Solve Voltage and Current Problem in DC Circuit

[slakedlime]

[SOLVED] Voltage and Current

Homework Statement

A d.c. circuit has a 20ohm resistor and ammeter connected in series. A voltmeter is connected in parallel to the resistor.

After some time, the reading on the ammeter suddenly decreases to nearly zero, but the voltmeter reading remains at 4V. Suggest a possible reason for this.


2. The attempt at a solution
I know that a voltage reading can still be obtained if the switch is opened. However, the reading is said to decrease almost to zero. Can somebody please explain this?

 

[Mindscrape]

You exploded the ammeter and now it has a very high resistance, an effective open circuit.

 

[slakedlime]

Thank-you for your help! I understand now. :)

 

[TVP45]

Can you ask your instructor what else is connected?

And if he says "Nothing", ask if the meters are the old-style moving coil galvanometers.

 

[slakedlime]

There's a 20ohm resistor connected in parallel with the original resistor, but I don't think this could cause circuit damage/affect the ammeter. The question was in a book without answers, and I think we are free to assume if the meters are moving coil as long as the explanation is feasible.

 

[Integral]

You know that the circiut current is zero, yet the volt meter across the resistor is still reading the same? If as suggested above, the Ammeter were open what would the voltage drop across the resistor be?

How else could you get no current but still have a voltage drop across the resistor?

 

[slakedlime]

The current suddenly drops to almost zero but doesn't cease completely

 

[Integral]

OK, now what that you know about in this circiut could cause this?

What else would be implied?

 

[slakedlime]

The only things that I can see occurring are 1) a large current causing the wire to melt and 2) the switch being opened.

I haven't much idea on what has to be implied. I've searched my books and the web but to no avail.

 

[TVP45]

OK, I don't quite understand the 20 Ohm in parallel with the original 20 Ohm, but that doesn't really change anything. So, here's a suggestion: draw your circuit using equivalent circuits for the meters rather than just a circle with a V or A.
A moving coil galvanometer looks like (approximately enough for this) a 50 Ohm resistor in series with a 10 mH inductor. If the meter is used as a voltmeter, it will have maybe a 1000 Ohm resistor in series with it. If it is used as an ammeter, it will have maybe a 0.01 Ohm resistor in parallel with it.
Now, electrical components usually fail either open (infinite resistance) or shorted (zero resistance), at least in introductory courses. So, take the 3 components (20 Ohm resistor, voltmeter, ammeter), one at a time, and fail them, first shorted, then open, and ask yourself each time: what happens to the voltage reading and to the current reading. And, remember that "almost zero" - it's important.

 

[Integral]

Do you know what voltage is read across an open?

 

[TVP45]

Integral,
Are you asking me or the OP?

 

[slakedlime]

Sorry about this very late reply; my computer's been down for a while. I see that the individual meters could fail, but why exactly? The current flowing through the circuit is quite small so I don't think the wires would melt. And the circuit's d.c. so there isn't much chance of current fluctuations.

However, I could suppose that the circuit wire is frayed. Then it acts like several parallel connections of very low resistance. The current flowing through the circuit is very high but the frayed wires are very thin so they melt. This could cause the current to drop to almost zero.

Any comments on this hypothesis would be useful. Thank you again for all your help, even if my reply is very late.

 

[TVP45]

Sorry about this very late reply; my computer's been down for a while. I see that the individual meters could fail, but why exactly? The current flowing through the circuit is quite small so I don't think the wires would melt. And the circuit's d.c. so there isn't much chance of current fluctuations.

However, I could suppose that the circuit wire is frayed. Then it acts like several parallel connections of very low resistance. The current flowing through the circuit is very high but the frayed wires are very thin so they melt. This could cause the current to drop to almost zero.

Any comments on this hypothesis would be useful. Thank you again for all your help, even if my reply is very late.

Don't go into it so deeply that you ask "why" at the start. Standard trouble-shooting at a beginning level should just look at each component being either open or shorted. And, I'll give you a start: the trouble is with the 20 Ohm Resistor (or maybe it's two resistors - I never did get clear on that) that is in parallel with the Voltmeter. So, which of the failure modes, open or shorted, in the resistor would account for the results? It also helps in any kind of circuit analysis if you go ahead and put a battery in the circuit so you have a complete circuit to look at.

 

[slakedlime]

It's two 20 Ohm resistors in parallel with the voltmeter. I should have done this earlier, but here's a diagram of the circuit:

I think the circuit would have to be open. If the wire at the junction before the two resistors has worn out, its resistance would be very high. This is effectively an open circuit (similar to if the switch was open) and very little current could pass through.

I can't see any opportunities for a short-circuit.

Mindscrape said something about the ammeter exploding. Unless the ammeter was initially faulty, I can't see a 0.4A current causing it to overheat and bust.

 

[TVP45]

OK. You're very warm. I'm guessing you haven't done a lot of circuit building - wire doesn't normally "wear out", it breaks. So, if that were the case (break the wire just to the left of the right hand junction dot on the resistor network so that both resistors are now disconnected), how would the current and voltage read? Why?

No, the ammeter didn't explode. If you had accidently shorted out the resistors, the current would have gone very high and the ammeter might have slammed against the right hand stop and bent the needle, but that would give you only a slightly low reading, not "nearly zero" unless you really, really, really bent it.

By the way, don't discount the possibility that one of the other people in the lab, jealous of your good grades, turned the zero dial on the ammeter when you weren't looking.

 

[slakedlime]

Actually, I've never built a circuit. My school doesn't have a lab so all our work is theoretical, which is why I have so much trouble with problems like this. I've never even seen a real circuit... :(

If the wire suddenly broke to the left of the right hand junction dot then current would stop passing through the circuit. However, the voltmeter reads the potential difference between the two sides of the junction - the one connected to the battery and also the disconnected one. So the p.d. is read as 4V.

This way the current reading would be zero. But if the wire was partially broken the thinner, broken wire would have much greater resistance (as resistance is inversely proportional to cross-sectional area) but would still allow a small amount of current to flow through. This way the current reading would drop to almost zero but not quite.

I don't know if this is right. I was wondering why the wire would suddenly break, but maybe I could suppose that it was a very old wire.

 

[TVP45]

OK, you're right about the Voltmeter reading essentially the supply voltage. In my first post, I said a Voltmeter could be modeled as a resistor of about 1000 Ohms (this varies from model to model and how it's set up for full-scale but the actual value doesn't matter). So your circuit current would be something like 4V/1000 Ohms = 0.004 A, nearly zero.

Wires really do break all the time, particularly when your professor is mentally messing with them. They sometimes get nicked when you're soldering them, etc. and later, when the circuit is moved a little, it breaks.

Good job sticking with this.

 

[chroot]

A voltmeter with an input resistance of 1k ohm would be pretty terrible. Modern voltmeters have input resistances on the order of megohms.

If you break the circuit in such a way that the resistors become disconnected (on either side), you're left with a purely series circuit: a 4V battery driving the very high resistance of the voltmeter. The voltmeter would continue to read 4V, the voltage across the battery, but the current through the circuit would drop to nearly zero.

- Warren

 

[TVP45]

Right you are. That's why I questioned earlier whether this was the old style moving coil galvanometer, some of which really were that bad.

Slakedlime,
I realized after posting that, if you have no circuit building experience, you probably don't know that resistor leads are typically solid wire about 22AWG, so one break is the whole kit and kaboodle.

 

[slakedlime]

Okay, all this makes sense now. I'm still wondering about one thing... The resistor lead is the bit of wire used to attach a resistor to a circuit, right?

Thank-you all for your kind help.

 

[Integral]

A voltmeter with an input resistance of 1k ohm would be pretty terrible. Modern voltmeters have input resistances on the order of megohms.

If you break the circuit in such a way that the resistors become disconnected (on either side), you're left with a purely series circuit: a 4V battery driving the very high resistance of the voltmeter. The voltmeter would continue to read 4V, the voltage across the battery, but the current through the circuit would drop to nearly zero.

- Warren

As little as 20yrs ago some University level Physics labs used just such meters. It is actually instructional to use a low quality meter. You really do not need high precision and they provide interesting errors for the students to work with.

 

[TVP45]

As little as 20yrs ago some University level Physics labs used just such meters. It is actually instructional to use a low quality meter. You really do not need high precision and they provide interesting errors for the students to work with.

Many years ago (Nixon was in the White House), I taught a lab for secondary science teachers and they would blow up two or three such meters each lab. In addition to being cheap, the darn things could usually be repaired.

 

[TVP45]

Okay, all this makes sense now. I'm still wondering about one thing... The resistor lead is the bit of wire used to attach a resistor to a circuit, right?

Thank-you all for your kind help.

Go to the Wikipedia article on electronic color code and the first picture shows the most likely kind of resistor that you would have for 4V and 20 Ohms.

 

[Mindscrape]

Hmm, I've never blown up a resistor. I always figured that when they blew up the resistive material just melted, and made a short circuit. I've never blown up an ammeter either, do they just get a really high impedance? The only thing I've ever exploded was an op amp because it had a non-standard pinout. Maybe I should start being irresponsible with my circuits so I can instruct better.

On the other hand, I work with those 22 AWG wires that come with evaluation breadboards all the time, and those always either just flat out break or work. I don't know if I buy the argument of the wire being mostly broken. Maybe if it were stranded.

The diagram helps a lot, especially since I originally had no idea there was another resistor in parallel, but I still think this question should have more information.

 

[TVP45]

Hmm, I've never blown up a resistor. I always figured that when they blew up the resistive material just melted, and made a short circuit. I've never blown up an ammeter either, do they just get a really high impedance? /QUOTE]

Long ago, when I did MTBF calculations for circuits, I knew which way which resistors failed - some go high, some go low - but I've forgotten which is which. I've never blown up an ammeter either, but I confess I've blown up several Multimeters, including one where I dumped 0.2 Farad at 6kV into it; it made lovely confetti. Usually I measure the resistance of energized 120VAC wall sockets - that's always good for a repair job.