The known resistance R = 6 Ω is connected in parallel with the unknown resistance R2, and the measured current passing through the circuit is 4a, if R1 R2 is connected in series to the same circuit, and the voltage at both ends of R2 is 6V. The power supply voltage and R2 are calculated

(1) In parallel, we can know the power supply voltage, 6 * 4 = 24 v
(2) Connected in series to the same circuit, the voltage at both ends of R2 is 6V, the voltage at both ends of R1 is 18V, and the current is 18g6 = 3A
R2=6/3=2Ω

The current group is very small. The resistance of an ammeter is 0.02 ohm. The maximum allowable current is 3a. Can this ammeter be directly connected to a 2V battery

2V/0.02=100A
The upstairs said that the battery would be short circuited, but it would not damage the battery, because the current overload capacity of the battery is absolutely greater than that of the ammeter, so the ammeter was burned in the end,

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