The internal resistance of the voltmeter is very large. The ordinary small bulb is equivalent to a wire. Why is it not short circuited when the voltmeter measures the voltage of the small bulb?

This requires you to know the voltmeter, with whom the voltage changes in parallel, there is the same principle of parallel voltage, the voltmeter is the total voltage of the parallel circuit and the branch voltage
When the voltmeter is connected in series with the bulb, because the resistance of the voltmeter is infinite (ideal voltmeter), the principle of series voltage division shows that the voltmeter will get all the voltage, and the voltage of the bulb is almost 0

A voltmeter in series is equivalent to a short circuit in a circuit. Why can a voltmeter in series be used in the circuit when measuring the internal resistance of the voltmeter with half bias method

Series connection of voltmeter in circuit is not equivalent to short circuit. The internal resistance of voltmeter is large enough to be equivalent to open circuit
But even if it is equivalent to open circuit, it can also measure the internal resistance
A voltmeter can be understood as a component composed of a small range ammeter and a large resistance
If it is connected in series at both ends of the battery, there is no open circuit, but the internal resistance is large (when there is no wire at both ends of the battery, there is an open circuit, but in fact the two stages are connected by air, and the air resistance is very large, which is equivalent to the open circuit), So it can be connected in series in the circuit. At this time, the current in the circuit is very small

There are two conductors with resistance of 10 Ω and 30 Ω respectively. If they are connected in series, the current passing through the 10 Ω resistor is 0.5 A. what is the supply voltage? What is the voltage on the two resistors? What is the relationship between the resistance and the voltage in the series circuit?

10 ohm resistance terminal voltage: U1 = IR = 0.5x10 = 5V
30 ohm resistance terminal voltage: U2 = IR = 0.5x30 = 15V
Power supply voltage: u = U1 + U2 = 5 + 15 = 20V
Total resistance of series circuit: r = R1 + R2 = 10 + 30 = 40 ohm
In series circuit, the voltage drop is higher when the resistance is large, and lower when the resistance is small. The voltage drop of two resistors is equal to the supply voltage

How many amperes can a 10 ohm resistor reduce

Because you don't know the internal resistance of your load, you can't calculate it completely. Suppose that your internal resistance of load is constant, set it to R, then the current I = 8 / r before adding 10 ohm resistance. After adding resistance, the current is 8 / (10 + R), so, and the reduced current is 8 / R-8 / (10 + R). You can calculate it by yourself. Of course, if the internal resistance of load is uncertain, it will be difficult,

The internal resistance of the voltmeter is very large. The ordinary small bulb is equivalent to a wire. Why is it not short circuited when the voltmeter measures the voltage of the small bulb?