Combination circuit help

[gar]

130721-1112 EDT

hurk27:

Your analysis provides an explanation for what happened.

An original question did not include the 22 mA statement. All the resistance values and the 50 V are consistent with a total current of 95.56 mA.

Then someone decided to change the problem, and made the decision that 22 mA would be the current thru the 242 ohm resistor, retained the same total current of 95.56 mA, and from this said the current was 73.56 mA thru the 180 ohms resistor. This is not a possible solution. If the resistance values are correct, then with 22 mA thru the 242 ohm resistor the current thru the 180 has to be 22*242/180 = 29.58 mA.

The author of the revised question was not thinking or does not understand circuit analysis. Probably was a quick judgement without thinking to make a change in the question so that it was somewhat different from a previous question.

.

 

[Eddy Current]

Answer I got is 73.6ma

I figured the total circuit resistance by finding the R of the two parallel resistors as was done in post 13, which is 103.22 ohms, adding this to the 420 ohms in series gave me 523.22 ohms for my Rt, then dividing this into the 50 volts gave me a current of 95.6ma for my It, removing the 22ma of the 242 ohm resistor gave me a current of 73.6ma for the 180 ohm resistor, using these figures it comes out as 13.248 volt on the two parallel resistors and taking this away from the 50 volts leaves 36.752 volts on the 420 ohm resistor.

But as others said as far as ohms law goes it doesn't compute which would give you totally different answers:?

To me the 242 ohm resistor should be 602.2 but that would still not clear up ohms law? but then that would also throw the whole equation off again as the parallel resistors wouldn't add up to 103.22 ohms, so in knowing this it has to be the 22ma given for the 242 ohm resistor that is in error.

Using the 523.22 ohms as the Rt and the 50 volts as the Et with 0.0956 as the circuit's It we could fill in the correct I value for the 242 ohm resistor.

Let's make this simpler to follow, lets number the 242 ohm resistor as R1 the 180 as R2 and the 420 as R3

In using the above total circuit figures here is what it comes out to:

E1= 9.848v
R1= 242Ω
I1= 41ma

E2= 9.848v
R2= 180Ω
I2= 54.7ma

E3= 40.152v
R3= 420Ω
I3= 95.6ma

As we can see R2 still doesn't line up with any of the answers???

So the problem did have to be simplified to a series circuit?

 

[Smart $]

So the problem did have to be simplified to a series circuit?

Not for the question as posed in the OP.

Being that the correct answer is not an avaliable choice, the best that can be done is speculate...

 

[Eddy Current]

Answer I got is 73.6ma

I figured the total circuit resistance by finding the R of the two parallel resistors as was done in post 13, which is 103.22 ohms, adding this to the 420 ohms in series gave me 523.22 ohms for my Rt, then dividing this into the 50 volts gave me a current of 95.6ma for my It, removing the 22ma of the 242 ohm resistor gave me a current of 73.6ma for the 180 ohm resistor, using these figures it comes out as 13.248 volt on the two parallel resistors and taking this away from the 50 volts leaves 36.752 volts on the 420 ohm resistor.

But as others said as far as ohms law goes it doesn't compute which would give you totally different answers:?

To me the 242 ohm resistor should be 602.2 but that would still not clear up ohms law? but then that would also throw the whole equation off again as the parallel resistors wouldn't add up to 103.22 ohms, so in knowing this it has to be the 22ma given for the 242 ohm resistor that is in error.

Using the 523.22 ohms as the Rt and the 50 volts as the Et with 0.0956 as the circuit's It we could fill in the correct I value for the 242 ohm resistor.

Let's make this simpler to follow, lets number the 242 ohm resistor as R1 the 180 as R2 and the 420 as R3

In using the above total circuit figures here is what it comes out to:

E1= 9.848v
R1= 242Ω
I1= 41ma

E2= 9.848v
R2= 180Ω
I2= 54.7ma

E3= 40.152v
R3= 420Ω
I3= 95.6ma

As we can see R2 still doesn't line up with any of the answers???

I am to assume the answer is 73.6 then?

I'm still not seeing how you got that, what does "removing the 22 of 242 resistor" mean?

Out of all the review questions in this particular section none of them have used Kirchoffs Current Law yet, is that how you got 73.6?

 

[gar]

130721-2137 EDT

Eddy Current:

You write on the test --- none of these --- the answer is 29.58 mA.

Then you complain to the teacher or author of the test and explain why none of the answers are correct.

.

 

[hurk27]

So the problem did have to be simplified to a series circuit?

Yes you always break down a series parallel circuit into its simplest form to get to a single resistance to find the Et, Rt, and It of that circuit, so you have to make the parallel part into a single resistor so you can add it to the other series resistor to get the total resistance of the circuit, you correctly started doing this in your post 13 but I think you got side tracked because figuring the rest of the circuit is in error as it will not come out to any of the answer choices you have.

I am to assume the answer is 73.6 then?

I'm still not seeing how you got that, what does "removing the 22 of 242 resistor" mean?

Out of all the review questions in this particular section none of them have used Kirchoff's Current Law yet, is that how you got 73.6?

The only way I was able to get any answer that you have in your choices of answers was to find the total current of the circuit by finding the total resistance of the circuit as I explained above, by finding the total resistance of the 242Ώ resistor along with the 180Ώ resistor which as you had done in post 13 which gave me 103.22Ώ then adding this to the 420Ώ resistor will give you the circuit total resistance of 523.22Ώ then following ohm's law by dividing the circuit voltage by this will give you the 95.6ma total of the circuit and knowing that Kirchoff's Current Law states in a series circuit that current will be the same in all parts of the series part of the circuit so from this we know we have 95.6ma on the 420 ohm resistor as well as the whole of the two parallel resistors so knowing that both parallel resistors have a total current of 95.6ma flowing through them it is only a matter of subtracting the known current of 22ma which they gave you for the 242 ohm resistor from this total to get the current flowing in the other resistor, to 95.6ma-22ma gives you 73.6ma for the 180 ohm resistor.

The problem as was pointed out is that the whole circuit does not follow ohms law if you use the 22ma figure as it would not be correct if you figured out the whole circuit as I done in post 20 that was done from the given resistors values and the circuit voltage, as gar pointed out someone most likely complained thinking it was too hard and added the 22ma figure thinking they were making the question easier, then changed the available answers to to make the question possible, but they erred because the 22ma was not the correct figure for the 242 ohm resistor as I pointed out in my analysis of the circuit which showed that the 242 ohm resistor should have had a 41ma current value to make the circuit fit ohms law, which would have given you a 54.7ma current value for the 180 ohm resistor and the correct answer of the question if the ohms values of each of the resistors are correct, which is not one of your choices, if they would have said to mark the closest answer given in the answers then the 59.4 would have been the correct choice if we were to ignore the 22ma for the 242 resistor as giving the 22ma for that resistor was un-necessary to analyze the circuit following ohms and Kirchoff's Law.

The problem of the people writing these test many times are office help who doesn't know ohms law, and no one proof checks them to make sure they are correct.

As for you second question all circuits use Kirchoff's Law whether or not we think about it, even a single component circuit uses it because of the resistance of the wiring feeding the single load we just don't think about Kirchoff's Law in these cases but it's there.

 

[gar]

130722-0826 EDT

Eddy Current:

It seems you are still confused by this question.

First, keep in mind some basics.

The sum of the voltages around any closed loop is zero.
The sum of the currents at any point is zero.
Ohms law applies to linear invariant resistances.
Ohms law does not include any analysis of power.
Voltage is measured between two points.
Current is measured at a point.

Picking apart the original question.

A 242 is in parallel with a 180 resistor, and a 420 resistor is in series with the combination in a 50 volt circuit. A current of 22mA flows throw the 242 resistor. The current through the 180 resistor is?

There are three resistors mentioned. Call these R1 (242), R2 (180), and R3 (420).
There is one defined current I1 (22 mA) thru R1.
There is one voltage mentioned, but where it is measured is undefined.

One question is asked --- what is the current I2 in R2.
The presented circuit does not explicitly define the 50 V as being the voltage across the series combination of R1 and R2 in parallel and in series with R3. You can guess that, but it is not stated.

Without assumptions you can determine the current thru R2 and it is 29.58 mA.

If you assume the series parallel combination is across a 50 V source, then there is a conflict in values, and something is not valid.

In the real world you recheck your measurements, and recalculate the circuit to find out why the disparity.

In the school environment you complain to the instructor. In one such case I was taking a final exam. (I should point out that most of our tests were open book, and on the honor system. The honor system meant that you signed a pledge on the blue book that you had not received or given help to other students.) In this exam our prof A. B. Macnee was sick or out of town, and Dow (head of the department) was the person overseeing the exam. Once the test was started Dow returned to his office. I encountered a question that seemed inconsistent. I went to Dow for clarification. His only response was --- describe on the test what you think is wrong in the question. You would have to know Dow and his teaching philosophy to understand his response. And it was the correct response for several reasons.

Your original post question does not have an answer because the question is incomplete or flawed. Unfortunately this question has not been a good question for teaching electrical circuit theory, but possibly it may be teaching you how to analyze an inconsistent question. For someone trying to learn circuit theory and get positive reinforcement this is a very bad question, and the author needs to be criticized.

Had the question been written as:

A circuit consists of three resistors and a voltage source where R1 and R2 are in parallel, and this parallel combination is in series with R3, and the series parallel combination is in parallel with the voltage source V1, what are the voltages and currents associated with each resistor. For one question this could be simplified to ask for the current or voltage of any one of the resistors or the voltage source. So many variations on a single circuit could be used for the test.

A different question could be --- the current in R1 is (using the correct value) what is the current thru R2? Or the voltage source could be left out of the question, and any value of current specified.

.

 

[n1ist]

Remember that Ohm's law does hold true. I have attempted to break it a number of times, and it always smells bad when you try.
/mike

 

[ggunn]

A 242 is in parallel with a 180 resistor, and a 420 resistor is in series with the combination in a 50 volt circuit. A current of 22mA flows throw the 242 resistor. The current through the 180 resistor is?


Do i have drawn it out but do you not calculate everything in the parallel part of the circuit first?

29.6mA. The circuit voltage and the 420 ohm resistor are irrelevant. The 22mA current through the 242 ohm resistor gives you 5.32V across it and the adjacent 180 ohm resistor. 5.32V across the 180 ohm resistor gives you 29.6mA through it. No other information is relevant.

The circuit voltage is 50V but they haven't told you what else is in the circuit. This is a typical test question where they give you what you need to solve the problem along with some red herrings. Figuring out what is the essential information is the real test question; the calculation is secondary. As a matter of fact, what I would have done is look at the resistors and see that since the 180 ohm resistor is about 1/3 smaller than the 242 ohm resistor, it should have about 1/3 more current. If there is only one answer to choose from that is anywhere close to that, I'd choose it and move on. Speed is of the essence.

 

[Eddy Current]

This question was in a NCCER hand book. There is a sheet in the back you can fill out if you see an error in the book, should i fill it out?

 

[gar]

130722-2337 EDT

Self-evident. Obviously you fill it out.

.

 

[ggunn]

This question was in a NCCER hand book. There is a sheet in the back you can fill out if you see an error in the book, should i fill it out?

If there was no answer of or near 29.6mA, then yes.

 

[Sahib]

Answer I got is 73.6ma

I figured the total circuit resistance by finding the R of the two parallel resistors as was done in post 13, which is 103.22 ohms, adding this to the 420 ohms in series gave me 523.22 ohms for my Rt, then dividing this into the 50 volts gave me a current of 95.6ma for my It, removing the 22ma of the 242 ohm resistor gave me a current of 73.6ma for the 180 ohm resistor, using these figures it comes out as 13.248 volt on the two parallel resistors and taking this away from the 50 volts leaves 36.752 volts on the 420 ohm resistor.

But as others said as far as ohms law goes it doesn't compute which would give you totally different answers:?

There you made a mistake. If the total current It for a parallel circuit of resistances R1 and R2 known, then current through R1 is It*R2/(R1+R2) and current through R2 is It*R1/(R1+R2). If you apply this, you will have no violation of ohm's law.

 

[Sahib]

Then the current through 242 ohms resistance must be 95.6*180/422=40.78 mA and not 22 mA.

 

[ggunn]

Then the current through 242 ohms resistance must be 95.6*180/422=40.78 mA and not 22 mA.

22mA through the 242 ohm resistor was in the problem statement according to the OP. If that is true, that defines the voltage across it and the resistor in parallel with it, which in turn determines the current through that resistor. All the other information is irrelevant.

 

[Sahib]

22mA through the 242 ohm resistor was in the problem statement according to the OP. If that is true, that defines the voltage across it and the resistor in parallel with it, which in turn determines the current through that resistor. All the other information is irrelevant.

I did not discuss about the self-contradictory nature of the problem statement but about the defect in the line of reasoning by hurk with actually leads to a value of 40.78 mA and not 22 mA through the 242 ohm resistor.

 

[kwired]

I did not discuss about the self-contradictory nature of the problem statement but about the defect in the line of reasoning by hurk with actually leads to a value of 40.78 mA and not 22 mA through the 242 ohm resistor.

Read the OP it says 22mA through the 242 ohm resistor. I have not been keeping up with any of the math here, but based on the given information we have to assume there is 22mA through that resistor and find missing currents, voltages and resistances.

 

[Sahib]

Read the OP it says 22mA through the 242 ohm resistor. I have not been keeping up with any of the math here, but based on the given information we have to assume there is 22mA through that resistor and find missing currents, voltages and resistances.

What about the 50V source? It is also given in the problem. You have not taken into account. Have you?

 

[kwired]

What about the 50V source? It is also given in the problem. You have not taken into account. Have you?

I have not done a single calculation, but how can you change a constant that was given in the original problem? 22 mA is the current in the 242 ohm resistor.

 

[Sahib]

I have not done a single calculation, but how can you change a constant that was given in the original problem? 22 mA is the current in the 242 ohm resistor.

I am not changing any thing, because I did not set that question paper.

I am just trying to bring to your kind notice the self-contradictory nature of the problem statement.