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Why is it that the charger stops charging when the battery is full?

I answered myself that the battery is like a variable resistor and when it empty the resistor is low and when it full of energy the resistor is high but I cant solve the problem that why the current pass when both battery and charger have the same volt "we know that the current pass from high to low volt"

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    \$\begingroup\$ Actually the voltages are not equal. Most chargers use a constant current supply and it maintains a voltage slightly higher than the voltage of the cell. \$\endgroup\$
    – chamod
    Jul 23 '16 at 2:08
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    \$\begingroup\$ The battery emulates a capacitor more than a resistor. \$\endgroup\$
    – user105652
    Jul 23 '16 at 2:43
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The voltage is not the same.

When the battery is charging, the charger output will be just slightly higher than the battery voltage in order to drive current through the wire between them. However, the voltage difference might be only a few microvolts, so it will read the same on a multimeter that doesn't have very good resolution.

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It will become much clearer if you use a model for your battery: a perfect voltage source in series with a perfect resistor.

For simplicity (note: I don't think there's any charger on the market that works like that), I'll assume that your charger is a perfect voltage source.

Starting with a discharged battery, the charger will provide a voltage above the "perfect" battery voltage - but the voltage you'd measure "outside" the battery would match the charging voltage - the difference is the result of V=RI applied to the parasitic resistor. As charging completes, the "perfect" battery voltage increase. Since the charger voltage stays the same, the current lowers, so V=RI stays true on the resistor. When the battery voltage reaches the charging voltage, there isn't room for any voltage across the resistor... and charging stops.

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