In this type of battery connections, two or more sets of batteries or cells are connected in parallel and then in series to the other set(s). This type of configuration gives larger charge [ampere hour (Ah)] capacity which is from the parallel batteries. Therefore, the series connection boosts the voltage to up to the required input voltage of the external load.

In this configuration, we sum up the power of the each of the sets of the batteries or cells to give the output power capacity of the configuration. Most of the laptop computers’ batteries use this type of configuration and are embedded with a unique cells or batteries monitoring, charging and controlling systems (circuits). The number of the sets that are in parallel are written followed by “P” and in series as “S.” For example if there are three (3) sets of batteries or cells in parallel and two (2) in series, it is written “3P2S.”

###### Right Configurations

The below configuration is the right way of connecting three batteries (two batteries in parallel and one series). Because, all the batteries have the same specs. Hence, the total capacity of the arrangement is 24V80Ah.

###### Wrong Configurations

Batteries are said to be wrongly configured if they are of different specifications i.e. their voltage (V) or ampere-hour (Ah) are different, and also if they are connected in such a way that the terminals for external load connections are alike i.e. if both are positives or negatives.

The below configurations are wrong.

###### Output Power

The output power capacity of this configuration is the sum of the power in parallel and series sets

Formula

*P _{(Series-parallel)} = P_{parallel} + P_{series}*

Example

Find the total output power of the configuration of Two 3.7V, 2Ah batteries in parallel and One 3.7V, 2Ah is connected in series to them.

###### Solution

*P _{(Series-parallel)} = P_{parallel} + P_{series}*

From the question, 2 batteries each has 3.7V, 2Ah specifications are connected in parallel, therefore solve for it first.

Batteries in parallel

*P _{parallel} = V × Ah_{1 }+*

*Ah*

_{2 }*=3.7×2+2 *

*=3.7×4 *

*=14.8VAh or Wh*

Then solve for the single one which also is a 3.7V, 2Ah battery connected in series to the parallel ones.

Battery in series

*P _{series }= V_{1} × Ah (only one battery)*

*=3.7×2 *

*=7.4VAh or Wh *

Therefore, add the parallel power to the series power

*P _{(Series-parallel)} = P_{parallel} + P_{series}*

*=14.8+7.4 *

*=22.2VAh or Wh*

Also, read Batteries in Series and Parallel Calculations or Batteries or Cells maintenance tips