# On the utilization rate of new and old batteries

In the study of new and old batteries, suppose the new battery's electromotive force is E, the internal resistance is r, the old battery's electromotive force is Ec, the internal resistance is rc, and the load resistance is R. When two new batteries are connected in series, the road end voltage is U = IR = 2ER + 2rR (1), the efficiency of the whole circuit is G = I2RI2 (R + 2r) = RR + 2r (2), when an old battery is connected in series with a new battery, the voltage obtained by the load is Uc = E + EcR + r + rcR (3), the efficiency of the power supply is Gc = RR + r + rc (4), because the electromotive force of the old battery becomes smaller, the internal resistance becomes larger, that is Ecr. Comparing equations (1) and (3) and equations (2) and (4), it can be seen that the mixed use of new and old dry batteries reduces the output voltage and efficiency of use compared to full-use new batteries.

However, if there is only one new battery, the output voltage is U0 = ER + rR, and its efficiency is G0 = RR + r. After adding an old battery, the voltage increase across the load is \$ U = Uc-U0 = E + EcR + r + rcR-ERR + r = Ec (R + r) -Erc (R + r + rc) (R + r) R (6), then the output power of the old battery is \$ P = IcEc-Ic2rc = Ic (Ec-E + EcR + r + rcrc) = Ic1Ec (R + r) -ErcR + r + rc2 (7). It can be seen from (6) and (7):

When Ec (r + R) When Ecrc = ER + r, \$ U = 0, \$ P = 0, the old battery has no effect on the voltage and power of the load, and adding the old battery has no effect.

When Ecrc> ER + r, the household ladder \$ U> 0, \$ P> 0, and adding a series of old batteries will increase the output voltage and useful power of the battery pack, and it is effective to use new and old batteries together.

The efficiency of the power supply is equal to the ratio of the voltage obtained by the appliance to the electromotive force of the power supply. When only a new battery is used for power supply, the efficiency is G0 = U0 / E0. As mentioned earlier, the new and old batteries are mixed, and the efficiency of the entire battery pack is reduced. However, if only the efficiency of the new batteries is considered, then Gc0 = UcE0, when Ecrc> ER + r, \$ U> 0, that is \$ Uc> U0, so Gc0> G0. This shows that at this time, the old battery is used with the new battery, which actually improves the utilization rate of the new battery.

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