# The effect of battery in parallel

1. Nov 24, 2009

### czaroffishies

Here is a diagram of the circuit I am analyzing:

http://www.sixtheorem.com/images/1.png [Broken]

I observe that bulb 2 does not light when the switch is closed. Could someone please explain what is happening here?

And, what happens if I reverse the direction of battery B?

Last edited by a moderator: May 4, 2017
2. Nov 24, 2009

### Redbelly98

Staff Emeritus
Is this homework, or a question for a class you are taking?

3. Nov 24, 2009

### czaroffishies

It would certainly help my understanding in class, but it's not homework... I mostly was just wondering more specifically what is happening here. Should I move it?

4. Nov 24, 2009

### Redbelly98

Staff Emeritus
It can stay here if it's not homework.

Are you aware that circuit elements in series always have the same current, and elements in parallel always have the same voltage?

Note that the open switch is in series with the Battery B and Lamp 2.

p.s. Welcome to Physics Forums!

5. Nov 24, 2009

### SystemTheory

Do you recognize the difference between a closed circuit and open circuit?

6. Nov 24, 2009

### czaroffishies

Hah, yes of course. Just made the edit. :P

7. Nov 24, 2009

### SystemTheory

The sketch shows the voltage at three nodes X, Y, and Z. The voltage on a node is uniform since lines represent perfect conductors in a circuit sketch.

Ohm's Law applies on the lit bulb and the unlit bulb:

V = IR

where V is the voltage difference across the two terminals and resistance R is fairly small for a conventional bulb. The bulb needs plenty of current I to light up, a small current flow will pass through but produce no visible light. Too much current will burn up the fillament.

If you reverse battery B (in the sketch) draw a new set of nodes and find the voltage on each bulb to answer your own questions.

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8. Nov 24, 2009

### mikelepore

If A and B are exactly equal, there is no voltage across lamp B. If they are unequal, there is a voltage across lamp B. (Assuming that the switch is closed.)

Lamp 1 has nothing to do with the question you asked. The behavior of lamp 2 is the same whether lamp 1 exists or not.

Your topic title "The effect of battery in parallel" is wrong. This is a series connection.

9. Nov 24, 2009

### Staff: Mentor

Are you saying you have actually built this circuit and are seeing that it doesn't light bulb 2? Because it looks to me like lighting the circuit should either light bulb 1 brightly and bulb 2 dimly or blow bulb 1 and kill the whole circuit.

Last edited by a moderator: May 4, 2017
10. Nov 25, 2009

### czaroffishies

But why is that, if you don't mind me asking?

11. Nov 25, 2009

### czaroffishies

Yep, though it's probably as System said... a small current, but not enough to observe light.

12. Nov 25, 2009

### diazona

The image that SystemTheory posted is getting at the right idea...

Basically, the whole bottom section of wire has to be at the same electrical potential. (Since it takes no energy for current to flow through an ideal wire, there can be no change in potential when moving from one part of the circuit to another part that is connected by only wires) Let's call that potential 0 - it's the part labeled X in SystemTheory's diagram.

The purpose of a battery is to fix the potential difference between its two ends. So, again with reference to SystemTheory's diagram, the top section of wire will be at a fixed potential Z because it is separated from the bottom section by a battery, namely battery A. If the voltage of battery A is 9 volts, then Z - X = 9 volts.

The same reasoning applies to the section labeled Y, between the bulb and battery B. If the voltage of battery B is 9 volts, then Y - X = 9 volts.

Now, can you see, if batteries A and B have the same voltage, what is Z - Y? That is the voltage across the bulb, and a bulb that has zero (or nearly zero) voltage across it will not light up.

13. Nov 25, 2009

### mikelepore

The standard procedure is to travel mentally around a closed loop, calling voltage rises positive terms and calling voltage drops negative terms, and add the terms. You will find the two batteries pointing in opposite directions, so that a combined voltage of (+A)+(-B) is applied across the lamp. That's A-B, which is zero (if A=B).

Same thing with a typical flashlight that takes two 1.5 volt batteries, to apply 1.5 + 1.5 = 3 V across the bulb. If you put one of the bulbs in backwards, the combination will become 1.5 - 1.5 = 0 V.

14. Nov 25, 2009

### SystemTheory

If you're actually building these circuits to learn I suggest buying an inexpensive digital multimeter. Avoid using high voltage batteries and be sure not to create a short circuit across the batteries!

I built a device that works on two 1.2V recharagable cells in series, but inside a case that accepts 4 cells. I inserted the jumper wrong and shorted the batteries 2.8V plus to minus terminals. The batteries became hot and one of them melted a bit near the negative terminal within less than 1 minute, due to high current flow. They would not accept a recharge and I discarded them for safety.

Do a google search for some tutorials on basic electricity. One OK reference is here:

http://www.allaboutcircuits.com/vol_1/chpt_1/2.html

Note the distinction between electron flow and conventional current flow if you study this reference. Mostly you want to use conventional current flow but understand that electons actually move in the other direction.

15. Nov 25, 2009

### czaroffishies

Wonderful! A lot of things make more sense now, thanks a lot!

16. Nov 27, 2009

### SystemTheory

There are some great animations showing potential and flow analogy at the bottom of this page:

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/DC-Current/WaterFlowAnalog.html [Broken]

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