I am new to the world of electronics, but keen to learn. The below all relates to the 12 V circuit of my van.

I have picked up a campervan which has a second "leisure" battery which powers items in the rear, such as LED lights etc. These lights run off a caravan-style "power management system" and this system is fed from the leisure battery via a 25 A fuse.

When driving, the van alternator (90 A) charges the starter battery like a normal car. There is a split charge relay (CB1FM-12v) which closes the relay when the ignition is on and this allows the current into the +ve terminal on the leisure battery also. This +ve terminal on leisure battery also has a wire off to the PMS system in the rear via a 25 A fuse.

My question is this: if the alternator can produce 90 A then when the relay switch is closed (ignition is on), why doesn't the relay fry as the relay is rated at 40 A?

Also, why doesn't the fuse going to the PMS blow as the fuse is only 25 A?

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  • 2
    \$\begingroup\$ You are confusing the ability of a source to provide current (the 90A in your situation), with that of a load drawing current. \$\endgroup\$
    – mike65535
    Commented Sep 4, 2019 at 15:17

1 Answer 1


Alternators (or any kind of electrical generator) work by providing a voltage between their terminals. The current provided by such alternator depends on what you connect to it (and the generated voltage, of course). The 90 A rating means you can draw at most 90 A without frying it, it does not mean that the generator is forcing 90 A through the circuit. So you can bet that if the relay and the fuse are not blowing, its because the current through them is below their rating. Hope that helps :)

  • 1
    \$\begingroup\$ If does, thanks. So in this context the "draw" would be the leisure battery charging itself. The current from the +ve starter battery is going through the relay and into the +ve of the leisure battery. How would the "draw" of the leisure battery charging be limited? I understand the flow is dictated by the higher voltage which would be coming from the alternator/starter battery, but i don't know how a battery being charged manages the amps it draws? \$\endgroup\$
    – marc
    Commented Sep 5, 2019 at 15:09
  • \$\begingroup\$ Can anybody elaborate on this last question of marc? I was wondering the same... \$\endgroup\$
    – Max
    Commented Sep 17, 2019 at 23:19
  • \$\begingroup\$ Batteries are weird beasts. Accurate models are kinda complex, but you can model them considering them as a variable voltage source in series with a small resistor. The voltage source changes its value depending on the state of charge of the battery. The series resistor models the internal resistance and acts as a limiting factor to the current draw. Batteries also have a self-discharge rate, which can be modeled as a big parallel resistor. \$\endgroup\$
    – Aldiom
    Commented Sep 26, 2019 at 14:15

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