Two methods for detecting battery power

To check if the power of ordinary zinc-manganese dry batteries is sufficient, there are usually two methods. **The method is to estimate the internal resistance of the battery by measuring the instantaneous short-circuit current of the battery, and then determine whether the battery is sufficient. The second method is to use a current meter in series with a resistor with appropriate resistance to calculate the battery by measuring the discharge current of the battery. Internal resistance, to determine whether the battery is sufficient.
**The advantage of the method is that it is simple. The large current file of the multimeter can directly judge the power of the dry battery. 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 dry battery to a certain extent. Service life. The advantage of the second method is that the test current is small, the safety is good, and generally does not adversely affect the service life of the dry battery, and the disadvantage is that it is troublesome.
The MF47-type multimeter was used to test and compare a new No. 2 dry battery and an old No. 2 dry battery by the above two methods. Suppose ro is the internal resistance of the dry battery, RO is the internal resistance of the ammeter. When using the second test method, RF is an additional series resistance with a resistance of 3Ω and a power of 2W.
The measured results are as follows. The new No. 2 battery E=1.58V (measured with 2.5V DC voltage), the internal resistance of the voltmeter is 50kΩ, which is much larger than ro, so it can be approximated that 1.58V is the electromotive force of the battery, or open circuit voltage. When using the ** method, the multimeter sets 5A DC current file, 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. With 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 is 0.6Ω, so ro=4-3-0.6=0.4Ω.
When measuring the old No. 2 battery with ** method, first measure the open circuit voltage E=1.2V, the internal resistance of the meter is RO=6Ω, the reading is 6.5mA, the multimeter sets the 50mA DC current file, ro+RO=1.2V÷0.0065A ≈184.6Ω, ro=184.6-6=178.6Ω. Using the second method, the measured current was 6.3 mA, ro + RO + RF = 1.2 V ÷ 0.0063 A = 190.5 Ω, and ro = 190.5-6-3 = 181.5 Ω.
Obviously the results of the two test methods are basically the same. The slight difference in the final calculation results is caused by many factors such as reading error, resistance RF error and contact resistance. This small error does not affect the judgment of the battery power. If the battery to be tested has a small capacity and a high voltage (for example, a 15V, 9V laminated battery), the resistance of the RF should be increased.