After the resistor R with a resistance value of 10 Ω is connected in series with the lamp L with a rated voltage of 2.5 V, it is connected to both ends of the 4 V power supply, and the switch S is closed. The electric power of the lamp L is measured to be 0.4 W Find the current through R and the rated power of the bulb

After the resistor R with a resistance value of 10 Ω is connected in series with the lamp L with a rated voltage of 2.5 V, it is connected to both ends of the 4 V power supply, and the switch S is closed. The electric power of the lamp L is measured to be 0.4 W Find the current through R and the rated power of the bulb

When the switch is closed, the current passing through the lamp is I (because it is in series, so the current is over R), then the voltage at both ends of the lamp is 0.4gi, and the voltage at both ends of the resistor is 10 * I
Therefore, the equation 10 * I + 0.4/i = 4 gives I = 0.2 a
Therefore, the resistance of the bulb is p / I & # 178; = 10 Ω, and the rated power of the bulb is 2.5 & # 178; / 10 = 0.625w

After connecting the customized resistor R1 = 10 ohm in series with the constant value resistor R2 to the power supply with constant voltage of u, the electric power consumed by R1 is
The power consumption of R1 is 1.6W, and R3 is known to be 4r2. Therefore, the power supply voltage U = V, R2 =
The formula 6.4w = u & # 178 / 10 Ω + R2 1.6W = u & # 178 / 10 Ω + 4r2 can't be calculated

The resistance and power of R1 are known, and the calculated current is 0.8A and 0.4A respectively
U=0.8*(10+R2)=0.4*(10+R3),R3=4R2
R2 = 5 ohm, u = 12V

Connect the resistor with 2 ohm R1 and R2 in series and connect them to a certain power supply. The electric power consumed on R2 is 10 watts, and the voltage at both ends of R2 is 4V
1) Resistance of R2
2) Power consumed on R1
3) Voltage of power supply
P.s

Since the total voltage at both ends of the series circuit is equal to the sum of the voltages at both ends of each part of the circuit, there is a function between the consumers?

For series circuit, there is voltage division between consumers and shunt between consumers
The current of the series circuit is equal, that is, the sum of the resistances of two or more consumers is divided by the total voltage, that is, I = u / R1 + R2... RN
The voltage of consumer in series circuit is divided according to its resistance. For example, the terminal voltage of consumer 1 is U1 = u * R1 / (R1 + R2 +... RN), other calculations are the same
The current of consumer in parallel circuit is divided according to its resistance, and its calculation is relatively simple, such as the calculation of consumer 1, the current of consumer I1 = u / R1, and other calculations are the same

In a series circuit, if one of the consumers has an open circuit, does the other consumer have voltage or current?
In a series circuit, if one of the consumers is short circuited, does the other consumer have voltage or current?

A series circuit is a circuit. If one of the electrical appliances is disconnected, the resistance of the electrical appliance will become infinite, then the circuit current will become infinitesimal, so other electrical appliances will have no current and voltage

When there is only one consumer in the circuit, is the voltage of the consumer equal to the supply voltage
ditto

The voltage at both ends of the appliance is equal to the supply voltage

Research: in parallel circuit, what is the relationship between the voltage at both ends of each electrical appliance (such as small lamp) and the total voltage at both ends of the power supply?
(1) My guess is:——————————
(2) Experimental verification:——————————
1. Equipment required:————————————
2. Design experiment circuit diagram (draw circuit diagram)
3. The experimental steps were as follows————————
4. Design experimental data record table——————
(3) Conclusion (or possible conclusion)——————

Each end of parallel circuit is power supply voltage. In series circuit, the voltage at both ends of each device varies with load resistance. The total voltage is power supply voltage

When a light bulb is connected in series in the circuit, if the lamp is open circuit, why is the voltage at both ends of the lamp power supply voltage

Ha ha, right. When you connect the voltmeter to the two ends of the lamp, it is equivalent to directly connecting to the power supply. Of course, the voltage of the power supply is measured

Why is the voltage at both ends of the appliance equal to the supply voltage when there is only one consumer in the series circuit?
Shouldn't electrical appliances be able to share a little voltage when they are connected in series? My question can be changed to: Why are the voltmeters equal when measuring the power supply voltage directly with a voltmeter and when there is only one electrical appliance in the circuit and measuring the voltage at both ends of the electrical appliance with a voltmeter?
This is the problem of physics in the third year of junior high school. I hope Dashen can tell me in simpler words, and I haven't learned the course after voltage. I'd like to explain it to you as much as possible. Thank you very much

If there is only one consumer in the circuit, there is no need to divide it into series and parallel. Series and parallel must be two or more consumers. At this time, the positive pole of the battery is connected to one end of the consumer, and the negative pole of the battery is connected to the other end of the consumer. What is the difference between the two ends of the power supply and the two ends of the consumer,

When exploring the voltage relationship in series circuit, the power supply voltage is always greater than the sum of the voltages at both ends of each electrical appliance. Why

In series circuit, according to Kirchhoff's second law, the power supply voltage should be equal to the sum of the voltages at both ends of the electrical appliances. If the power supply voltage is greater than the sum of the voltages at both ends of the electrical appliances in the actual circuit, it can only be said that the voltage drop of the line is not taken into account. Considering the line voltage drop, the power supply voltage should be equal to the sum of the voltage of the electrical appliances and the line voltage drop