The power supply voltage remains unchanged. R0 is a constant value resistor of 8 Ω. If the rated voltage of a and B is 4 V, what is the rated power and the lightest bulb?

The power supply voltage remains unchanged. R0 is a constant value resistor of 8 Ω. If the rated voltage of a and B is 4 V, what is the rated power and the lightest bulb?

Do you have a diagram? If the resistor is connected in parallel with the bulb, the voltage at both ends of the bulb is also 4 V, and the working current is the maximum. The power is the highest, and the two degrees are the brightest

Does the resistor have rated power? The 220 V power supply is equipped with a 6W bulb. As a result, I connected a 2000 ohm resistor in series, and the resistor was broken down. So I asked if the resistor has power. What kind of resistor should I choose when the 220 V power supply is equipped with 6W bulb

Resistance certainly has power
P=U*I=I*I*R
The thicker the resistance wire is, the larger the current can pass through. When the resistance is fixed, the greater the power is
For 6W / 220V bulb, the resistance is about 8000 ohm and the current is about 0.27 a
If you use this 2000 ohm resistor string in a 6W bulb, the resistance should be at least 1W
We usually use 1 / 4W on PCB

In the figure, the electromotive force of the power supply is e = 12V, and the internal resistance is r = 0.5 Ω. After a bulb with rated voltage of 8V and rated power of 16W is connected in parallel with a DC motor with coil resistance of 0.5 Ω, it is connected with the power supply. The bulb just lights normally and is energized for 100min? (2) How much work does the current do to the bulb and the motor? (3) How much heat is produced by filament and motor coil? (4) What is the efficiency of the motor?

(1) According to the meaning of the question, if the terminal voltage of the circuit is u = 8V, then the internal voltage is uine = E-U = 12v-8v = 4V; the current in the circuit is I = u, and the internal R = 4v0.5 Ω = 8A; therefore, the energy provided by the power supply is w = uit = 8 × 8 × 100 × 60j = 3.84 × 105J; (2) the work of the current on the filament is w = Pt = 16 × 100 × 60j = 9.6 × 104j

Why does a bulb with high rated power generate more heat per unit time when its resistance is smaller? But according to Joule's law, the greater the resistance is, the more heat per unit time is generated. Are they contradictory?

In fact, according to Joule's law, the greater the resistance is, the more heat is generated per unit time. This statement is also accurate. It should be said that when the current is the same, According to Joule's law, the heat generated by electricity is directly proportional to the square of the current and the size of the resistance. That is to say, the size of the current has a greater thermal impact on electricity than the resistance. The voltage of the power supply is certain (220 V). The smaller the resistance, the greater the current. This can be understood by Ohm's law
The electric power of light bulb can be calculated by two formulas. 1. P = the square of U / R. from this formula, we can know that u is fixed, and the smaller R is, the greater P is
2. The formula P = the square of I * r shows that the size of P is determined by both I and r