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A junior high school physics problem about ohm's law
To find out the short circuit, we must first find out the wires, batteries, etc. There is resistance. The resistance of the battery is about 1 ohm, and the internal resistance of the wire is very small and can be ignored. Only in junior high school, for the convenience of calculation, these were ignored. If the dry battery is short-circuited, the resistance of the whole circuit is the power supply resistance plus the wire resistance, so the voltage of the external circuit is still very small, because many voltages are distributed to the internal resistance of the power supply. Of course, this is still very small, because the voltage is basically distributed to the internal resistance of the power supply, so the battery is very hot and easy to burn out. At the same time, the conductor is also very hot because the current is relatively large.
When two dry batteries are connected in series, the total voltage is the sum of two batteries, which is twice that of one battery. Parallel connection, in essence, is to increase the current, that is, if one dry battery can only output the current of 1A at the highest (because of internal resistance, the current during short circuit is the maximum current, of course, the current is far less than this number without damaging the battery), then the maximum current provided by two dry batteries will be doubled when they are connected.
Because many things in junior high school have been simplified, you can't see the difference. When I get to high school, I will understand this very well.
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