What are the methods for detecting battery power?

There are usually two methods to check whether the power of the ordinary zinc-manganese dry battery 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 to determine whether the battery is sufficient; the second method is to use an ammeter in series with a resistor of appropriate resistance, and calculate the battery by measuring the discharge current of the battery Internal resistance to determine whether the battery is sufficient. The biggest advantage of the first method is its simplicity. It can be directly judged by the high current gear of the multimeter. The disadvantage is that the test current is very large, which far exceeds the limit value of the allowable discharge current of the dry battery, which will affect it to a certain extent. Service life of dry battery. The advantage of the second method is that the test current is small and the safety is good. Generally, it will not adversely affect the service life of the dry battery. The disadvantage is that it is more troublesome. The author uses the MF47 type multimeter to test and compare a new No. 2 dry battery and an old No. 2 dry battery respectively with the above two methods. Assuming that ro is the internal resistance of the dry battery, and RO is the internal resistance of the ammeter, when using the second test method, RF is an additional series resistance with a resistance value of 3 and a power of 2W. The measured results are as follows. The new No. 2 battery Eu003d1.58V (measured with 2.5V DC voltage file), the internal resistance of the voltmeter is 50k, which is much larger than ro, so 1.58V can be approximated as the electromotive force of the battery, or open circuit voltage. When using the first method, the multimeter is set to 5A DC current, the internal resistance of the meter is ROu003d0.06, and the measured current is 3.3A. So ro+ROu003d1.58Vu0026pide;3.3A≈0.48, rou003d0.48-0.06u003d0.42. When using the second method, the measured current is 0.395A, RF+ro+ROu003d1.58Vu0026pide;0.395Au003d4, and the internal resistance of the current 500mA range is 0.6, so rou003d4-3-0.6u003d0.4. When the old No. 2 battery is measured by the first method, first measure the open circuit voltage Eu003d1.2V, the internal resistance of the meter ROu003d6, the reading is 6.5mA, the multimeter is set to the 50mA DC current range, ro+ROu003d1.2Vu0026pide;0.0065A ≈184.6, rou003d184.6-6u003d178.6. Using the second method, the measured current is 6.3mA, ro+RO+RFu003d1.2Vu0026pide;0.0063Au003d190.5, rou003d190.5-6-3u003d181.5. Disclaimer: Some pictures and content of articles published on this site are from the Internet. If there is any infringement, please contact to delete. Previous post: Will lithium batteries be replaced by solid state batteries?