A calculation problem of electric power in junior high school physics The "220 V 15 W" and "220 V 60 W" bulbs are connected in series with the 220 V power supply to form a closed loop (with the filament resistance unchanged), what is the total power of the two lamps?

The resistance R1 = U2 / P1, R2 = U2 / P2 of the two lamps can be obtained respectively
Because the two lamps are connected in series, the total resistance is r = R1 + R2 = U2 / P1 + U2 / P2
The total power of two lamps is ptotal = U2 / (R1 + R2) = U2 / (U2 / P1 + U2 / P2)
=P1P2/(P1+P2)=15WX60W/(15W+60W)=12W

When an electrical appliance is connected to a power supply, its power is 100 watts. If it is connected to the same power supply with a resistor R string, the electrical power consumed by the resistor R is 9 watts. (r > R). To calculate the power of the electrical appliance at this time? (assuming that the voltage of the power supply remains unchanged) it is required to write out the solution process,

Power supply u = √ PR = 10 √ R
The electric power of resistance R is 9 = I ^ 2R
I = (10 √ r use) / (r use + R)
Substitute, simplify
9R uses ^ 2-82r uses R + 9R ^ 2 = 0
That is, (9R with - R) (R with - 9R) = 0
So r = 9R or 1 / 9R (rounding off)
So p = 9 * 9 = 81 watts

Application and calculation of electric power
According to China's regulations, color TV sets with standby power higher than 9W cannot be sold in the market. Xiaoming wants to know whether the color TV sets at home meet the regulations. He connects a 100W light bulb into the circuit alone, observes that the energy meter turns five times in one minute, and then makes the color TV set connect into the circuit alone and in standby mode. The energy meter turns two times in five minutes to judge the standby power of the color TV set, How much energy does Xiaoming's house consume every month (30) days due to color TV standby?

Electric energy consumed by electric lamp in one minute W1 = p1t1 = 100 * 60 = 6000l
The energy consumed by each revolution of the meter is 6000 / 5 = 1200J
Power consumption of color TV standby W2 = 1200 * 2 = 2400j
Color TV standby power P2 = W2 / T2 = 240 / (5 * 60) = 8W
Total electric energy consumed in one month w = Pt = 8 * 30 * 24 * 3600 = 20736000j = 57.6kwh

How to calculate the electric heating power is I & # r?

Pure resistance can be p = I ^ 2R, P = (u ^ 2) / R
Non pure resistance P = UI

The "number of units" of an air conditioner is a rough identification of the input power of the air conditioner. Because the refrigeration efficiency of different brands of air conditioners with the same number of units is quite different, for comparison, the efficiency of air conditioners is often expressed as "cooling capacity". When selecting an air conditioner, besides the cooling capacity and heating capacity (for cold and hot air conditioners), the "energy efficiency ratio" should also be considered. The energy efficiency ratio of air conditioner = cooling capacity (W) input At present, the energy efficiency ratio of air conditioners in the market varies greatly, generally between 2 and 3, and the highest can reach about 3.5. The air conditioners with high energy efficiency ratio should be selected as far as possible. The cooling capacity and heating capacity required for a room can be calculated according to the following formula: cooling capacity Q = s (room area) × 140w-180w; heating capacity Q = s (room area) × 180W- 240W (this formula is applicable to the room with a floor height of 2.5m, if the floor height is too high, the power should be increased appropriately). The following table is a part of the operation manual of a certain type of air conditioner Xiao Mingjia is going to buy. The rated cooling capacity / W & nbsp; 5000 indoor unit noise DB & nbsp; (a) ≤ 45 & nbsp; rated heat / W 5300 & nbsp; outdoor unit noise DB & nbsp; (A) External dimension of indoor unit ≤ 57 power supply 220 V 50 Hz (length × width × height) / mm 195 × 1024 × 320 working voltage / V & nbsp; external dimension of 220 outdoor unit (length × width × height) / mm & nbsp; 253 × 800 × 646 & nbsp; rated cooling input current / A & nbsp; & nbsp; 8.6 rated cooling input power / W & nbsp; & nbsp; 1850 rated heating input current / A & nbsp; 8.7 &Heat input power of heat pump / W & nbsp; & nbsp; 1880 please answer the following questions by calculation: A. what is the energy efficiency ratio of this air conditioner. B. the living room area of Xiaoming's new building is about 30m2, and the floor height is 2.7m. He wants to install the air conditioner shown in the table above in the living room. Please explain whether this choice is reasonable by calculation. C. If Xiaoming uses this air conditioner for refrigeration The compressor of the air conditioner actually works for 6h every day. How much is the electricity charge for only air conditioner in a month? (local electricity price is 0.5 yuan / kWh)

a. The energy efficiency ratio of this air conditioner is: energy efficiency ratio = refrigerating capacity (W) input power (W) = 5000w1850w ≈ 2.7; B. the required refrigerating capacity is: Q cooling = s × 140w-180w = 30 × 140w-180w = 4020w; the required heating capacity is: Q heating = s × 180w-240w = 30 × 180w-240w = 5160w

Formula of electric work, electric power and circuit
My summary is not complete. Thank you

Electric power: P = UI = u ^ 2 / r = I ^ 2 * r
Electric work: w = Pt = uit = u ^ 2 / R * t = I ^ 2 * r * t
There are two kinds of circuits, series and parallel. Your teacher must have a summary
I think these formulas have changed a lot. Just remember the key ones, three
W=Pt,P=UI,U=IR
Formula is only an auxiliary function, the key is to understand the circuit itself

A calculation problem of electric power in junior high school physics The "220 V 15 W" and "220 V 60 W" bulbs are connected in series with the 220 V power supply to form a closed loop (with the filament resistance unchanged), what is the total power of the two lamps?