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The rated voltage is 220 V, the power is 40 W, and the resistance is calculated
P=U*U/R
R = u * U / P = 220 * 220 / 40 = 1210 ohm
It should be noted that the resistance of the bulb is closely related to the temperature, that is, the applied voltage. 1210 is the resistance value under the voltage of 220 v. if the voltage is low, the resistance value will also be significantly reduced
The resistance of the filament of a bulb is 4 Ω. When the voltage at both ends is 3V, the current through the bulb is calculated
According to Ohm's law, the current through the bulb is I = ur = 3v4 Ω = 0.75a. A: the current through the bulb is 0.75a
The rated power of an electric furnace is 2.2kW, and the rated voltage is 220V. When it is connected to a circuit, the measured current through it is 8a, and the resistance and actual work of the furnace are calculated
First calculate the resistance of the electric furnace, r = u ^ 2 / P = 48400 / 2200 = 22 Ω
The actual power is p = UI = I ^ 2 * r = 8 ^ 2 * 22 = 64 * 22 = 1408w = 1.4kw
The resistance of electric furnace r = 220 & # 178; / (2,2 × 1000) = 22 Ω;
The actual power after power on is p = 8 & # 178; × 22 = 1408w, 1408w = 1.408kw.
P = 2200W, u = 220V, rated current I = P / u = 10a, r = u / I = 22 Ω, at present, I = 8a, P = I ^ 2 * r = 8 * 8 * 22 = 1408w
According to r = u square / P, r = 48400 / 2200W = 22 Ω; actual voltage U = IR = 8A * 22 Ω = 176v, then the actual power is p = IU = 8A * 176v = 1408w = 1.408kw. Answer: the resistance of electric furnace is 22 Ω, at this time the actual power is 1.408kw
Electric furnace can be regarded as pure resistance, but in practice there will be inductive reactance, so Ohm's law ～
Given the rated power and voltage, the rated resistance can be calculated
According to P = I ^ 2R (obtained from Ohm's law of electric power formula), the resistance of electric furnace r = 22 Ω can be obtained
Know the resistance of electric furnace, know the actual current, and then use p = I ^ 2R to calculate the actual power P = 1.4kw
1. Calculate the rated current and filament resistance of a 220 V, 40 W bulb. 2. Connect an induction cooker with a resistance of 20 to a 220 V power supply, and its electric power?
The answer formula of the first floor is correct, but the calculation result of the first part of the first question is seriously wrong. The correct result should be 40 / 220 = 0.1818a
The first floor is right
1. I = P / u = 5.5A r = u * U / P = 1210 Ω P = u * U / r = 2420w
1. R = u * U / P = 220 * 220 / 40 = 1210 Ω
I = P / u = 40 / 220 = 0.1818a
2。 P = u ^ 2 / r = 220 * 220 / 20 = 2420w
In the experiment of "measuring the resistance of small light bulb", the power supply voltage is 3V, the rated voltage of small light bulb is 2.5V, and the resistance is about 10 Ω. The sliding rheostat is marked with "20 Ω & nbsp; 1A". "1) please complete the circuit connection with a stroke line instead of a wire in the figure. (2) before the switch is closed, the slide of the sliding rheostat should move to the right position______ (fill in the "left" or "right") end. (3) after closing the switch, it is found that the small bulb does not light up, the ammeter has indication, and the voltmeter has no indication. When the slide P is moved, the indication of the ammeter changes with & nbsp;, then the cause of this fault is______ (4) after troubleshooting, adjust the slide p to make the small bulb light normally. If the ammeter shows the number as shown in the figure, the resistance of the small bulb light normally is______ Ω (accurate to 0.1)
(1) The rated voltage of the bulb is 2.5V, the range of the voltmeter is 0 ~ 3V, and the voltmeter is connected in parallel with the bulb. The circuit current is about: I = ur = 2.5v10 Ω = 0.25A < 0.6A, so the range of the ammeter is 0 ~ 0.6A, and the ammeter is connected in series with the bulb
The rated power of an electric furnace is 2.2kW, and the rated voltage is 220V. When it is connected to a circuit, the measured current through it is 8a, and the resistance and actual power of the furnace are calculated
Explain the formula,
The resistance of this furnace is: u ^ 2 / w = 220 * 220 / 2200 = 22 (Ω)
The actual power of this electric furnace is: w = I ^ 2 * r = 8 * 8 * 22 = 1408 (W) = 1.408 (kw)
An electric iron is marked with the words of 220V 40W. When it works normally, the heat generated in 1min is?
When the electric iron works normally, the power P = 40W
t=1min=60s
If all the electric energy consumed by the electric iron is converted into heat, then
Q=W=Pt=40×60=2400（J）
A: slightly
40W generates 40 joules of energy per second, and 40 * 60 = 2400 joules in 60 seconds
3. Two series connected dry batteries are used to supply power to the small bulb. The voltage is 3V, and the current passing through the small bulb is 0.3A. At this time, the resistance of the small bulb is?
At this time, the resistance of the small bulb r = u / I = 3V / 0.3A = 10 Ω
20 ohm
How much heat does a "220V 1000W" electric furnace produce per minute when it is connected to a 220V circuit? What is the resistance of the furnace wire when the electric furnace works normally?
Like the title,
W = Pt = 1000W × 60s. We can calculate the resistance by substituting the formula P = U2 / R
The specification of a household electric iron is "220V & nbsp; 1100W". Under the rated voltage, what is the working current of the electric iron? How much heat is produced by working for one minute?
Under the rated voltage, the electric power of the electric iron: P = P value = 1100W, ∵ P = UI, ∵ I = Pu = 1100w220v = 5A; ∵ the work done by the electric iron is used to generate heat, ∵ q = w = Pt = 1100W × 60s = 6.6 × 104j. A: under the rated voltage, the current of the electric iron is 5a, and the heat generated by working for 1min is 6.6 × 104j
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