There are two methods to detect whether the power of ordinary zinc manganese dry battery is sufficient. One 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 an appropriate resistance in series with an ammeter, and calculate the internal resistance of the battery by measuring the discharge current of the battery, so as to judge whether the battery power is sufficient
The major advantage of the above methods is simple. The power of the dry battery can be directly determined by using the large 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 service life of the dry battery to a certain extent. The advantage of the second method is that the test current is small, the safety is good, and the service life of dry batteries will not be adversely affected. The disadvantage is that it is more troublesome
Use MF47 multimeter to test and compare a new No. 2 dry cell and an old No. 2 dry cell with the above two methods respectively. Assuming that ro is the internal resistance of the dry cell and RO is the internal resistance of the ammeter, RF is an additional series resistance with a resistance value of 3 Ω and a power of 2W when using the second test method
The measured results are as follows. The new No. 2 battery E=1.58V (measured with 2.5V DC voltage range), the internal resistance of the voltmeter is 50k Ω, which is far greater 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 this method, the multimeter is set to 5A DC current gear, the internal resistance of the meter RO=0.06 Ω, and the measured current is 3.3A. So ro+RO=1.58V ÷ 3.3A ≈ 0.48 Ω, ro=0.48-0.06=0.42 Ω. When using the second method, the measured current is 0.395A, RF+ro+RO=1.58V ÷ 0.395A=4 Ω, and the internal resistance of 500 mA gear is 0.6 Ω, so ro=4-3-0.6=0.4 Ω
When the old No. 2 battery was measured in one way, the open circuit voltage E=1.2V, the internal resistance of the meter RO=6 Ω, and the reading is 6.5mA. The multimeter was set at 50mA DC current, and ro+RO=1.2V ÷ 0.0065A ≈ 184.6 Ω, ro=184.6-6=178.6 Ω. Using the second method, the measured current is 6.3mA, ro+RO+RF=1.2V ÷ 0.0063A=190.5 Ω, ro=190.5-6-3=181.5 Ω
Obviously, the results of the two test methods are basically consistent. The slight difference in the final calculation results is caused by reading error, resistance RF error, contact resistance and other factors, which will not affect the judgment of battery power. If the capacity of the battery to be tested is small and the voltage is high (such as 15V, 9V laminated battery), the RF resistance should be appropriately increased