Hi all,

I have a general question.

When we have a circuit containing heavy loads, (1) why does our battery voltage drop for sometime? (2)How long does it take to recover?

I have learnt that during the circuit turn on, heavy currents are drawn and this drops a lot of the voltage across the internal resistance of the battery and that's why the battery voltage drops. (3)Is this right?

Can someone provide better light on my understanding? Can someone provide answers for all for my understanding

Thanks.

 

Battery drops its voltage the same way as a reservoir drops its level when drained.
A battery can recover some of its chemical reaction capability to restore its voltage after some time that depends on the current that was drawn.

 

It took nearly 45 seconds to find this. You are encouraged to cultivate your web search skills.

https://www.quora.com/Why-does-voltage-of-a-battery-drop

 

These references might help you with a little more insight
into cause and effect of battery chemistry/electricals.

Google "battery handbook pdf", several reference sources.

Regards, Dana.

 

Battery drops its voltage the same way as a reservoir drops its level when drained..

That's an incorrect analogy ... the reservoir level does not drop (appreciably) when momentary high flow rates occur. The correct analogy is to say the pipe work from the reservoir (int res of battery) reduces flow to the user who has his own pipe work (load).

This is all about internal resistance .... if the int. resistance is the same as the load resistance then the voltage across the battery will be half open circuit voltage.

The voltage should only drop for as long as the load is applied ... the battery should recover instantly to it's original voltage (minus the small drop in voltage to account for consumption while load was applied)

 

Hello,

You also might want to read this page of the eBook:
https://www.allaboutcircuits.com/textbook/direct-current/chpt-11/battery-construction/

Bertus