Can a high internal resistance 9v battery be applied in parallel to 3vDC battery D cell amps pack to run a 10 amp 6V DC headlight bub at full brightness? Neither source power by itself can run bulb. Superimposed 2 sources to operate load bulb resistance? 9 volts DC added to separate power pack. A variable power resistor + amp meter can substitute for bulb, as a higher current reading obtained.
IsaiSporer
- Feb 19, 2025
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Hello Geometry Dash!
I think a 9V battery with high internal resistance can be used in parallel with a 3VDC D portable amp pack to run the 10 amp 6V DC headlight BUB at full brightness. Why not? More convincing reasons.
I think a 9V battery with high internal resistance can be used in parallel with a 3VDC D portable amp pack to run the 10 amp 6V DC headlight BUB at full brightness. Why not? More convincing reasons.
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Hello! That setup sounds interesting, but using a 9V battery in parallel with a 3VDC pack could cause voltage imbalance and possible damage to your circuit. It’s better to match voltages or use a regulator for safety. face rater sometimes testing the balance first can save your components from burning out.
Boost the LV supply, boost switching regulator.
Then implement current sharing on outputs ?
Or implement outputs as current sources, scaling each source for its appropriate
current manageable load ?
When you say 9V battery what is its current rating ? Same for D cell pack ? How long
you want to run headlight off these batteries ?
Then implement current sharing on outputs ?
Or implement outputs as current sources, scaling each source for its appropriate
current manageable load ?
When you say 9V battery what is its current rating ? Same for D cell pack ? How long
you want to run headlight off these batteries ?
Last edited:
Delta Prime
- Jul 29, 2020
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Asmost certainly not. Assuming the 3 V battery has a low ESR, it will "short out" the 9 V source, holding its output (and the voltage across the lamp) to 3 V.
You don't way what the "high internal resistance" is. If you assign values to the two battery's output resistances, such as 1 ohm and 10 ohms, you can draw the complete circuit and solve for the lamp voltage with Ohm's Law.
Also, it's gonna take a lot of D cells to source 10 A.
ak
eithernebulous
- Nov 12, 2025
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No — that won’t work. Paralleling a 9V and a 3V source to try to make a 6V, 10 A supply is asking them to do something they weren’t designed for: their voltages don’t match, so they’ll fight each other and dump current into one another; a typical 9V or small 3V pack can’t supply anywhere near 10 A without huge voltage sag or overheating; internal resistance will collapse the voltage under load so the headlight won’t reach full brightness; and you risk damaging the batteries or causing heat/fire. If you need 6 V at 10 A, use a proper low-impedance 6 V power source or a correctly rated battery pack or converter instead.
No. That is like shorting out the weaker cell. (weak = high Rs)
Parallel voltage sources that differ in voltage will result in circulating currents where the source with the lower impedance dominates the superimposed parallel voltage in cells.
I made a simulation with my estimate of the series resistance by recalling from memory the short circuit current of an example of random 9V cell and two D cells and chose the series resistance.
Using Kirchoff's Laws the voltage & current can be computed just as performed in this simulation.
Press the dummy remote switch so you don't mess up the connections for a momentary connection between cells and see the effects of the "Superposition Principal". Just like jump starting a car unless the voltages are close you get a big spark on connection.
I used an ideal voltage source and added R to make a simple battery models. They are actually a little more like complex Supercaps with an activation voltage with thousands of Farads.
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