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# There are usually two ways to detect whether the power of ordinary zinc-manganese dry batteries is sufficient

There are usually two ways to detect whether the power of ordinary zinc-manganese dry batteries is sufficient.
The first method is to estimate the internal resistance of the battery by measuring the instantaneous short-circuit current of the battery, and then judge whether the battery power is sufficient.
The second method is to connect a resistor with an appropriate resistance in series with an ammeter, and calculate the internal resistance of the battery by measuring the discharge current of the battery, thereby judging whether the battery power is sufficient.
The biggest advantage of the first method is that it is simple. The power of the dry battery can be directly judged by the high current range of the multimeter. The disadvantage is that the test current is very large, far exceeding the limit value of the allowable discharge current of the dry battery, which affects the use of the dry battery to a certain extent. life. The advantage of the second method is that the test current is small and the safety is good. Generally, it will not have an adverse effect on the service life of the dry battery. The disadvantage is that it is more troublesome.
It is understood that the above two methods are used to test and compare a new AA battery and an old AA battery with an MF47 multimeter. Assuming that ro is the internal resistance of the dry cell, and RO is the internal resistance of the ammeter, when using the second test method, RF is an additional series resistance, the resistance value is 3Ω, and the power is 2W.
The measured results are as follows. The new No. 2 battery E=1.58V (measured with 2.5V DC voltage range), and the internal resistance of the voltmeter is 50kΩ, which is much larger than ro, so it can be approximately considered that 1.58V is the electromotive force of the battery, or the open circuit voltage. When using the first method, the multimeter is set to 5A DC current gear, the internal resistance of the meter is RO=0.06Ω, and the measured current is 3.3A. So ro+RO=1.58V÷3.3A≈0.48Ω, ro=0.48-0.06=0.42Ω. When using the second method, the measured current is 0.395A, RF+ro+RO=1.58V÷0.395A=4Ω, and the internal resistance of the current 500mA gear is 0.6Ω, so ro=4-3-0.6=0.4Ω.
When measuring the old No. 2 battery with the first method, first measure the open circuit voltage E=1.2V, the internal resistance RO=6Ω, the reading is 6.5mA, the multimeter is set to 50mA DC current gear, ro+RO=1.2V÷0.0065A ≈184.6Ω, ro=184.6-6=178.6Ω. Using the second method, the measured current is 6.3mA, ro+RO+RF=1.2V÷0.0063A=190.5Ω, ro=190.5-6-3=181.5Ω.
Obviously, the results of the two test methods are basically the same. The slight difference in the final calculation result is caused by various factors such as reading error, resistance RF error and contact resistance, and this small error will not affect the judgment of battery power. If the capacity of the battery under test is small and the voltage is high (such as 15V, 9V laminated battery), the resistance value of RF should be adapted to increase.

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