-78)
First, the surface resistance and volume resistance are measured with a high resistance meter, and the appropriate measurement voltage is selected
Surface resistivity = 81.6 times the surface resistance, in ohm
Volume resistivity = 2.1237 times volume resistance ＜ Ω ＞ and then divided by thickness ＜ mm, in Ω & # 8226; M
How many joules is 1.5 kwh?
How many joules does 1.5 kwh equal? How many kwh does 9 x 10's sixth power Joule equal?
1 kwh = 3.6 * 10 ^ 6 joules, so 1.5 kwh = 3.6 * 10 ^ 6 * 1.5 = 5.4 * 10 ^ 6 (^ is the power) (9 * 10 ^ 6) / (3.6 * 10 ^ 6) = 2.5
380V asynchronous motor changes star connection into triangle connection, driven by 220 V inverter, what is the difference between output power and speed?
Of course, the inverter power is a level of motor
If the three-phase 380V asynchronous motor was originally star connected, the phase voltage of each phase winding is 220V, and the I line of the motor = I phase, motor power P = 3 * I phase * u phase * cos ф * η = √ 3 * I phase * u line * cos ф * η = 660 * I phase * cos ф * η
How much power can 4 square aluminum wire bear
It is related to the actual heat dissipation level
Experience formula: 5 under 10, 2 over 100 (5 a per square meter for 10 square meters)
So 4 square safe current carrying capacity is 20A
If 220 V single phase, 4800 w
380V three phase, 10kW (motor load)
For reference
Connect two resistance wires a and B of the same thickness made of the same material to two points a and B in the circuit as shown in the figure: when a and B are indirect resistance wires a, the reading of ammeter is 0.2A; when a and B are indirect resistance wires B, the reading of ammeter is 0.3A______ The resistance is high______ The resistance wire is long
It can be seen from the circuit diagram that the circuit is a simple circuit of resistance wire. The voltage in the circuit measured by the ammeter remains unchanged, that is, the voltage at both ends of resistance wire a and B remains unchanged. According to I = ur, if the current through a is less than that through B, the resistance of resistance wire a is greater than that of resistance wire B. the resistance depends on the material, length and cross-sectional area of the conductor When other conditions remain unchanged, the longer the conductor, the greater the resistance of the conductor. When the material and thickness are the same, the greater the resistance of a, the longer the resistance wire of A