When an ordinary household incandescent lamp lights normally, the current through it is about () A. 2 × 101 a B. 2 × 102 a C. 2 × 10-1 a D. 2 × 10-2 a

When an ordinary household incandescent lamp lights normally, the current through it is about () A. 2 × 101 a B. 2 × 102 a C. 2 × 10-1 a D. 2 × 10-2 a

Ordinary household lighting incandescent bulb, generally "220 V, 40 W", normal work, voltage is 220 V, power is 40 W, from the formula P = UI I = Pu = 40 W, 220 V ≈ 0.2A = 2 × 10-1a

Connect the light marked with 220 V 25 W into the home circuit,
The resistance of the lamp is____ Oh
Solving process, thank you!

Because P = u ^ 2 / R, r = u ^ 2 / P = 220 * 220 / 25 = 1936 ohm

There are 11 220 V, 60 W electric lamps connected to a family circuit. The total resistance of the transmission line is 2 Ω, which is all positive
When 11 "220 V, 60 W" electric lamps are connected to a family circuit, and the total resistance of the transmission line is 2 Ω, all of them light normally
(1) What is the total voltage input to the circuit?
(2) How much heat is generated on the transmission line in one hour?
(3) What size of fuse should be used in this circuit?
I want to calculate the process

The normal lighting of the lamp means that the lamp reaches the rated power, and the voltage at both ends of the lamp is the rated voltage of 220 v
According to the formula, the power calculation formula P = UI = I2R = U2 / R (the three formulas are equivalent, the first is the basic formula, and the second is the square, you can understand it, and you don't bother to use the formula editor) -
I=P/U；R=U2/P;
have to
Rated current of lamp ie = 60W / 220V = 0.27A
When the filament is normally lit, the resistance re = 220 V * 220 V / 60 W = 806.67 Ω
11 lights on at the same time, total current I = 11 * ie = 3A
Then the voltage drop on the transmission line is u = IR = 2 * 3 = 6V
The total input voltage is U + UE = 226v (6V is divided by the resistor, and the remaining 220V is added to both ends of 11 parallel lamps, so the lamp lights normally)
Conductor heating power PQ = I2 * r = 3A * 3A * 2 Ω = 27W
One hour calorific value of conductor q = PQ * t = 0.027kw. H
Finally, the current has reached 3A during normal operation. According to the habit, 3A * 1.3 = 4A, you should choose 4A fuse (multiply 1.3 to leave a certain margin)

A 220 V, 25 W incandescent lamp is connected to the 220 V line. How much is the current? How much is the resistance? If you use 4H every day, how much is the monthly electricity?

According to the formula P = UI, the current I = P / u = 25 / 220 = 0.1136a
According to the formula u = IR, the resistance R = u / I = 220 / 0.1136 = 1936.62 Ω
Four hours a day
W = 0.025 * 4 = 0.1 kWh
That's 0.1 kilowatt hour a day