Please bear with me if this is a fundamental question.

Let's suppose we have an RC circuit, charging a capacitor up to the voltage of a battery of 9V. The capacitor is currently at 9V, we switch the 9V battery with a 3.3V one.

Because of the difference in potential of 5.7V between the capacitor and the battery, does current now flow into the battery?

What are the dangers when this happens? Am I trying to charge it at this point?

  • \$\begingroup\$ Discharge current I max = ΔV/ΣESR(cap+bat)[mΩ] while Duration Tau=Ccap * ΣESR(cap+bat) Battery is 10kF ~>=50mΩ, Cap is what? 100uF 10mF 1F? ESR? probably harmless unless a dendrite moves and causes internal battery short and starts to heat up, as you toss it in the dirt outside before it reach thermal runaway \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 20 at 18:33

Yes, current will flow into the battery, and in a sense you are trying to charge the battery.

The possible danger is that the battery is not designed to be recharged, or that the value of \$R\$ is not high enough to limit the current to a safe level. In either case there is the possibility of damaging the battery, perhaps causing a fire.

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