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# What are the detection methods of the battery of the battery?

2022/04/08

Author ：Iflowpower – Portable Power Station Supplier

Detecting whether the power of ordinary zinc manganese cuffs is sufficient, there is usually two methods. 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 table in series with an appropriate resistance, and calculate the battery by measuring the discharge current of the battery. Internal resistance, thus judge whether the battery power is sufficient.

The greatest advantage of the first method is simple. It can directly determine the power of dry batteries with a large current gear of the multimeter. The disadvantage is that the test current is large, far exceeds the limit value of the dry battery to allow the discharge current to a certain extent.

Dry battery life. The advantage of the second method is that the test current is small and safe. It generally does not affect the useful life of the dry storage battery, the disadvantage is more troublesome.

The author uses the new No. 2 dry storage battery and the old No. 2 dry battery by the author to test comparison with the above two methods.

It is assumed that RO is the internal resistance of the dry storage battery, and RO is a current table internal resistance. When using the second test method, RF is an additional series resistance, the resistance value of 3.W.

The measured results are as follows. New 2 battery E = 1.58V (measured with 2.

5V DC voltage file), the internal resistance is 50K, which is much larger than RO, so it can be approximately considered to be 1.58V is the electric power of the battery, or the open circuit voltage. When using the first method, the multimeter is used 5A DC current gear, the mesh internal resistance RO = 0.

06, the measured current is 3.3a. So RO + RO = 1.

58V & Pide; 3.3A≈0.48, RO = 0.

48-0.06 = 0.42.

When using the second method, the current is 0.395a, RF + RO + RO = 1.58V & PIDE; 0.

395A = 4, the current 500mA file internal resistance is 0.6, so RO = 4-3-0.6 = 0.

4. When the first method is measured in the first method, the open circuit voltage E = 1.2V is measured, the internal resistance RO = 6, the reading is 6.

5mA, the meanometer 50mA DC current gear, RO + RO = 1.2V & Pide; 0.0065A≈184.

6, RO = 184.6-6 = 178.6.

With the second method, the current was 6.3 mA, RO + RO + RF = 1.2V & Pide; 0.

0063A = 190.5, RO = 190.5-6-3 = 181.

5.