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Know a thing power is 1500W, voltage is 220V, how to find its resistance? Fine formula!
P=UI=U*(U/R)=U^2/R
=>R=U^2/P
R = 220 ^ 2 / 1500 = 32.27 ohm
P=U^2/R
P = UI = u times u / r = u squared / R, r = u squared / P = 220 squared / 1500 Ω = 32.26 Ω
What item, if it can only generate heat, can be divided by R with P = u
The relationship between circuit resistance and bulb brightness
For a circuit with only one small bulb, the smaller the bulb is, the brighter the slider moves
When connected in series, the sliding rheostat moves to the direction with small resistance, and the brighter the bulb is
In parallel, the sliding rheostat moves in the direction of high resistance, and the brighter the bulb is
How many kilowatts of power can a 5 * 10 square cable carry
According to the calculation of 2A current per kilowatt, the rated power of 25kW can be loaded. But this is the continuous working current at the rated time. The starting current is 6-8 times of the rated current. Under the premise of ensuring safety, there must be margin, and the power below 20kW should be selected
How much work does a 220 V 40 W incandescent lamp do per minute under rated voltage? What is the focal resistance? What is the power of this lamp connected to 220 V voltage
Work per minute under rated voltage w = Pt = 40wx60s = 2400j
Resistance: r = u ^ 2 / P = 220 * 220 / 40 = 1210 Ω
Connect the lamp to the voltage of 220 V and the power is 40 W
Under the rated voltage, the power of the lamp is equal to its rated power, so the work done is 40 joules per second, and 2400 joules per minute. The resistance can be calculated according to P = u ^ 2 / R.
W=Pt=2400
R=U2/t=1210
P=40
When the voltage of 4V is applied to both ends of a constant resistance, the current passing through the resistance is 200mA. What is the resistance of the appliance?
R=U/I=4V/0.2A=20Ω
R=U/I=4V/0.2A=20Ω
20 euro
4/0.2=20Ω
The resistance is equal to the voltage at both ends of the resistance (V), divided by 0.2A
2kW. 4kw. What size of air leakage switch is used. All the appliances use 220V
The current of 220 V 2 kW is about 9 A, and the leakage protection switch is 16 a. the current of 220 V 4 kW is about 18 A, and the leakage protection switch is 32 a
A 220 V 60 W electric lamp in a distant construction site is supplied by a 220 V power supply. Due to the resistance of the conductor, the actual power consumed by the bulb is 55 W,
What is the resistance consumed by the wire?
If the transformer is not involved
R = u ^ 2 / P = 806.67
I = under radical (P real / R) = 0.2611
Total resistance = u / I = 842.54
R wire = total resistance - r = 35.87
When the voltage at both ends of a resistor increases from 3V to 3.8V, the current through the resistor increases by 0.2A. What is the resistance?
I ′ & nbsp; − I = u ′ R − ur = 3.8vr − 3VR = 0.2A, so: r = 4 Ω answer: the resistance is 4 Ω
How big is the leakage switch for household 220 V
First of all, we need to figure out a problem. Do you want an air switch or an air switch with leakage protection? The general function of an air switch is to trip off when the current is too high and protect the circuit, while the leakage switch is to cut off the circuit when there is leakage in the circuit to protect the safety of equipment and people
Two 220 v.40 w electric lamps are connected in parallel to a power supply with a voltage of 200 v. if the line resistance R is 2 Ω, what is the voltage and power P of the two lamps?
Resistance of each lamp r = u ＾ 2 / P = 1210 Ω
Parallel resistance R = 605 Ω
Total resistance R = 607 Ω
Current I = 200V / 607 Ω = 0.329a
The voltage of the two lamps is 0.329a * 605 Ω = 199.3v
The total power of the two lamps is 199.3v * 0.329a = 65.57w
The power of each lamp is 65.57w/2 = 32.8w
Lamp resistance R1 = 220 ^ 2 / 40 = 1210 Ω
Resistance of two lamps: r = 1210 / 2 = 605 Ω
Voltage of two lamps = 200 * 605 / (605 + 2) = 199.3v
Power of two lamps P = 199.3 ^ 2 / 605
^2 is the square.
RELATED INFORMATIONS
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