First, the weather in Weifang

9.(20 10. Laiwu) At present, there are two circuit elements, namely constant resistance R 1 and thermistor R2 (resistance changes with temperature and is very sensitive to temperature). The current and voltage at both ends of them are shown in Figure A. Connect these two circuit elements into a circuit as shown in Figure B, and the power supply voltage is 6V. When the ammeter reads 70 mA, find: (6544) (2) the current through thermistor R2; (3)R2' s resistance; (4) The total power of the power supply.

Solution: (1) (Take any point in the image and make a calculation example:)

Check the image correctly. When the voltage across R 1 is U 1=2V, the current I1= 20ma = 0.02a..

(2) the current through R 1

The water flowing through R2 is

(3) Check the I-U diagram of R2. When the current through R2 is 10 mA, U2=4V… 1.

R2' s resistance at this time.

(4) The total current in the circuit is I = 70 Ma = 0.07a.

The total power of the power supply is

5.(20 10. Jingmen) In an extracurricular activity, Xiaohua made relevant observation and research on the induction cooker at home, and recorded the relevant data of the induction cooker and her electric energy meter, as shown in the right table. (the specific heat capacity of water c= 4.2× 103J/(kg? c)。 Please solve the problem according to the relevant information obtained by Xiaohua during the activity:

(1) heat absorbed by water;

(2) the actual power of the induction cooker;

(3) The efficiency of the induction cooker for boiling water.

( 1)

(2) 2. Electric energy consumed within 5min

or

(3)

6.(20 10. Jinhua) The position of the switch can't be seen at night, so Xiaoming designed the circuit as shown in the figure. When the single-pole double-throw switch S is turned to B, the lighting lamp is on and the indicator lamp is off to avoid wasting electric energy. When S turns to A, the lighting goes out and the indicator lights up to indicate the switch position (S can only be in these two states). He completed the connection circuit between the indicator lamp and the resistance wire in the picture box, and the indicator lamp specification was "10V, 1W".

(1) What is the resistance of the lamp when it works normally?

(2) If the lighting works normally for 6 hours, how much electricity does the lighting need to consume during this time?

(3) In order to make the indicator light shine normally, what resistance wire should be connected?

(1)P amount =U amount I amount =U2 amount /R

r = U2/P =(220v)2/ 100 w = 484ω

(2)W = Pt = 100 W×6h = 600 W = 0.6kwh

(3) The lighting lamp and the resistance wire are connected in series, so the current flowing through the lighting lamp is equal to the current flowing through the resistance wire.

The lighting lamp shines normally, reaching the rated power.

I=P photo /U photo =1w/10v = 0.1a.

The electrical industry of series resistance wire is U resistance =U- U lighting =220V- 10V=2 10V.

R= U resistance/I = 210v/0.1a = 2100ω.

7.(20 10. Wenzhou) The Wenzhou Municipal Government plans to build 40 charging piles for electric vehicles this year, so as to provide a complete set for the further development of electric vehicles in Wenzhou.

Facilities construction. The following table is some relevant data of a certain brand of electric vehicles:

(1) When the vehicle is empty, if the total contact area between the wheels of the electric vehicle and the ground is 400cm2, what is the pressure of the vehicle on the ground? (g = 10N/kg)

(2) If the car travels at a constant speed of 40 km/h for 2.5 hours, how many kWh will it consume?

(3) Electric vehicles are powered by iron batteries. As shown in the figure, the discharge curve of iron battery and zinc-manganese battery when supplying power to the same appliance. What are the characteristics of iron battery compared with zinc-manganese battery? (Write two points).

(1) solution: p = f/s = (1600kg×10n/kg)/400×10-4m = 4×105pa.

(2)S = v . t = 40k m/h×2.5h = 100Km W = 100Km×0. 12kWh/Km = 12kWh

(3) The discharge time is longer, and the voltage is higher when discharging.

28.(20 10. Guilin) Xiaohua bought a home hair dryer, as shown in Figure A, its simplified circuit is shown in Figure B, and its main technical parameters are shown in the right table.

Please answer the following questions:

(1) When the blower blows cold air, what is the electric energy consumed by normal operation for 5 minutes?

(2) When the selector switch is turned to where, what is the maximum power for the normal operation of the hair dryer? What is the total current in the circuit at this time?

(3) When the supply voltage of the circuit is 200V, what is the actual electric power of the heater of the blower?

(One decimal place is reserved for the calculation result)

Solution: (1) When the blower blows cold air, there is P=W/t:

Power consumption: w = pt =120 w× 5× 60s = 3.6×104j.

(2) When the selector switch is turned to point B and point C, the electric power of the blower is maximum.

By: P=UI

The current in the circuit I = p/u =1000 w/220v = 4.5a.

(3) Rated power of heating wire: p =1000 w-120 w = 880 w.

From: P=UI and I=U/R:

Resistance of heating wire: r heat =U2/P quantity =(220V)2/880W=55.

Actual power of heating wire: pReal = uReal2/RHEAT = (200 V) 2/55 = 727.3W.

Answer: When the hair dryer blows cold air, the electricity consumed in normal operation for 5 minutes is 3.6×104j; When the selector switch is turned to point B and point C, the electric power of the blower is maximum. When the current is 4.5A and the actual voltage in the circuit is 200V, the actual power of the blower is 727.3 W. 。

34.(20 10. Jixi) Xiaohua wants to buy an instant electric water heater (commonly known as "superheated water heater") to replace the original box-type electric water heater, which can make the water temperature reach a suitable temperature in a few seconds.

(1) The nameplate of "superheated water" reads "220V, 6000 W". What is the current through it during normal operation?

(2) 0.5 yuan/kW for domestic electricity in Jixi City? H, the ticket price of local bathing place is 4 yuan. If he takes a bath with this "superheated water", the cumulative power-on time is 10 minutes each time. Please calculate how much money he can save every time he takes a bath. (The water consumption cost is calculated at 0. 1 yuan).

⑶ What should I pay attention to when installing and using because the "overheated" working environment is humid and powerful?

(It is known that the maximum rated current marked on his air switch is 20A)

Recently, Xiaohua found that many neighbors are using "solar" water heaters. If you are a small flower, do you choose "solar energy" or "overheating"? Tell me the basis of your choice.

Answer: (1)

(2)

0.5 yuan per kWh/kW? h × 1 kW？ H =0.5 yuan

Cost saving: 4 yuan -0.5 yuan -0. 1 yuan =3.4 yuan.

(3) For safety reasons, leakage protection devices should be installed (three hole socket and triangular plugs can also be used);

Carry out circuit transformation (answer the specific transformation method, or indicate that the maximum rated current of the electrical appliance is inconsistent with the original circuit, and you can also score)

(4) The solar water heater is clean, pollution-free, energy-saving, renewable, low-carbon and relatively safe; "Overheating" has the advantages of faster heating speed, convenient use, small size, saving space and being unaffected by the weather (if you choose between "solar energy" and "overheating", you can score as long as the reason is reasonable).

22.(20 10. Bengbu II) As shown in the figure, the power supply voltage and the fixed resistance RX are unknown, and R 1 is a resistance box with a large enough adjustment range, and the resistance values of R2 and R3 are 3R and 6R respectively. When the switch S is closed, the ammeter reads 0 when the resistance value of the resistance box is adjusted to R, and 0 when the resistance value of the resistance box is adjusted to 6R.

It is found that: (1) When the resistance value of the resistance box is adjusted to r, the electric power of R3 is calculated;

⑵ Resistance of power supply voltage and fixed value resistor RX;

(3) If a voltmeter is connected between point A and point B in the circuit and the resistance of the resistance box is constantly changed, a series of readings of the voltmeter and ammeter will be obtained. Please draw qualitatively the graphs reflected by the readings of voltmeter and ammeter in the coordinate system U-I in Figure B. 。

22.⑴

⑵

⑶

(If the graph can reflect: ① linear ② subtraction function, it is correct not to extend to two coordinate axes))

(20 10. Yantai) Xiaoming uses three ammeters (voltmeter or ammeter), two small bulbs, a switch, wires and a power supply to connect into the circuit as shown in the figure. After the switch is turned on, both light bulbs are on. The readings recorded by Xiao Ming from three meters are as follows:

6V； 1.5a； 0.5A。

(1) Draw the experimental circuit diagram, and point out the readings of three instruments, A, B and C, respectively.

(2) Calculate the resistance of Ll and the total power consumed by the circuit.

Solution: (1) See the figure (the connection sequence of the circuit diagram should be consistent with the physical diagram).

A is an ammeter with a reading of1.5a.

B is an ammeter with a reading of 0.5A

C is a voltmeter reading 6V.

(2) according to ohm's law, I = u/r.

r 1 = U/I 1 = 6V/0.5A = 12ω

According to P=UI

P total = UI = 6V×1.5a = 9W.

(20 10. Suzhou) As shown in the circuit, R 1= 10, R2=20, indicating the ammeter after the switch is closed.

0.3A Find:

(1) the voltage across the resistor R 1;

(2) the current through resistor R2;

(3) The total power consumed by the circuit.

Solution: (1) resistance r1u = I1r1= 0.3a×10ω = 3v.

(2) Current I2 through resistor R2 = u/R2 = 3v/20ω = 0.15a.

(3) The total power consumed by the circuit P = UI = U (I1+I2) = 3v× (0.3a+0.15a) =1.35w.

(20 10. Guangzhou) An electric iron has two heating wires, R 1 and R2. Its simplified circuit is shown in Figure 24, and it can work at 220V or 1 10 V.

Normal operation: set S to 1 10V, and set S to 2 at 220V. In both cases, iron power.

They are all1100 w.

(1) How much power does the electric iron consume when it works normally for 5 minutes?

(2) What is the normal working current of the electric iron at 220 volts?

(3) What is the resistance ratio of two heating wires?

Solution: (1) Because the electric iron works normally, so

(2) from,

(3) from,

therefore

(20 10. Jiaxing) Expo 20 10 will fully use a new type of energy-saving lamp, called led lamp. Compared with the old light bulbs, using LED lamps can save energy by 90%, and the color of the lamps will change with the voltage (see the table below). Figure B is the mascot of the Expo-"Haibao". Under the illumination of LED lights, it can show different colors at night, and the image is very cute.

(1) According to the data in the table, when the LED lamp emits red light, it consumes 0.036 watt of electric power.

(2) Connect the LED lamp to the circuit shown in Figure A above. It is known that the power supply voltage is 5 volts and remains unchanged, and the resistance r is 90 ohms. After closing the switch, the milliammeter (an ammeter that is more accurate than the ammeter) reads 20 mA. By calculating what color light the LED light emits at this time?

(3) There is a sea treasure with a mass of 2.0× 103kg, which is placed on the horizontal ground with a contact area of 0.4m'. Find the pressure of this treasure on the ground (take g = 10N/kg).

Solution: (2) I = u/r, ur = IR = 0.02A× 90ω =1.8V.

Uled = 5v- 1.8v = 3.2v。

The LED light of the watch is blue.

(3) f = g = mg = 2.0×103kg×10n ton /kg = 2.0× 104 cattle.

Answer: (2) At this time, the LED light emits blue light. (3) The pressure of this "Haibao" on the ground is 5× 104 Pa.

(20 10. Zhanjiang) Grandma Xiaoming lives in Xuwen village, and the light bulbs in her corridor are easy to burn out in the middle of the night. In order to solve this problem, Xiaoming designed the circuit as shown in the figure, in which L 1 and L2 are both "220V40W". Please answer the following questions: (Assuming the bulb resistance is constant, ignoring the change of power supply voltage)

What is the resistance of (1) B bulb? When switches S and S 1 are closed, L 1 normally emits light. What is the power consumption after working for 5 hours?

(2) Turn off S 1 and turn on S, so that both bulbs will glow, and the bulbs will not burn easily in the middle of the night. Please calculate the voltage across L 1 at this time.

(3) Please calculate the total power consumed by the circuit when both bulbs are on.

Solution: the resistance of (1) B bulb r = U2/p = (220v) 2/40w =1210ω.

L 1 electric energy consumed in normal operation for 5h W=Pt=0.04kW×5h=0.2 kW? h

(2) when s1is off and s is on, the current I=U/R? Total = 220v/(1210ω+0210ω) =11a.

L 1 Voltage at both ends = IR1=11a×1210Ω =1/0v.

(3) When both bulbs are on, the total power consumed by the circuit is Ptotal = UI = 220v×11/a = 20w.

A: Omit.

1 1, (20 10. Luzhou) Solar electric vehicle is powered by solar cells. The right picture is a simplified diagram of its circuit. Light energy is converted into electric energy and stored in solar cells, which provides the power needed for motor rotation. The speed of motor can be adjusted by changing the speed regulating resistor. As we all know, the voltage of solar cells is 600 volts. Please answer the following questions:

(1) When the slider P of the speed regulating resistor is at the extreme right, the resistance of the speed regulating resistor connected to the circuit is10Ω. When the switch s is closed, the voltage is 500V v. What is the electric power of the motor at this time?

(2) When the slider P of the speed regulating resistor slides to the far left, the electric power of the motor reaches 18KW. What is the current representative?

1 1. solution: (1) when the slider p is at the extreme right, the motor is connected in series with the speed regulating resistor, and the voltmeter measures the two carbonyl voltages of the motor, and the voltages at both ends of the speed regulating resistor are u-u = 600 v-500 v =100 v.

I = U/R = 100V/ 10ω= 10A。

Motor electric power P= U machine I=500V× 10A=5kW.

(2) When the slider P of the speed regulating motor is at the far left, only the motor is connected to the circuit, and the voltage at both ends of the motor at this time.

U machine1= u = 600 v.

P= U machine 1×I represents the current: I=P/ U machine1=18000 w/600 v = 30a.

Answer: When the slider P is at the extreme right, the electric power of the motor is 5kW; when the slider P slides to the extreme left, the indicator of the ammeter is 30A.

26.(20 10. Nanjing) The circuit diagram shows that the sliding rheostat is marked with "50ωlA", the power supply voltage is constant at 6V, the ammeter range is 0-0.6A, and the voltmeter range is 0-3V. When the switch S is closed and the slider of the sliding rheostat moves to a certain position, the current representation is 0.2A and the voltage representation is 2V. Find:

Resistance value of (1) resistor R 1

(2) The power consumed by the sliding rheostat at this time

(3) In order to ensure the safety of the circuit, the allowable power consumption range of resistor R65438 is +0.

26. solution: (1) r1= u1/I = 2v/0.2a =10ω.

(2)U2=U-U 1=6V-2V=4V

P2=U2I=4V×0.2A=0.8W

(3) When the voltage representation number is 3V, the power consumed on R 1 is the largest.

p 1da =(u 1da)2/r 1 =(3v)2/ 10ω= 0.9w。

When the resistance of sliding rheostat is the largest, the power consumed on R 1 is the smallest.

The minimum current I in the circuit is small =U/R total = 6V/60Ω = 0.1a.

P 1 small =I minimum 2r1= (0.1a) 2×10ω = 0.1w.

The allowable power consumption range of R 1 is 0.1w-0.9w-0.9w.

Answer: The resistance of the resistor R 1 is10Ω, and the power consumed by the sliding rheostat is 0.8W In order to ensure the safety of the circuit, the allowable power consumption range of the resistor R 1 w is 0.1w-0.9W..

24.(20 10.Shanghai) In the circuit shown in Figure 13(a), the power supply voltage is 6V, which remains unchanged, and the sliding rheostat R 1 is marked with the words "100ω2A". After the S key is turned off, the indicators of ammeter A 1 and A2 are respectively.

(1) Resistance of resistor R2.

(2) When the current is energized 10 second, the work done by the current through the resistor R2.

(3) Is it correct to move the slider P of rheostat R 1?

There is an angle that makes the hands of two ammeters deviate from the zero scale.

Same situation? If it exists, request the rheostat R 1 at this time.

Power consumption p1; If it does not exist, please explain why.

Solution: (1) I2 = I-I1=1.2a-0.4a = 0.8a.

R2 = u/I2 = 6V/0.8A = 7.5Ohm.

(2) W2 = UI2t = 6V× 0.8A× 10 second = 48 joules.

(3) If the range of ammeter A 1 is 0~0.6 A, and the range of ammeter A2 is 0~3 A, then when the two ammeter hands deviate from the zero scale by exactly the same angle, there are: i? =5 I 1？

I 1？ +I 2=5 I 1？

I 1？ +0.8 A = 5i 1？

I 1？ = 0.2 amp

P 1=U I 1？ = 6V× 0.2A = 1.2W。

Answer: The resistance of resistor R2 is 7.5 ohms. The work done by the current through the resistor R2 is 48 Joules. At this time, the power consumed by rheostat R 1 is 1.2 watts.

18.(20 10. Wuhu) Household appliances often fail to reach the rated voltage and power when working. A classmate did the following experiment at home: he first checked his electric energy meter, which was marked "3600 r/kW?" h”； Then he turned off all the electrical appliances in the house, leaving only an electric kettle marked "220V, 12 10W" filled with water to continue working. At the same time, he observed that the dial of the watt-hour meter turned 180 in 3 minutes. Assuming that the resistance of the electric kettle remains the same, ask:

(1) electric kettle resistance;

(2) The actual electric power when the electric kettle works;

(3) The actual voltage of the classmate's home.

Solution: (1) times p amount = U2 amount /R

The resistance of the kettle is r = U2 = (220v) 2/1210w = 40Ω.

(2)t=3min= 180s

The electric energy consumed by the electric kettle at t time w = 180/3600 = 0.5 (kW? h)= 1.8× 105J

The actual electric power p of the electric kettle is w/t =1.8×105/180 =1000 (w).

(3) The actual voltage of the classmate's home, U = = 200 (V)

Answer: The resistance of the electric kettle is 40 Ω, the actual electric power when the electric kettle works is 1000W, and the actual voltage of the classmate's home is 200 V. 。

26.(20 10. Weifang) 20 10 The economic construction of Bohai Economic Development Zone rose to the national strategic height. Weifang responded to the call of the country to develop a low-carbon economy and built wind power plants in coastal areas, as shown in the figure. The following table gives the power of each wind turbine at different wind speeds. (The results are kept to two decimal places. )

(1) If the wind speed is 6m/s in a certain period of time and a generator works 1 hour, how many hours can Xiaoming's air conditioner (220V 5A) and rice cooker (220v40ω) work at the same time?

(2) If the annual average wind speed in Weifang coastal area is 8m/s, how many tons of coal can 50 such wind turbines burn completely in one day? (calorific value of coal is 3× 103 J/kg)

Answer: (1) Electric energy generated by wind turbines: w = Pt =16kW×1h =16kwh,

The power of the air conditioner is: p1= UI = 220v× 5a =1100w =1.1kw.

Electric heating power:

Working hours are:

(2) The power generation in one day is: w = Pt = 3.8×104 w× 50× 24× 3600s =1.6416×1kloc-0/.

The mass of burning coal is:

Shaoxing (20 10) The Shanghai World Expo with the theme of "Better City, Better Life" kicked off on May 1 day, 2065438.

(1) Xiao Wang and Xiao Lin, both working in a factory in S city, want to go to the World Expo by train together. We know that the opening time of the World Expo is from 9: 00 to 24: 00. See table 1 for the local train schedule and table 2 for the working schedule.

① If they want to visit the Expo Park all day (9: 00- 19: 00), they can choose to take the __D5552__ train on Saturday. In order to ensure the smooth progress of the work, it is required to have 4 hours to get on the bus after work and 4 hours to rest after returning to S city. )

Table 2: Work Schedule

Monday Tuesday Wednesday Thursday Friday Saturday Sunday

Xiao Lin has a day off, night shift, middle shift and a day off.

Xiao Wang's middle shift, middle shift, day shift, day shift and night shift rest.

Note: Working hours

Day shift:

7:O0—— 15:OO

Middle shift:

15:O0——23:OO

Night shift:

23: 00- 7: 00 the next day

② As can be seen from the train timetable, the average speed of the train D3 10 117.9 _ km/h (with one decimal place reserved) running between the two places.

(2) As shown in the figure, the theme pavilion is located in the southwest of the Expo Cultural Center.

(3) Installing solar power generation facilities on the roofs and glass curtain walls of China Pavilion and Theme Pavilion is expected to generate 5 million kWh of electricity every year. Try to calculate how many "220V 1000W" lighting lamps can be used all the year round 15 hours. (365 days a year)

Answer: W=Pt, that is, 5×106 kwh =1kw× n×15× 365h.

The solution is n=9 13.

Answer: The 9 1000W lamp can be used for one year.

(4) Reducing emissions is the basic requirement of Expo's "low carbon" concept. As shown in Figure A, the resistance of tin oxide sensor changes with the change of carbon monoxide content, and whether the carbon monoxide content exceeds the standard can be judged by observing the voltage representation. The technical data of the tin oxide sensor shows that the relationship between the reciprocal of its resistance (1/R) and the carbon monoxide content is shown in Figure B, so the voltage representation number U in Figure A is equal to the voltage representation number of the carbon monoxide content.