How to easily measure battery power

There are usually two ways to detect whether the ordinary zinc-manganese dry battery is sufficiently charged. 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 sufficiently charged; the second method is to use an ammeter in series with a resistor of appropriate resistance, and calculate the internal resistance of the battery by measuring the battery discharge current, so as to determine whether the battery is sufficiently charged.

The biggest advantage of the first method is that it is simple. You can directly determine the battery power by using the high current range of the multimeter. The disadvantage is that the test current is very large, far exceeding the limit 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 it generally does not have an adverse effect on the service life of the dry battery. The disadvantage is that it is more troublesome.

The author used the MF47 multimeter to test a new No. 2 dry cell and an old No. 2 dry cell using the above two methods. Assuming ro is the internal resistance of the dry cell, RO is the internal resistance of the ammeter, when using the second test method, RF is the 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 a 2.5V 直流电'); companyAdEvent.show(this,'companyAdDiv',[5,18])"> DC voltage range), 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 the 5A DC current range, the internal resistance of the meter RO=0.06Ω, and the measured current is 3.3A. Therefore, 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, the internal resistance of the current 500mA range is 0.6Ω, so ro=4-3-0.6=0.4Ω.

When the old No. 2 battery is measured by the first method, the open circuit voltage E=1.2V is measured first, the internal resistance of the meter RO=6Ω, the reading is 6.5mA, the multimeter is set to 50mA DC current range, ro+RO=1.2V÷0.0065A≈184.6Ω, ro=184.6-6=178.6Ω. Using the second method, the current is measured to be 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 results is caused by many factors such as reading error, resistance RF error and contact resistance. This slight error will not affect the judgment of the battery power. If the capacity of the battery being tested is small and the voltage is high (such as 15V, 9V laminated battery), the resistance value of RF should be increased accordingly.