When a resistor is applied with 1V voltage at both ends, the current passing through is 200mA, and its resistance value is______ Ω; if 2V voltage is applied at both ends of it, its resistance is______ Ω, the current passing through is______ A．

When a resistor is applied with 1V voltage at both ends, the current passing through is 200mA, and its resistance value is______ Ω; if 2V voltage is applied at both ends of it, its resistance is______ Ω, the current passing through is______ A．

Resistance R = UI = 1v0.2a = 5 Ω, when 2V voltage is applied at its two ends, the resistance will not change, and the current I1 = u1r = 2v5 Ω = 0.4A. So the answer is: 5; 5; 0.4
2kw.4kw.6kw what size of air leakage switch is used. All the appliances use 220V
Single phase 220 V electrical apparatus 1kW is about 4.5a, 2; 4; 6kW, the rated current of electrical apparatus is about 9A; 18a; 27a, so use 10A; 20A; 32A air leakage switch respectively; use 2.5mm2; 4mm2; 6mm2 for wires respectively
In the series circuit, the voltage is 220 V, the lamp is 220 V and 100 W, the rheostat is marked with 121 Ω 1 A, and the actual power variation range of lamp L is calculated
Maximum power, sliding resistance 0, light power 100W
Minimum power, maximum sliding resistance 121 ohm, bulb resistance = 220 * 220 / 100 = 484 ohm
So the partial voltage of the bulb is 4 / 5 * 220 = 176v
Power = 176 * 176 / 484 = 64W
R lamp = 220 × 220 / 100 = 484 Ω
I / 220 = 5 / 100
When r = 0 Ω, I total = I amount = 5 / 11a
When the voltage at one end is 3V, the current passing through is 200mA. What is the resistance of the conductor? First increase the voltage to 6V. If the ammeter has two ranges of 0 ~ 6a and 3a, which range should be selected to measure the current in the conductor?
Resistance: r = u / I = 3 / 0.2 = 6 (Ω)
When the voltage reaches 6V, the current: I = u / r = 6 / 6 = 1 (a)
So choose a table of 0-3A
The total power is 60kW, the voltage is 220V, and the incoming line is supplied by three-phase four wire power supply
Rated current of each phase:
I＝P/1.732/U/COSφ＝60/1.732/0.38/0.8≈114(A)
The leakage switch ≥ 120a shall be selected
The coil resistance of 220 V 66W electric fan is 20 ohm. After the voltage of 220 V is applied, the power consumed by the electric fan is converted into the power of mechanical energy and heating power;
U total = 220 V, P total = 66 W, r = 20 Ω. When connected with the voltage of 220 V, the electric fan can work normally, so the electric power it consumes is p total = 66 W. by P total = u total * I total, the current passing through the coil of the electric fan is I total = P total / u total = 66 / 220 = 0.3 A. then the heating power of the coil resistance is
Current in normal operation: I = P / u = 66 / 220 = 0.3A
Heating power: P1 = 0.3 * 0.3 * 20 = 1.8W
The power consumption of the electric fan is 66W
Power converted into mechanical energy: P2 = 66w-1.8w = 64.2w
When a voltage of 3V is applied at both ends of a conductor, the current passing through it is 200mA, and the resistance of the conductor is () a
When 3V voltage is applied at both ends of the conductor, the current passing through it is 200mA and the resistance of the conductor is () A. when 9V voltage is applied at both ends of the conductor, the resistance of the conductor is () Ω and the current passing through the conductor is () A. when the voltage at both ends of the conductor is zero, the resistance of the conductor is () Ω
15, Euro
0.6A
15 euro
3V / 200mA = 3V / 0.2A = 15 Ω; 9V / 15 Ω = 0.6A; 15 Ω
How to prevent the electrical appliances from burning out when the 220 V is wrongly connected to the 380 V power supply? Is there any switch that can trip when the voltage rises?
Yes, Yimin electrical protector can not only protect electrical appliances, but also protect itself from high voltage burning
The coil resistance of "220 V 66 W" electric fan is 20 Ω. (1) after connecting 220 V voltage, the power consumed by the electric fan is converted into mechanical energy power and heating power; (2) if the fan blade is stuck and cannot rotate after connecting the power, the power consumed by the motor and heating power are calculated
(1) From P = UI, we can get the following current: I = Pu = 66220 = 0.3A; (2) coil resistance heating power: PQ = I2R = 1.8W; mechanical power: pmechanical = p-pq = 64.2w; (3) when the blade does not rotate, as pure resistance, according to Ohm's law, I = ur = 11a; P = UI = I2R = 11 × 11 × 20 = 2420w
The voltage at both ends of a circuit is 3V, and the current does 20j work in 5S. The electric power consumed by this circuit is (), and the resistance of this circuit is ()?
The voltage at both ends of the circuit is 3V, and the current does 20j work within 5S. The electric power consumed by this circuit is (4W), and the resistance of this circuit is (2.25 Ω)
P=W/t=20J/5s=4w
If P = UI = u & sup2 / R, r = u & sup2 / P = (3V) & sup2 / 4W = 2.25 Ω
Electric power 4W, resistance 2.25 ohm
Electrical rate P = 20j / 5S = 4W
Resistance R = u ^ 2 / P = 9 / 4 = 2.25 Ω