There are two resistors, R1 = 2 Ω and R2 = 3 Ω. If they are connected in series in the circuit, the total resistance is______ Ω, the ratio of voltage at both ends of R1 and R2 U1: U2=______ ．

There are two resistors, R1 = 2 Ω and R2 = 3 Ω. If they are connected in series in the circuit, the total resistance is______ Ω, the ratio of voltage at both ends of R1 and R2 U1: U2=______ ．

Because two resistors are in series, the total resistance after series connection is r = R1 + R2 = 2 Ω + 3 Ω = 5 Ω; ∵ R1: R2 = 2 Ω: 3 Ω = 2:3 ∵ U1: U2 = R1: R2 = 2:3

Resistor R1 and R2 are connected in series in the circuit. If the voltage ratio U1: U2 at both ends is 2:3, then R1: R2=____ If the power supply voltage is 12V, both ends of R1
Voltage U1=____ V.

Resistor R1 and R2 are connected in series in the circuit. If the voltage ratio U1: U2 at both ends is 2:3, then R1: R2 = 2:3. If the power supply voltage is 12V, then the voltage at both ends of R1 is U1 = 4.8V

If two resistors with resistance value of R1 = 20 and R2 = 30 are connected in series in the circuit, the voltage ratio U1: U2 at both ends of the two resistors=_______ Through their circuits
Ratio I1: I2=______ If two resistors are paralleled in the circuit, the voltage ratio U1: U2=______ Through their circuit ratio I1: I2=______ .
Along with the problem-solving formula and so on. The teacher will ask questions in class tomorrow

The voltage of series circuit is proportional to the resistance, that is: U1 / U2 = R1 / r2 = 20 / 30 = 2 / 3, the current of series circuit is equal, I1: I2 = 1:1
The voltage of each branch in parallel circuit is equal, that is U1: U2 = 1:1, I1: I2 = R2: R1 = 30:20 = 3:2

The electric power consumed by an electric appliance must be equal to its rated power

No. It depends on the environment and conditions of use, as well as the type of electrical appliances. For example, when the motor load is small, the actual power is less than the rated power. When the power supply voltage is reduced, the actual power is less than the rated power

For two vehicles with the same traction power, the ratio of the distance along the horizontal highway in the same time is 4:5, then the following conclusion is correct:
A the ratio of engine traction of two vehicles is 4:5
B the ratio of engine traction of two vehicles is 5:4
C the ratio of engine work of two cars is 4:5
D the ratio of engine work of the two cars is 5:4

Choose a according to P = f / s, because P is equal, the ratio of S is 4:5, so the ratio of F is 4:5

As shown in the figure, a fixed pulley is used to lift an object with a weight of 500N. If the work done by a person's pulling force is 360j when the object rises 0.6m, then the person's pulling force is 360j______ N. The extra work he did was______ J. What is the mechanical efficiency of the fixed pulley______ ．

It is known that: g = 500N, H = 0.6m, wtotal = 360j, find: (1) tension F, (2) the amount of additional work he does, (3) the mechanical efficiency of the fixed pulley η (1) in the fixed pulley, the distance of the rope end moving is equal to the distance of the object rising, i.e. s = H = 0.6m, according to w = FS, we can get: Tension f = w, total S = 360j0.6m = 600N; (2

Pengshui power supply company started to use electronic energy meter in 2012. A student found that there was no turntable, but the red indicator light was flashing. The dial of his electronic energy meter at the beginning and end of last month is shown in the figure. The dial is marked with "220V & nbsp; 20A & nbsp; According to the information on the dial, answer the following questions: (1) how much kW · h did the student share last month? (2) How much is the total power of electrical appliances used at the same time? (3) If only one electric heater marked with "220V & nbsp; & nbsp; 484w" is allowed to work for 3min and the indicator light of the electric energy meter flashes 32 times, what is the actual voltage of the student's home? (let the resistance of the electric heater remain unchanged)

(1) Energy consumption in this month: w = 12614.6kw · h-12519.6kw · H = 95kw · h. (2) because "220V" represents the rated voltage of the meter, "10A" represents the calibration current of the meter, the maximum power of the meter in normal operation is: P = UI = 220V × 10A = 2200W. (3) because the indicator light of the meter flashes 32 times, the energy consumption of the electric heater is: w = 321600k W · H = 0.02kw · H = 7.2 × 104j, t = 3min = 180s, so the actual power of the electric heater is: P = wt = 7.2 × 104j180s = 400W, which can be obtained from P = u2r, resistance of the electric heater: r = U2, P = (220V) 2484w = 100 Ω, so the actual voltage of the student's home is: u = P, r = 400W × 100 Ω = 200V (3) the actual voltage of the student's home is 200V

How to judge the short circuit of electrical appliances?

If a wire is connected in parallel with the consumer, the consumer is short circuited because the charge passes through the wire with almost no resistance and does not flow through the consumer

In the first lesson of physics electricity, how to judge which electrical appliance is short circuited? What is the basis?

In parallel with a voltmeter at both ends of the electrical apparatus, observe whether there is indication on the needle. If there is a reading, it proves that the electrical apparatus is working normally. If there is no reading, it means that the electrical apparatus is short circuited. The reason is that when the electrical apparatus is short circuited, it is equivalent to a wire, and the voltage at both ends of the wire is very small, which can be ignored. Therefore, if there is no reading, it means that the electrical apparatus is short circuited

How to judge the short circuit of electric appliance

Short circuit of electrical appliances: short circuit of electrical appliances is defined as connecting the two ends of electrical appliances with wires (Note: there can be no power supply between the two ends of wires). Feature (1) when electrical appliances are short circuited, most of the current passes through the wires, and only a small current passes through the electrical appliances. At this time, it can be approximately considered that no current passes through the electrical appliances