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Replacing the Battery
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<blockquote data-quote="jarance" data-source="post: 351252" data-attributes="member: 21"><p>Nice write up.. Actually the starter can draw a current of <u><strong>200 Amp</strong></u> for cranking.</p><p></p><p>In terms of power, the higher the voltage, the lower the current will be draw for the same power rating. <span style="color: Blue"><em>Again using Power = Voltage x Current theory. </em></span></p><p></p><p>however, 48V means 12VDC X 4 numbers of battery in series or 1.2 VDC X 40 cells. Even if you reduce the current drawn, you still need to compact 40 cells into 1 battery bank. The reason that it is not picking up is because there is no place to put 4 nos of 12 VDC battery in the engine compartment or the boot. </p><p></p><p>In addition, most electronic like cmos & IC are operating at 5 VDC or 3.84VDC. Imagine you have to put a lot of voltage regulators to regulate the 48VDC to 5 VDC supply for all these component to work properly. It will be a nightmare for car manufacturer and battery manufacturer.</p></blockquote><p></p>
[QUOTE="jarance, post: 351252, member: 21"] Nice write up.. Actually the starter can draw a current of [U][B]200 Amp[/B][/U] for cranking. In terms of power, the higher the voltage, the lower the current will be draw for the same power rating. [COLOR="Blue"][I]Again using Power = Voltage x Current theory. [/I][/COLOR] however, 48V means 12VDC X 4 numbers of battery in series or 1.2 VDC X 40 cells. Even if you reduce the current drawn, you still need to compact 40 cells into 1 battery bank. The reason that it is not picking up is because there is no place to put 4 nos of 12 VDC battery in the engine compartment or the boot. In addition, most electronic like cmos & IC are operating at 5 VDC or 3.84VDC. Imagine you have to put a lot of voltage regulators to regulate the 48VDC to 5 VDC supply for all these component to work properly. It will be a nightmare for car manufacturer and battery manufacturer. [/QUOTE]
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