Why does the bulb with high resistance share more voltage in series circuit?

Why does the bulb with high resistance share more voltage in series circuit?

In series circuit, the current is equal everywhere. According to Ohm's law u = IR, when the current is constant, the voltage is proportional to the resistance, so the greater the resistance, the greater the voltage of the bulb

There was a light bulb in the series circuit. Now add a resistor. Will the current change? What about the voltage in the parallel circuit

Adding a resistor in series will increase the total resistance and decrease the current, but the current of the two consumers is the same
Parallel circuit increases resistance, voltage unchanged, still equal to supply voltage
Note: junior high school understanding
High school about parallel circuit will be different
I don't know what to ask

There are two lamps marked with 6v2w and 12v8w, which are connected in series in the circuit. The maximum allowable voltage at both ends of the circuit is () a6v b12v
C 18V d9v (process)

The bulbs are usually used in parallel with the same voltage. Your problem may be to connect the bulbs in series for the purpose of analyzing the problem. Because the bulbs are used in series, the current on the bulbs should be equal. According to the original rated data of the bulbs, it can be calculated that the resistance of each bulb is 18 ohm, so in the series circuit of two bulbs, the resistance is the same and the current is the same, The voltage on both ends of the bulb should be the same. The bulb with low voltage is rated to withstand 6V, so the maximum applied voltage is 12V (the bulb with high voltage actually bears lower than the rated voltage)