Xiaoqiang designed an electric kettle, the circuit is as shown in the figure, s is the temperature control switch, R1 is the heater. When the switch S is connected to a, the electric kettle is in the heating state, and the electric power of the heater is 1000W; when the water is boiling, the switch S automatically switches to B, and the electric kettle is in the insulation state, and the electric power of the heater is 160W (1) Please calculate the resistance value of R1 and R2 (regardless of the influence of temperature on the resistance value) (2) in order to make full use of electric energy, on the premise of not changing the circuit connection method, please put forward reasonable improvement methods for the design of Xiaoqiang

Xiaoqiang designed an electric kettle, the circuit is as shown in the figure, s is the temperature control switch, R1 is the heater. When the switch S is connected to a, the electric kettle is in the heating state, and the electric power of the heater is 1000W; when the water is boiling, the switch S automatically switches to B, and the electric kettle is in the insulation state, and the electric power of the heater is 160W (1) Please calculate the resistance value of R1 and R2 (regardless of the influence of temperature on the resistance value) (2) in order to make full use of electric energy, on the premise of not changing the circuit connection method, please put forward reasonable improvement methods for the design of Xiaoqiang

(1) When s is connected to a, R2 is short circuited. According to the data in the question, the resistance R1 is: R1 = u2p plus = (220V) 21000w = 48.4 Ω; when s is connected to B, R1 and R2 are connected in series. At this time, the power of the heater is: P = I22 ·, R1 = (ur1 + R2) 2 ·, R2 = 72.6 Ω The internal energy converted from electric energy is used for heat preservation and energy saving

There are 10 220 V 40 W fluorescent lamps in a classroom. If the lamp is turned on less than one hour every day, the electricity consumption can be saved in one month (30 days)_____ degree

40W * 30 = 1200 = 1.2kwh = 1.2 degree

On a problem of electric power in Physics
In home circuit, the junction of wires is more likely to generate heat, accelerate aging and even cause fire?
I know that the resistance has increased. However, wires belong to the category of electric heaters. In a parallel circuit, if q = u squared divided by R times T, the resistance should not increase,
But from q = I square RT, the larger R is, the larger q is
The answer to this question is that the greater the resistance, the hotter it is. But isn't this parallel connection?

There is a certain truth in what you said, but it doesn't solve the question in the owner's mind. Why can we only use q = I & sup2; RT instead of q = u & sup2 / R. these two formulas can be used, but it depends on how you use them. It's not simply that this one can be used, that one can't be used

A watt hour meter is marked with "220 V, 10 a" and "3000 R / kWh" to indicate the maximum total power that can be connected

2200w