What is the electrical appliance with rated current closest to 5A? A, refrigerator B, TV C, energy saving lamp D, electric pressure cooker

What is the electrical appliance with rated current closest to 5A? A, refrigerator B, TV C, energy saving lamp D, electric pressure cooker

According to the calculation formula of power, the power of household appliances with rated current closest to 5A is about P = u * I = 220 * 5 = 1100W, while the power of general electric pressure cooker is about 900W, so the electric pressure cooker should be the one with rated current closest to 5A

Calculate 1. (a + b) & # 178; - 2 (a + b) (a-b) + (a-b) & # 178; 2. (a-2b + 3C) (a + 2b-3c)
Use the formula
Simple calculation

1, the original formula = A & # 178; + 2Ab + B & # 178; - 2A & # 178; + 2B & # 178; + A & # 178; - 2Ab + B & # 178; = 4B & # 178; 2, the original formula = [a - (2b-3c)] [a + (2b-3c)] = A & # 178; - (2b-3c) &# 178; = A & # 178; - 4B & # 178; + 12bc-9c & # 178;

The normal working current of ordinary household incandescent lamp is about

I also answer according to physics, because the specifications of electric lamps are different, so it depends on the wattage. My formula is electric power P = UI (U is the working voltage, I is the working current)
As the voltage of household appliances is 220 V, so
It depends on the wattage of the light
30 / 220 of 30W = 0.13a
50 / 220 of 50W = 0.23a
100 / 220 of 100W = 0.45a

62.5*0.16+2640*0.0625+5.2*6.25+0.625*20

62.5*0.16+2640*0.0625+5.2*6.25+0.625*20
=6.25×1.6+26.4×6.25+5.2×6.25+6.25×2
=6.25×(1.6+26.4+5.2+2)
=6.25×35.2
=220

As shown in the figure, the power supply voltage remains unchanged, the resistance R1 = R2 = R3 = 10 ohm, if R2 and R3 are connected in parallel, the switch shall be closed --------, at this time, the indication of the ammeter is I1, if R1 and R2 are connected in series, the switches S1 and S2 shall -------- (closed / open), at this time, the indication of the ammeter is I2, then I1: I2=-----------
[the first layer is R1 and R2 in series, S2 connects the first layer and the second layer, the second layer is S1 and R3 from left to right, and the third layer is ammeter and power supply]

If R2 and R3 are connected in parallel, the switch should be closed ------ S1S2 ------. At this time, the indication of ammeter is I1. If R1 and R2 are connected in series, the switches S1 and S2 should be opened ------ (closed / opened). At this time, the indication of ammeter is I2, then I1: I2 = --- 4:1-------
The parallel resistance is 5 ohm, the current is I1 = u / 5 ohm, the total resistance is 20 ohm in series, and the current is I2 = u / 20 ohm
I1: I2 = u / 5 Ω: U / 20 Ω = 4:1

A simple algorithm of 4.8 × 0.25

=1.2×4×0.25
=1.2×1
=1.2

Detailed test process and points for attention in measuring resistance by voltammetry

The main problems should be paid attention to according to the order of measurement;
1. Selection of range of ammeter and voltmeter
2. Is the ammeter connected internally or externally
3. Is the sliding line rheostat continuous partial pressure type or current limiting type
4. The resistance value of the access circuit of the loop rheostat is the largest

8×3/5×12.5 75*0.5+15*0.5
Simple calculation

First calculate 8 * 12.5 = 100, then multiply by 3 / 5 = 60
The second is 75 + 15 = 90, then multiply by 0.5 = 45

How to convert American weather unit f into Chinese centigrade?
Currently:34 F
Condition:Few Clouds
Wind Chill:24 F
Dewpoint:23
Humidity:64 %
V isibility:10 mi
Wind:WSW/15 mph

The conversion formula F = (C × 9 / 5) + 32;
C = (f-32) × 5 / 9, where f is Fahrenheit temperature and C is centigrade temperature·
The conversion formula between centigrade temperature and Kelvin temperature (absolute temperature) is k = C + 273.16, where K -- Kelvin temperature, C -- centigrade temperature, Fahrenheit temperature and Rankine temperature scale

Ask for help! Physics in life
21. If you connect one of the festival lights in series
"Jump bubble", it will make the circuit constantly on, off, make the whole string of lights
A while on, a while off, a flash, very good-looking
The structure of the jumping bubble is shown in FIG. 12. At room temperature, the middle of the jumping bubble is vertical
The bimetallic sheet is next to the side inverted L-shaped lead, as shown in Figure 12
As shown in figure a, after being electrified, the current flows into the filament through the bimetallic sheet to make the jump start
Because the filament in the jumping bubble emits light, the bimetallic sheet bends when heated and leaves its original position, thus cutting off the circuit, as shown in Fig. 12b. Without current passing through, the bimetallic sheet cools and returns to its original shape, and then turns on the circuit. A series of such festival colored lights are connected to the power source (the power supply voltage is fixed), and suddenly several of them go out, The rest of the lights are still on and off, but they flash faster than before
(1) what are the reasons for the failure of those off lights? (2) what are the reasons for the accelerated flash rate of other lights?
Festival lights are all mixed together!!! This is common sense... I don't understand without pictures...
The second floor is not right. After the bulb is broken, it is equivalent to a resistor with larger resistance in series in the circuit. The resistance should be larger

1. Jumping bubbles are connected in series in the circuit
2. The lamps are connected in parallel, or a group of lamps are connected in series first and then connected in parallel in the circuit, otherwise one of the lamps will not work when it breaks down
3. If one of the lights in a group is damaged, all the lights will not work
4. If one or more groups are not on, the parallel resistance of the circuit will increase and the load will decrease
5. As the resistance increases, the discharge current decreases
6. When the current is reduced, the discharge energy and temperature will be reduced
7. When the temperature is low, the deformation of the sheet metal is small and the recovery is fast
8. The flashing frequency will increase