Two test methods for battery power

The first 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 a suitable resistance to 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 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 use of the dry battery to a certain extent. 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 author used the MF47 multimeter 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 3 ohms 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 50k ohm, 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. In the first method, the multimeter is set to 5A DC current, the internal resistance of the meter is RO=0.06 ohm, and the measured current is 3.3A. So ro+RO=1.58V÷3.3A≈0.48 ohms, ro=0.48-0.06=0.42 ohms. With the second method, the measured current is 0.395A, RF+ro+RO=1.58V÷0.395A=4 ohms, and the current 500mA internal resistance is 0.6 ohms, so ro=4-3-0.6=0.4 ohms.

When the old No. 2 battery was measured by the first method, the open circuit voltage E=1.2V was measured first, the internal resistance of the meter was RO=6 ohm, the reading was 6.5 mA, and the multimeter was set to 50 mA DC current file, ro+RO=1.2V÷0.0065 A ≈ 184.6 ohms, ro = 184.6-6 = 178.6 ohms. Using the second method, the measured current was 6.3 mA, ro + RO + RF = 1.2 V ÷ 0.0063 A = 190.5 ohms, and ro = 190.5-6-3 = 181.5 ohms.

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 capacity of the battery under test is small and the voltage is high, the resistance of the RF should be adjusted to increase.

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