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# Part 4: Battery Banks

Battery banks in small power systems normally have nominal voltages of either 12V or 24V however, lead acid batteries are available from 4V up to 24V. Batteries can be combined in series (figure 7a) so that their voltages are added together: two 12V batteries in series will provide 24V. Although voltages are add the same current will flow though each battery, so that two identical batteries 12V in series supplying 5A to a load each supply 5A: therefore the Ah capacity of two identical batteries in series is the same as one battery on its own. The total internal resistance (RC) of batteries in series will equal the internal resistances of the individual batteries added together. Figure 7: (a) 3 batteries and their internal resistances in series; (b) 3 batteries and their internal resistances in parallel.

Example
(a) A 12V battery with an internal resistance of 0.3Ω is connected to a load with a resistance of 4Ω.What Current will flow? $I=\dfrac{E}{R+R_C}=\dfrac{12}{4+0.3}=2.8A$

(b) What current will flow in the same load if the current is supplied by two similar 12V batteries connected in series? $I=\dfrac{E}{R+R_C}=\dfrac{12\times 2}{4+\left (2\times 0.3\right )}=5.2A$

When batteries are connected in parallel (figure 7b) they all operate at the same voltage and only identical batteries should every be connected in parallel. With this arrangement the total current being provided is split equally between the batteries so that two 12V batteries supplying 5A contribute 2.5A each, therefore the total capacity of these two batteries is twice the capacity of one battery supplying 2.5A (which in turn will be greater than the capacity of one battery supplying 5A). The internal resistances must be summed as if they are resistors in parallel; that is that the reciprocal of the total resistance equals the sum of the reciprocals of each resistor.

Example
From the previous example:
(c) If three of the same 12V batteries are connected in parallel to the 4Ω what current flows?

Total RC: $\dfrac{1}{\text{Total}~R_C}=\dfrac{1}{3}\times 3=10\Omega$

Therefore: $I=\dfrac{E}{R+R~\text{Total}~R_C}=\dfrac{12}{4+0.01}=3.0A$

Battery banks may be constructed from several strings of batteries in series connected in parallel (figure 8); note that all of the batteries must be identical and of course all of the series strings must contain the same number of batteries. The EMF of such a bank is equal to the number of batteries in series multiplied by the battery EMF, the Ah capacity is equal to the capacity of one battery (at the appropriate rate) multiplied by the number of string in parallel and the total internal resistance is given by: $\dfrac{R_C\times \text{number of batteries in a string}}{\text{number of strings in parallel}}$ Figure 8: Batteries their internal resistances in a bank containing 3 strings of 6 batteries each.

Example
Continuing the previous example:
(d) If a battery bank consists of three strings of two batteries each what current will flow?

The EMF of the battery bank is: $2\times 12=24V$

The total internal resistance is: $\dfrac{2\times 0.3}{3}=0.2\Omega$

Therefore: $I=\dfrac{E}{R+\text{Total}~R_C}={24}{4+0.2}= 5.7A$

Note that this is about the same current that is supplied by two batteries in series however, since there are three strings in parallel so the bank will be able to supply this current for more than three times as long (more than three times because the discharge rate has reduced). If the batteries have a capacity of 50Ah each at the 20 hour rate the bank will have a capacity of (3 x 50) 150 meaning that it can supply 7.5A for 20 hours, therefore it could supply 5.7A for slightly longer than 20 hours.