In the circuit as shown in the figure, the power supply voltage is 6V constant, the range of ammeter is 0-0.6a, the range of voltmeter is 0-3v, the lamp L1 and The specifications of L2 are "6V, 1.8W" and "6V, 1.2W" respectively. The specifications of sliding rheostat are "50ohm, 1.5A". Only close S1S2 to find the range of ammeter and voltmeter (the safety of components in the circuit)

In the circuit as shown in the figure, the power supply voltage is 6V constant, the range of ammeter is 0-0.6a, the range of voltmeter is 0-3v, the lamp L1 and The specifications of L2 are "6V, 1.8W" and "6V, 1.2W" respectively. The specifications of sliding rheostat are "50ohm, 1.5A". Only close S1S2 to find the range of ammeter and voltmeter (the safety of components in the circuit)

Since you can raise such a question, it shows that your foundation is good. I'll solve it for you
I'm a physics teacher. I often download questions that need gold coins
First, you should know: when you just close S1S2, the circuit is a sliding rheostat in series with L1
Second, let's consider the maximum current in the circuit, that is, the minimum value of the rheostat: Although the range of the ammeter is 0.6A, we can see that the normal working current of L2 should be 0.3A, so the maximum current can only be 0.3A. At this time, the resistance value of the sliding rheostat is its minimum value. If it is smaller, the bulb will be burned, Then subtract the bulb L1 resistance 20 ohm, (the number of this topic is interesting), and it happens to work out that the resistance of the sliding rheostat is 0, that is, in order to protect the circuit, it can be adjusted to the minimum of 0
Third, let's consider the voltmeter, because the larger the sliding rheostat is, the larger the voltmeter will be, so when the voltmeter is 3V, it is the maximum resistance of the sliding rheostat. We use the power supply voltage minus 3V to find out the actual voltage of the bulb is 3V, and then find out the current in the circuit: divide 3V by the bulb resistance 20 ohm, find out the current is 0.15A, and then divide 3V by the current, The maximum value of rheostat is 20 ohm
To sum up, in order to protect the whole circuit, the resistance range of sliding rheostat is 0 - 20 Ω

In a circuit, the power supply voltage is 6V, the maximum range of the voltmeter is 3V, the maximum range of the ammeter is 0.6A, and the rated power of the small bulb is 1.2W. Assuming that the resistance of the voltmeter is small enough, Q: if the power supply, the ammeter, the voltmeter, the small bulb and a 0-100 ohm resistance box are connected in series, when the resistance of the resistance box is how much, the circuit works normally (the ammeter, the voltmeter does not exceed the maximum range, Small light bulb lights normally)

When the power supply voltage is 9V and the resistance is how much, the voltage will become 6V and the current will become 0.5A
What is the resistance of sliding rheostat in the circuit?

(9-6) / 0.5 = 6 (Ω). That is to say, when the internal resistance R of the power supply is 6 Ω, the power supply voltage will change to 6V. At this time, the resistance R of the external circuit is 12 Ω

When 3V voltage is applied at both ends of a conductor, the current passing through the conductor is 0.2A, and the resistance of the conductor is______ When a voltage of 6V is applied at both ends of the conductor, the resistance of the conductor is equal to______ Ω．

(1) The resistance of the conductor: r = UI = 3v0.2a = 15 Ω; (2) because the resistance has nothing to do with the voltage at both ends of the conductor, when the voltage at both ends of the conductor is 6V, the resistance of the conductor is still 15 Ω