A bulb is marked with 220 V 100 W. B bulb is marked with 220 V 40 W. A and B are connected in a 220 V circuit in series. What is the voltage ratio between the two ends of the lamp Such as the title

A bulb is marked with 220 V 100 W. B bulb is marked with 220 V 40 W. A and B are connected in a 220 V circuit in series. What is the voltage ratio between the two ends of the lamp Such as the title

Series in the circuit, the current of two lamps is equal, the ratio of voltage = the ratio of resistance, the resistance of a lamp is
220 * 220 / 100 = 484, B lamp resistance is 220 * 220 / 40 = 1210, resistance ratio = 2:5
Power 2200W, voltage 380V or power 2200W, voltage 220V, electricity consumption will be the same?
If the equipment is rated power PN and rated voltage UN, the theoretical power consumption per unit time is the same. If there are two devices with resistance power of 2200W and voltage of 380V, one is connected to 380V and the other is connected to 220V, the power consumption of the electrical equipment connected to 220V is less than 380V
The voltmeter measures that the voltage at both ends of a small bulb is 3.8V. Use the ammeter to measure that the current passing through the small bulb is 0.3A. Calculate the resistance of the small bulb
R = u △ I = 3.8V △ 0.3A = 12.6Ω
67 ohm.
3.8 divided by 0.3 is 12.67
[urgent] "220 V 100 W" and "200 V 40 W" bulbs are connected in series and then connected to the circuit of 220 v. which bulb is brighter
What is the voltage at each end? What is the actual power on each bulb?
The resistance of a bulb R1 = u ^ 2 / P = 220 ^ 2 / 100 = 484 Ω
The resistance of the other bulb R2 = u ^ 2 / P = 40000 / 40 = 1000 Ω
After series connection, the voltage of the first bulb is 220 * 484 / (1000 + 484) = 71.75v
The voltage of the other bulb is 220-71.75 = 148.25v
The actual power of the first bulb is p = u ^ 2 / R1 = 71.75 ^ 2 / 484 = 10.64w
The actual power of the other bulb is p = u ^ 2 / r2 = 148.25 ^ 2 / 1000 = 21.98w
The second one is on
The series current is the same, the power is equal to I * I * r, the resistance of 40W is large, so it is on
The calculated resistance of 100W bulb is r = 220 V * 220 V / 100W = 484 Ω;
The calculated resistance of 40 W bulb is r = 220 V * 220 V / 40 W = 1210 Ω;
After series connection, the current I = 220 V / (484 + 1210) is about 0.13 a
The series current is equal
Voltage at both ends of 100W bulb = 0.13 * 484 = 63v; power = UI = 63 * 0.13 = 8.19w
Voltage at both ends of 40W bulb = 0.13 * 1210 = 15
The calculated resistance of 100W bulb is r = 220 V * 220 V / 100W = 484 Ω;
The calculated resistance of 40 W bulb is r = 220 V * 220 V / 40 W = 1210 Ω;
After series connection, the current I = 220 V / (484 + 1210) is about 0.13 a
The series current is equal
Voltage at both ends of 100W bulb = 0.13 * 484 = 63v; power = UI = 63 * 0.13 = 8.19w
40W bulb voltage at both ends = 0.13 * 1210 = 157v; power = UI = 157 * 0.13 = 20.4w
In fact, it's easy to calculate. You can do it yourself. Ask: Thank you! Hee hee
Power calculation method of 220 V and 380 V
There is only one power calculation method, P = V * A. for three-phase circuit, each phase is calculated separately and then added. For balanced load such as motor, one phase current p = 380 * a * 3 (triangle) or 220 * a * 3 (Y shape) is measured
Two resistors with resistance values of 10 ohm and 20 ohm are connected in series in the circuit with supply voltage of 9V. What is the current and the voltage at both ends of the resistor
The resistance is R1 and R2
I=U/(R1+R2)
=9/30
=0.3A
U1=IR1
=0.3*10
=3V
U2=IR2
=0.3*20
=6V
Current = u / R total = 9V / 30ohm = 0.3amp
U1=I*R1=3V
U2=I*R2=6V
Two "220V 100W" bulbs are connected in series to the circuit, and the voltage at both ends of each bulb is at the same level when the bulb is emitting______ 5. The current through the filament is_______ A.
Please explain why,
110V,0.23A
Because the specifications of the two bulbs are the same, the resistance is the same, r = u ^ 2 / P = 220 ^ / 100 Ω = 484 Ω
The resistance is equal because R1 / r2 = U1 / U2
So the 220 V voltage is divided equally
So the voltage is 110V
According to I = u / r = 110V / 484 Ω = 0.23a
Voltage 220 V power 10 kW power factor 0.37 how to calculate the current?
∵ power = voltage × current × power factor
Current = power △ voltage △ power factor = 10000 △ 220 △ 0.37 ≈ 122.85 (a)
=(10*1000/220)/0.37=122.85012285012285012285012285012
∵ power = voltage × current × power factor
Current = power △ voltage △ power factor = 10000 △ 220 △ 0.37 ≈ 122.85 (a)
If the voltage at both ends of a resistor is 4V and the current through the resistor is 400mA, how many ohm is the resistance?
In fact, I want to ask if the current here must be changed to a to calculate the resistance
The current must be changed to a to calculate the resistance
I=400mA=0.4A
R = u / I = 4 / 0.4 = 10 Ω
10 ohm
It must be
His unit of calculation is: voltage V, resistance Ω, current a
4V divided by 400mA is 0.01 kiloohm, which is 10 ohm.
For the bulb marked with 220 V 100 W, if the bulb works normally, what is the voltage at both ends? The current intensity through the filament? The filament resistance of the bulb
Its voltage at both ends u = rated voltage = 220 V
Current intensity I = P / u = 100W / 220V = (5 / 11) a = 0.45a
Filament resistance R = u / I = 220 / (5 / 11) = 484 Ω
Voltage at both ends u = 220 V
Power P = 100W
Current I = P / u = 0.45 a
Resistance R = u / I = 484 Ω
Effective voltage 220 V, maximum voltage 311 V (AC)