Effect of the supply internal resistance

Feb 07, 2009 26 Replies

No, you can't, because BAT is a device that is *defined* to have a voltage potential difference across it of 10.1V . You cannot arbitrarily choose to recognize and disregard this on a whim when you analyze the circuit.

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What's to be careful about? The circuit plainly shows the 1R resistor connected to the battery + terminal at one end, and R1 at the other. Nowhere in sight is there a hint of a direct connection of the 1R resistor to ground (which by common convention is taken to be the negative terminal of the battery for a simple circuit like this).

But, when I drew the circuit, BAT and 1R is an equivalent circuit for a battery. Consisting of a perfect battery (BAT) and an internal resistance (1R).

Face value you want to say, hey if BAT is a perfect battery with zero resistance, there's 10V across 1R. Of course that is not true. It's just one of these issues you get when dealing with perfect components in series with resistance. 10v is just a statement that that point is 10V above a refrence point. And there really is not a zero resistance across BAT. It's mixing a fiction with reality. The equivalent circuit of a batttery is a fiction.

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No, because then the 1R would be in parallel with BAT, not in series as is drawn. You don't get to arbitrarily assign the stated potential of the voltage source to another passive component (the internal resistance).

with

Sorry, but it's not a common problem. You seem to have your own unique set of confusions that lead to such inferences.

For all intents and purposes the BAT component has no resistance associated with it. The resistance of the "real" battery is lumped into the 1R resistor. Any circuit analysis that is performed on this circuit *must* consider the potential difference across the battery.

Note also that when you eventually come to analyse AC circuits that also have DC sources in them, the DC sources *can* look like short circuits (zero resistance) to the AC signal components. The +Vcc voltage supply in an audio amplifier circuit, for example, will "look" like a ground every bit as much as the actual DC ground does to the audio signal.

Really, the issue has been what to make of that fact that in the circuit as drawn, R1 is at a potential of + 10v and it's other side is "seemingly* connected through BAT to GND.

That can cause you to wonder how to explain why there is not really a 10V potential across R1. Not that one is unable to calculate the actual voltage across R1, which I can.

Not really, just trying to see where the error would be in thinking there was 10V across R1. That I don't think would be an uncommon thing to do.

Yes, I've noticed that in my ham radio experience.

Note that the battery and its internal resistance have both been placed in the same box (which is how it works in real life) and look at it like this: If you have a 10V battery with an internal resistance of 1 ohm and you measure its output voltage with a high-impedance voltmeter, then the current through the voltmeter will be miniscule as will the current through the battery\'s internal resistance(since current is everywhere the same in a series circuit) and the voltmeter will read 10.0V: . 10.0V . / . +-----E1

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There is no ambuiguity. One *end* of the 1R resistor is fixed at a potential of Vbat with respect to ground (taking the battery negative as ground). This says nothing about the other end of the 1R resistor until you look at the rest of the circuit and hence the current flowing through the 1R resistor.

If the circuit happens to be open (no path to ground via the other end of the 1R resistor), then *both* ends of the 1R will be fixed at Vbat above ground; there will still not be a potential of Vbat

*across* the 1R.

with

The error is in assigning the properties of a voltage source (battery) to a resistor. This is not a 'legal' operation. An ideal resistor is a passive component that contains no sources.

A resistor presents across its terminals a voltage that depends upon the amount of current flowing through it:

v(I) = I*R.

Without a current there is no voltage across the resistor.

An ideal voltage source presents the same voltage across its terminals irrespective of the current flowing through it:

v(I) = V

Shaun, if you are using a newsgroup aware program (as I do) it should regonize the address as a newsgroup (usenet) ID. All I need to do is double click it and a search takes place looking for the article, for example. First within my existing local database, then next online with my newsgroup server via NNTP (one of many protocols, this one called "network news transfer protocol," if memory serves.) If found online, the article is downloaded and stored locally, automatically, and then displayed for me. (It's possible that the server I use no longer stores the article, though.)

For example, the message ID for your own message (the one I'm responding to) was found by looking at the message headers for your message as:

Message-ID:

Every newsgroup message that gets posted, I think, has one of these fields present in the headers. I don't know how it is generated or who does the generation such that they are unique, but there must be an RFC (request for comment), or more than one, that documents it.

If I double-click the message ID above, for example, your message immediately shows in my newsgroup browser -- flipping me away from this message I'm composing. Then I just click back to the tab for this message and continue writing.

Jon

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