1. Connect R1 and R2 into the circuit, the voltage at both ends of R1 is known to be U1, and the voltage at both ends of R2 is known to be U2 Connecting R1 and R2 into the circuit, we know that the voltage at both ends of R1 is U1, the voltage at both ends of R2 is U2, and the current through R1 and R2 is I1 and I2 respectively. If U1 = U2, then P1: P2 = if I1 = I2, then P1 ': P2'=

electric current?
I＝U1/R1=U2/R2 U=U1+U2

In single-phase bridge rectifier circuit, if the effective value of input voltage is U2, the average voltage of load resistance RL is equal to

0.9u2

In the capacitor filter circuit, if the load resistance RL is disconnected, the output voltage will be reduced

In the capacitor filter circuit, if the load resistance RL is disconnected, the output voltage will rise to 1.414 times of the original average voltage! (peak value)

Figure 1-7-7 shows a series circuit composed of a battery, a resistor R and a parallel plate capacitor
Figure 1-7-7 shows a series circuit composed of battery, resistor R and parallel plate capacitor
Figure 1-7-7
A. There is no current in the resistance R. B. the capacitance of the capacitor becomes smaller
C. The resistance R has the current from a to B. the resistance R has the current from B to a
Can you explain why there is a current flowing from a through R to B in the circuit?
(not one of those options)
Yes: [there is current flowing from a to B through R in the circuit]

The voltage at both ends of the capacitor is equal to the power supply voltage, so the voltage U remains unchanged. The capacitance of the capacitor C = ε s / 4 π KD, increasing the distance d between the two plates of the capacitor, C becomes smaller, while the charge q = Cu, that is, q decreases, so the capacitor discharges. Because plate a is positively charged, there is a current flowing from a to B through R in the circuit

For a 5.5kW three-phase asynchronous motor, the resistance between U1 and U2 is 3 ohm. Is it normal

Normally, 5.5kW resistance is small

Judgment of short circuit of electric appliance

Method: check whether the terminals of an electrical appliance are connected by the same wire (including ammeter and switch). If so, it is short circuit

Normal working voltage and rated power of various household appliances

The rated voltage is generally 220 v
Electric power of household appliances
1. Air conditioning - about 1000W
2. Microwave oven - about 1000W
3. Electric furnace - about 1000W
4. Electric water heater - about 1000W
5. Vacuum cleaner - about 800W
6. Hair dryer - about 500W
7. Electric iron - about 500W
8. Washing machine - about 500W
9. Electronic computer - about 200W
10. TV set - about 200W
11. Range hood - about 140W
12. Refrigerator - about 100W
13. Electric fan - about 100W
14. Flashlight - about 0.5W

For bulbs with the same rated voltage, the higher the rated power, the smaller the resistance, and the more heat generated per unit time during normal operation. However, according to Joule's law, the greater the resistance, the more heat generated per unit time. There seems to be a contradiction between the two. What's the matter?

In parallel circuit, the two loads have the same voltage, the rated power P = u * U / R, and the resistance is inversely proportional to the power. The larger the resistance, the smaller the heat generated per unit time. In series circuit, the two loads have the same current, the rated power P = I * I * r, and the resistance is proportional to the power. The larger the resistance, the greater the heat generated per unit time

For bulbs with the same rated voltage, the higher the rated power is, the smaller the resistance is, and the more heat is generated per unit time. However, according to Joule's law, the greater the resistance is, the more heat is generated per unit time. There seems to be a contradiction between the two. What's the matter?
I know it is deduced by P = u ^ 2 / R, but what's the matter if this formula is not available in non pure resistance circuit? If the problem is not pure resistance circuit, how to do? I'm in a mess. I can't figure it out

The control variables are different. The former is voltage and the latter is current. Suppose that the parallel voltage of two bulbs is the same and they are all rated voltage, but the power of a is greater than that of B. according to the formula of the square of u, R A is less than R B. think about the electric furnace, where the furnace wire and wire are connected in series. So there are two variables that are not easy to compare, and we need to experiment to find out

Is there any contradiction between the higher the rated power of the bulb with the same rated voltage, the smaller the resistance and the more heat generated per unit time?

No contradiction, different conditions
The higher the rated power of the bulb with the same rated voltage, the smaller the resistance, and the more heat per unit time, P = u ^ 2 / R;
The larger the resistance is, the more heat is generated per unit time, P = I ^ 2 * r
One is the same voltage, and the other is the same current. Ohm's law can be used to replace it with the same form