There is a small bulb with a resistance of 10 Ω. When it lights normally, the voltage at both ends is 6V. Now there is only a 9V power supply Q: 1. How to connect a small light bulb to make it shine? 2. How much resistance do you need to give it? There are three ways to answer the second question!

Because the voltage of the small bulb is 6V when it works normally, and the power supply voltage is 9V, according to the principle of voltage division in series circuit, a resistor should be connected in series in the main circuit to divide the voltage. The current of the small bulb when it works normally is: I = u / r = 6 / 10 = 0.6A (1) the total voltage and current in the circuit are 9V and 0.6A, so the total resistance is r = u / I = 9 / 0

When a small light bulb lights normally, the resistance of the filament is 10 ohm. When it works normally, the voltage is 6V. If we only have a 9V power supply, how large a resistor needs to be connected in series to make the small light bulb work normally? Draw the circuit diagram

Small bulb, the current of normal lighting is 6 / 10 = 0.6A
9-6=3；3/0.6=5
A 5 ohm resistor needs to be connected in series
Do you still use drawing?

Physics problem: connect the 9V, 3W light bulb L1 and 12V, 6W light bulb L2 in series into the circuit, then what is the maximum allowable voltage at both ends of the circuit?
No need to explain, try to meet the scope of junior high school students

I 1 = P / u = 3W / 9V = 1 / 3A
I2=p/u=6w/12v=0.5A
Because I1

There is a small bulb with a resistance of 10 Ω. When it lights normally, the voltage at both ends is 6V. Now there is only a 9V power supply Q: 1. How to connect a small light bulb to make it shine? 2. How much resistance do you need to give it? There are three ways to answer the second question!