How consumers to properly test the dry battery do?

Correctly detect ordinary zinc-manganese dry battery power is sufficient, there are usually two ways:

1, the first method is by measuring the instantaneous short-circuit current to estimate the battery internal resistance of the battery, and then determine if the battery is fully charged. The biggest advantage of this method is simple, high-current file with a multimeter can directly determine the dry battery, the disadvantage is the test current is large, far exceeding the limits of battery discharge current is allowed to a certain extent, affect battery life .

2, the second method is to use ammeter in series with a resistor of proper value by measuring the discharge current of the battery internal resistance of the battery is calculated in order to determine if the battery is fully charged. The advantage of this approach is to test the current is small, good safety, generally will not adversely affect the life of the battery, the drawback is more trouble.

Specific methods are as follows:

With MF47 multimeter on a New No. 2 batteries and a battery old on the 2nd of the two methods were used to test contrast. Assuming the battery internal resistance ro, RO is the resistance meter, with the second test method, RF is an additional series resistance, resistance 3 ohms, power 2W.

Measured results are as follows: The new 2 batteries E = 1.58V (2.5V DC voltage profile with measurement), voltmeter resistance is 50k ohms, much larger than ro, it can be similar to that 1.58V is the battery electromotive force, or open-circuit voltage. Use the first method, the multimeter set 5A DC current file, meter resistance RO = 0.06 ohms, the measured current is 3.3A. So ro RO = 1.58V ÷ 3.3A ≈ 0.48 Ohm, ro = 0.48-0.06 = 0.42 ohms. Using the second method, the measured current of 0.395A, RF ro RO = 1.58V ÷ 0.395A = 4 ohms, current 500mA file resistance of 0.6 ohms, so ro = 4-3-0.6 = 0.4 ohms.

Old 2 batteries when measured using the first method, first the measured open circuit voltage E = 1.2V, the internal resistance RO = 6 ohm meter reading is 6.5mA, multimeter set to 50mA DC current file, ro RO = 1.2V ÷ 0.0065A ≈ 184.6 ohms, ro = 184.6-6 = 178.6 ohms. The second method, the measured current is 6.3mA, ro RO RF = 1.2V ÷ 0.0063A = 190.5 ohms, ro = 190.5-6-3 = 181.5 ohms.

To sum up: apparently the result of two test methods are basically the same. The end result of a slight difference is due to the reading error, the error and the contact resistance RF resistance, and many other factors, such small errors will not affect the judgment on battery power. If the measured battery capacity is small, high voltage, it should adapt to increasing the resistance of the RF