What happens if i connect 2 batteries, 12V & 6V, in parrallel to a 6V rating bulb? How much volt it receives ?

Answer 1

(12 + 6)/2 = 9 volts

But, the problem is that the current in each battery will skyrocket.

Easiest way to convince yourself of this is draw the circuit you are talking about and include a series resistance for each battery, call them both R, and let the light bulbs resistance be RL, and solve for Vout.

You'll get:

Vout = (RL/(2RL +R))(Vb1 +Vb2)

now take the limit as R -->0 you get

Vout = (Vb1 + Vb2)/2

NOW, calculate the currents

(Vb1 -(Vb1 + Vb2)/2)/R = (Vb1 - Vb2)/(2R)

again, the limit as R -->0 you get

(Vb1 - Vb2)/0 which is infinite, this means that the battery will be destroyed.

Answer 2

You will be perfect till you add the 12 volt battery. Before that the 6V battery would light the 6V lamp. Once you put the 12V battery across the 6V battery there will be 12V across the bulb and it will be gone in a one bright flash. Similarly the 6V battery will receive an over voltage and be overheated and ruined.

Answer 3

First of all, why would you want to connect two dissimilar batteries to each other in parallel? Series would be okay, but not in parallel. If you used a bucking arrangement where the positive terminal of the 12 volt battery were connected to the negative terminal of the 6 volt battery, then this series arrangement would give you a 6 volt output, but at a loss of the 6 volt battery, eventually due to heating from the current flow through it. Drop the 12 volt battery from what you are trying to do. Ignoring the internal resistance of the battery, especially if it is a lead acid, Ni-Cad, or Lithium battery since their internal resistance tends to be low because of their current density (ability to provide high load currents for a considerable length of time) you would get the rated voltage of the battery at a current level dependent on the resistance of the lamp filament.

Answer 4

Your imaginary Qn can be solved by applying elementary ohms law and Kirchhoff's law for various loops provided the internal resistance of the 2 batteries and the resistance or wattage of the bulb are known. But I wonder whether the asker knows the basic rule that the voltages of the batteries to be paralleled should have closely same voltages as any unequal voltage will result in a circulating current from the higher voltage battery to the lower one. If the circulating current limited only by the internal resistances becomes large, a burst can take place. Your imaginary Qn is meaning less and dangerous as an experiment.Pl avoid.

Answer 5

Battery charger voltage must match the battery that's being charged. A 12 V charger will melt a 6V battery, while a 6V charger won't charge a 12V battery.

Answer 6

If you connect a 12V. and a 6V. battery in parrallel,
the 12V. battery will discharge through the 6V. battery.
The voltage and current at the bulb will be dependent on the internal resistance of the batteries.

Answer 7

You have to know the internal resistance of each battery to answer this question as well as the wattage of the bulb. It can then be treated very simply by the superposition theory.

Answer 8

Dont do it, you will destroy both batteries in a few minutes. Or cause them to explode or vent, if they are Gel cells. Lithium could be worse (fires have started).

Answer 9

I think they will explode. It will receive about 7 volt, I guess.