When a bulb works normally, the voltage is 12 V and the resistance is 24 Ω. If the power supply voltage is 36 V, in order to make the bulb work normally, it is necessary to When a bulb works normally, the voltage is 12 V, and the resistance is 24 Ω. If the power supply voltage is 36 V, in order to make the bulb work normally, what is the resistance in series?

When a bulb works normally, the voltage is 12 V and the resistance is 24 Ω. If the power supply voltage is 36 V, in order to make the bulb work normally, it is necessary to When a bulb works normally, the voltage is 12 V, and the resistance is 24 Ω. If the power supply voltage is 36 V, in order to make the bulb work normally, what is the resistance in series?

The resistor connected in series must be divided into 24 V voltage, so his resistive bulb is twice as much as 48 ohm
The characteristics of equal series current: U1 = IR lamp, U2 = IR. Because U2 is twice of U1, R is twice of R lamp

A LED connected in series with a 470 resistor can be connected to a power supply of several volts

If the rated voltage of the LED is 2V and the current is 8Ma, the load voltage of 470 resistor in series shall be as follows:
U＝RI＝470×0.008≈4(V)＋2V＝6V

When the small bulb is connected in series with a 40 ohm resistor on a 6V power supply, it can light normally. At this time, the power of the small bulb is 0.2W. What is the rated voltage of the small bulb
Seek detailed problem-solving process

Let the resistance of the small bulb be r
Resistance (= =) R1 = 40 Ω small bulb power P = 0.2W total circuit voltage U = 6V
The total resistance in the circuit R = R + R1
The total current in the circuit = the current flowing through the bulb I = u / R total
Small bulb power P = I & sup2; R total. ③
R & sup2; - 100r + 1600 = 0
The solution is r = 20 Ω or r = 80 Ω
In this case, I = 0.1A or I = 0.05A
And u = IR
So the rated voltage of small bulb is 2V or 4V
Ah, I haven't done physics circuit for many years

When a small light bulb is connected in series with a 4 ohm resistor and connected to a 6V circuit, the bulb lights normally and the power is exactly 2W. Calculate the resistance, rated voltage and rated current of the small bulb. Be sure to be correct!
Are there really two answers

First, let the current be I and the resistance be r. according to the principle of equal current in series circuit and the power formula, we can get the formula (1) 4I + IR = 6
Formula (2) I ^ 2R = 2, I and R can be calculated from these two formulas, and u can be calculated from the formula u = IR