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Below is an explanation attached from my textbook:

enter image description here

It explained the batteries connected in series very well as seen. Honestly, I've been researching about batteries connected in parallel on my textbook. However, I couldn't see anything useful. Is there any chance that book didn't explain it? I'm out of my mind right now. What I mean is there should be something that explains what to do. Can you tell how to research or specify the reason?

My Kindest Regards!

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  • \$\begingroup\$ I'd be grateful If someone gives a tip. \$\endgroup\$ – Busi Mar 26 '18 at 18:56
  • \$\begingroup\$ the batteries in the diagram are in parallel .... they would be in series if one of them was reversed \$\endgroup\$ – jsotola Mar 26 '18 at 19:01
  • \$\begingroup\$ @jsotola Can you be more clear, please? \$\endgroup\$ – Busi Mar 26 '18 at 19:05
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    \$\begingroup\$ They are neither strictly in series nor in parallel, because they share no nodes. \$\endgroup\$ – Selvek Mar 26 '18 at 19:21
  • \$\begingroup\$ @Selvek How? why do they share no nodes? \$\endgroup\$ – Busi Mar 26 '18 at 19:25
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Your book likely doesn't mention batteries connected in parallel because it is using an idealized simplification of a "battery" that makes parallel battery connections meaningless.

In the problem above, the "batteries" are represented by ideal voltage sources. They have an exact output voltage, infinite capacity to source current, and no internal resistance.

If you put two ideal voltage sources of the same voltage in parallel, they will behave no differently than a single ideal voltage source. If you put two ideal voltage sources of different voltage in parallel, you will have a circuit with infinite current because the over-simplified ideal voltage source just doesn't represent reality with high enough fidelity to give you an idea of what really happens.

In reality, batteries include series resistance (voltage changes as a function of current) and discharge curves (voltage changes as a function of state of charge). To first order, putting two batteries of the same voltage in parallel will reduce the voltage drop across the series resistance (because each battery sources only ~half of the current) and double the battery life. However, those interactions are fairly complex, and I would recommend you ask a more specific follow-on question if you have something specific in mind.

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  • \$\begingroup\$ So are the parallel batteries meaningless? I didn't get what you mean ;) \$\endgroup\$ – Busi Mar 26 '18 at 19:27
  • \$\begingroup\$ @Busi "in parallel" implies that their two anodes are directly connected to each other, and their two cathodes are directly connected to each other. But, there are no such direct connections in the circuit in your picture. Those two batteries are not in parallel with each other because of the resistors that come between them. \$\endgroup\$ – Solomon Slow Mar 26 '18 at 19:59
  • \$\begingroup\$ I still didn't get what you are trying to mean. Why didn't the book explain it? \$\endgroup\$ – Busi Mar 26 '18 at 19:59
  • \$\begingroup\$ Related: Why are ideal voltage sources in parallel meaningless. \$\endgroup\$ – The Photon Mar 26 '18 at 20:09
  • \$\begingroup\$ @jameslarge, what if there are no actual resistors, but a really crappy battery holder? \$\endgroup\$ – jsotola Mar 26 '18 at 20:15
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The problem comes when trying to place two 'ideal' voltage sources in parallel. Ideal voltage sources have an internal resistance equal to 0 Ohms.

Let's think about it. These are two ideal voltage sources in parallel:

schematic

simulate this circuit – Schematic created using CircuitLab

The issue isn't obvious yet, but if \$V_1\neq V_2\$, as it is in practice, you have a contradiction.

In other words, if you put a voltmeter across \$V_1\$, is it going to give you the value of \$V_1\$? Or \$V_2\$? The ideal voltage sources enforce a value across the nodes they are placed. So in the previous case, you have two enforcing conditions for the voltage at the same node, which leads to a contradiction if the voltages are not the same.

The more general case looks like this:

schematic

simulate this circuit

Notice that if \$R_1=R_2=0\$, the above becomes the ideal case for voltage sources. Say, \$R_1\$ and \$R_2\$ are nonzero (probably \$\approx 0\$), then it doesn't matter what the values of \$V_1\$ and \$V_2\$ are—the current will flow from one source to the other for \$V_1\neq V_2\$ and we don't have a contradiction.

Now, back to your question, of "why the sources add up to the same". Imagine two perfectly matched sources (\$V_1= V_2\$ and \$R_1=R_2\$). Also notice that in the last circuit I have labeled two nodes: A and B. Then the voltage between A and B (what you'd think of 'parallel') is (after using KCL):

$$V_{AB}=V_1\dfrac{R_2}{R_1+R_2}+V_2\dfrac{R_1}{R_1+R_2} $$

Since we assume the sources are matched,(\$V_1= V_2\$ and \$R_1=R_2\$),

$$V_{AB}=V_1=V_2$$

If the sources aren't matched in some way, then \$V_{AB}\$ will be different. So when dealing with ideal voltage sources, you have to be careful as to not create a contradiction by placing two different enforcing conditions across the same nodes (similar argument can be said about ideal current sources). If you use the more realistic model, then the contradiction issue goes away but the parallel combination only adds up to the 'same' when the sources are matched.

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  • \$\begingroup\$ What a great answer you gave! :) Can you answer the other question too? Which is why the textbook didn't explain parallel batteries? \$\endgroup\$ – Busi Mar 26 '18 at 20:53
  • \$\begingroup\$ Why your book doesn't explain it? I am going to guess that for an introductory lesson on ideal sources, Kirchhoff's, etc, this may be a bit out of the scope, imho. And with two ideal voltage sources in parallel, you can't really work the math (KCL or KVL) unless you add the non-ideal parameters (resistors)...All you have is an equality that can lead to contradiction if V1 isn't equal to V2. You need to start from the general case to make sense of it. \$\endgroup\$ – Big6 Mar 26 '18 at 20:59
  • \$\begingroup\$ Would it be meaningless to know? Or there should be a better reason, right? \$\endgroup\$ – Busi Mar 26 '18 at 21:19
  • \$\begingroup\$ @Busi it's quite important, just out of the scope of an introductory course. Details become important as you take more advanced courses. \$\endgroup\$ – Big6 Mar 26 '18 at 21:22
  • \$\begingroup\$ Or It would explain it. I need to make sure that I'm on the right topic. How should I research it? \$\endgroup\$ – Busi Mar 26 '18 at 21:26

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