Physics problems related to electric power When the output voltage of a power plant is 5kV, the output current is 400A, and the resistance of the transmission line is 10 ohm, what is the power consumed on the line

Physics problems related to electric power When the output voltage of a power plant is 5kV, the output current is 400A, and the resistance of the transmission line is 10 ohm, what is the power consumed on the line

P = I squared R
P = 400A * 400A * 10 Ω = 1600000w = 1600kw
U = IR = 400A * 10 Ω = 4000V
P=UI=
It's the same

A light bulb is marked with "220 V, 40 W". One degree of electricity can be used to connect the light bulb to 110 V circuit and calculate its actual power. If the light bulb is connected to 380 V circuit, what is its actual power? How can it light normally?

1. T = 1000 / 40 = 25 hours
One degree of electricity will last the bulb 25 hours
2. The bulb is connected to a 110 volt circuit and the power will be reduced, assuming the resistance is constant
P=UU/R ,40 = 220*220/R,
P' = U'U'/R ,P' = 10W
3. If the lamp is connected to a 380 volt circuit, its actual power will increase and it will burn down
P'' = 380*380/R ,P'' =120W
4. Can step down (with transformer), can series resistance voltage,
Simple way, with two identical bulbs in series in 380V circuit, can be normal lighting

Some questions about physics in grade two of junior high school (electricity)
1. What is the physical quantity of current?
2. What is the relationship between current (I) and electric energy (W)?
3. Why is the battery only marked with current (a) and voltage (V), while the bulb only marked with rated power (W) and voltage (V)?
The brightness of 4.11w light tube is equivalent to that of 60W light bulb. What does it mean? What is the percentage of power saving?

Electric energy is the energy generated by the current flowing through the conductor. The most obvious example is that electric appliances often generate heat, which is a kind of energy. Measuring the energy of a battery depends on the voltage and current. The energy of a battery = voltage × current × time

When an electrolyte conducts electricity, both positive and negative charges move in it. It is measured that in one minute, the positive charge moving to the left through a cross section of the electrolyte is 4 Coulomb, and the negative charge moving to the right through the same cross section is 3 Coulomb. The current in the electrolyte is calculated

3 Coulomb negative charge moving to the right is equal to 3 Coulomb positive charge moving to the left, so it is equivalent to 7 Coulomb positive charge moving to the left. According to I = q / T, the current can be calculated. I = 7 / 60