As shown in the figure, it is a schematic diagram of the electric blanket circuit. R0 is the resistance wire in the electric blanket, R is the resistance in series with the resistance wire of the electric blanket, the electric blanket is marked with "220 V & nbsp; 100 W", and S is the switch that controls the electric blanket in the heating state or the heat preservation state. (1) use the learned common formula to explain whether the electric blanket is in the heating state or the heat preservation state when the switch S is disconnected? (2) If it is required that 60 J of electric energy is converted into internal energy per minute when the current passes through the resistance wire R0 during heat preservation, what is the resistance value of the resistance R?

As shown in the figure, it is a schematic diagram of the electric blanket circuit. R0 is the resistance wire in the electric blanket, R is the resistance in series with the resistance wire of the electric blanket, the electric blanket is marked with "220 V & nbsp; 100 W", and S is the switch that controls the electric blanket in the heating state or the heat preservation state. (1) use the learned common formula to explain whether the electric blanket is in the heating state or the heat preservation state when the switch S is disconnected? (2) If it is required that 60 J of electric energy is converted into internal energy per minute when the current passes through the resistance wire R0 during heat preservation, what is the resistance value of the resistance R?


(1) As shown in the figure, when switch S is off, R0 is connected in series with R, P = u2r, total = u2r0 + R, when switch S is closed, R is short circuited, at this time, P = u2r, total = u2r0. Therefore, the power of closing is greater than that of opening. Therefore, more heat is generated during closing in the same time, so when switch S is off, the electric blanket is in thermal insulation state. (2) according to "220 V & nbsp; R = (U) 2p = (220V) 2100W = 484 Ω, because R0 has 60j electric energy converted into internal energy every minute. According to the formula q = i2rt, I2 = QRT = 60j484 Ω × 60s, I = 122a, so the voltage on R0 is U0 = I · r = 122a × 484 Ω = 22V, so the voltage on R is U1 = 220v-22v = 198v, so the resistance of R is r = u1i = 198v122a = 4356 Ω



The electric heater works by electricity. When an electric heater works normally, the current through the electric heater is 5a, which is generated in 1min
The electric heater works by using. When an electric heater works normally, the current passing through the electric heater is 5a, and the heat of 6.6x10 fourth power J is generated in 1min, then the electric power of the electric heater is w, its normal working voltage is V, and the resistance of the electric wire is Ω
It's better to have a process


The electric heater works by using the thermal effect of current (when the current flows through the resistance, the resistance will generate heat). If the current passing through the electric heater is 5A and the heat of 6.6x10 is generated in 1min, the electric power of the electric heater is 220 W, w = u * I * t = P * t, so p = w / T = 66000 / 300 = 220 W, and its normal working voltage is 44



An electric water heater, connected to the lighting circuit, consumes 48000 joules of electricity per minute when heating. How many amperes is its working current


The title says that it is connected to the lighting circuit, which means that the voltage U = 220 V, t = 1 min = 60 s, the power consumption per minute when heating is 48000 joules, which means that q = 48000 J. then according to the formula P = q / T = 48000 J / 60 s = 800 W, according to the formula P = UI, I = P / u = 800 w / 220 V = 3.64 a