# Discharge Behavior of a Battery

I am working with a zinc air battery (emf 1.3 V). I have plugged a resistance of 100 ohm to this battery and I have recorded the voltage at the battery. When the circuit is opened, the voltage is 0. When I have closed the circuit, U(battery)=-U(resistance)=Ri=emf-ri so i=emf/(R+r) and U(resistance)=R*emf/(R+r). According to the last formula, the current should have a constant value but on my experimental graph, the current is slightly decreasing (as you can see on the graph)... could you explain this phenomenon ? • DId u open the air hole and is it free to breath? – Trevor_G May 31 '17 at 16:42
• the battery is free to breath – velleda May 31 '17 at 17:26
• Your small battery is performing correctly. The small flat part of your curve at the start is the constant voltage portion, The decay curve is the end of life. If you use a 1000 and a 10K resistor you should see the flat portion is longer and judge how little current you can draw. – KalleMP Jun 1 '17 at 5:01
• ok so the decreasing part is because I consume my battery. If I was using a generator, this current would be flat. – velleda Jun 1 '17 at 8:02

## 1 Answer

When you place a resistor across a battery to discharge it the voltage of the battery will decrease as the discharge proceeds. This translates directly to the voltage across the resistor and with a decreasing voltage the current through the resistor will also decrease. You will not achieve a constant current load on the battery by discharging with a simple resistor.

It is possible to discharge a battery at constant current (as long as the battery can supply at least the constant current load level) but it requires a special circuit that uses a BJT or MOSFET as the load across the battery. The drive signal to the Load device is controlled by an opamp that regulates the signal to the level needed to keep the load current constant. Load current in such circuit is typically monitored as the voltage drop across a low value current sense resistor.