Combination circuit help

[Eddy Current]

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?

 

[GoldDigger]

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?

Try replacing the parallel components by a single equivalent resistance and then working out the simple series circuit that results. Let us know if that gets you started.
Actually you have been given more information than you need in the problem, so there is also a shortcut you can use. Try to figure out what it is.
Hint:
What is the voltage across the 242?
What is the voltage across the 180?

 

[Eddy Current]

Try replacing the parallel components by a single equivalent resistance and then working out the simple series circuit that results. Let us know if that gets you started.
Actually you have been given more information than you need in the problem, so there is also a shortcut you can us50e. Try to figure out what it is.
Hint:
What is the voltage across the 242?
What is the voltage across the 180?

50/180 would give you the voltage of the whole parallel circuit right?

 

[GoldDigger]

50/180 would give you the voltage of the whole parallel circuit right?

No.

Apply E=IR to the 242 resistor all by itself and tell me what you get.

 

[jerryalan]

not enough info

not enough info

16 milliamps or thereabouts flow through the smaller resistor, but usually the total current is derived and then the voltage drops are calculated throughout. . .

 

[GoldDigger]

16 milliamps or thereabouts flow through the smaller resistor, but usually the total current is derived and then the voltage drops are calculated throughout. . .

Wrong answer. Try again. Why would the smaller resistor have a smaller current?

 

[jerryalan]

t z.most said

t z.most said

.029A, but i don't think there is enough information given . . .

 

[n1ist]

Something doesn't add up here. Combining all the resistors gives 523 ohms. At 50V, that's 95mA thru the 420R resistor. Voltage across the parallel pair is 9.8V, so you would have 40mA, not 22mA on the 242R resistor. The 9.8V across the 180R resistor would give 55mA, so Kirchhoff is happy.

If you ignore the 50V source and 420R resistor, and just consider the parallel combination of a 242R and 180R resistor, you get 5.3V across the pair and 29mA thru the 180R resistor.

/mike

 

[charlie b]

If you ignore the 50V source and 420R resistor, and just consider the parallel combination of a 242R and 180R resistor, you get 5.3V across the pair and 29mA thru the 180R resistor.

I knew, as soon as I saw both a voltage number and a current number, that too much information had been given in the problem. That told me that there was the possibility that the information had contradicted itself. And Mike has proven that it does. I agree with your math.

But I will take it one step further. Let's ignore the 50V source for a moment, but then let's consider the 420 resistor. You have 22 mA going through the 242 resistor and 29 mA going through the 180 resistor, meaning that there is a total of 51 mA going through the 420 resistor. It will then have a voltage drop of 21.7 volts. Add that to the 5.3 volts dropped across the parallel pair, and the total voltage drop in the circuit is therefore 27 volts. That is the part that contradicts the original problem's statement that it is a 50 volt circuit. So perhaps the problem's author forgot to mention the other 452 ohm resistor further downstream of the 420 resistor. :happyyes:

 

[Smart $]

I knew, as soon as I saw both a voltage number and a current number, that too much information had been given in the problem. That told me that there was the possibility that the information had contradicted itself. And Mike has proven that it does. I agree with your math.

But I will take it one step further. Let's ignore the 50V source for a moment, but then let's consider the 420 resistor. You have 22 mA going through the 242 resistor and 29 mA going through the 180 resistor, meaning that there is a total of 51 mA going through the 420 resistor. It will then have a voltage drop of 21.7 volts. Add that to the 5.3 volts dropped across the parallel pair, and the total voltage drop in the circuit is therefore 27 volts. That is the part that contradicts the original problem's statement that it is a 50 volt circuit. So perhaps the problem's author forgot to mention the other 452 ohm resistor further downstream of the 420 resistor. :happyyes:

Perhaps the author did not. I've ran across many a question in my career where the author intentionally throws in misleading, but not necessarily inaccurate information. Just because he stated the combination is in a 50V circuit does not mean the named resistors are the only components of the circuit. In fact, the 420 resistor is not necessary either.

 

[GoldDigger]

Perhaps the author did not. I've ran across many a question in my career where the author intentionally throws in misleading, but not necessarily inaccurate information. Just because he stated the combination is in a 50V circuit does not mean the named resistors are the only components of the circuit. In fact, the 420 resistor is not necessary either.

If the question included a drawing, it is definitely wrong.
And it is somewhat dishonest to say they named components are in a series circuit and leave out that there are other components in series too.

Anyway, I have also seen the same overspecified problem given with consistent numbers, so my SWAG is that the author of this question modified the other one but did not realize they were not free to change all the values independently.

 

[Smart $]

If the question included a drawing, it is definitely wrong.

That would depend on whether the drawing was of the complete circuit or just a portion of the circuit... and how it is annotated.

And it is somewhat dishonest to say they named components are in a series circuit and leave out that there are other components in series too.

The question as stated by the OP does not say it is a series or complete circuit. It only says a 420 resistor is in series with the combination, which is irrelevant.

A 242 is in parallel with a 180 resistor, and a 420 resistor is in series with the combination in a 50 volt circuit.

Anyway, I have also seen the same overspecified problem given with consistent numbers, so my SWAG is that the author of this question modified the other one but did not realize they were not free to change all the values independently.

And I'll counter that with saying the author did realize he was free to change all the values...

 

[Eddy Current]

So here is what i have so far and i still can't find a correct answer. Goldidgger i applied E=IR to the 242 resistor and came up with 242Rx22I=5324E, then i divided 5324E/180R= 29.5 which was not one of the multiple choice
answers.


So i tried using the reciprocal method to find the total resistance in the parallel part of the circuit. 1/((1/242+1/180))=103.22 and then i tried to do 50E/103.22=.48 and nothing came of that either.


I have also tried inserting all the multiple choice answers into the problem and i got two different voltages in the parallel part of the circuit on the supposed correct answer.


Im convinced that this circuit has to be simplified somehow down to a series circuit to find the correct answer.

 

[Eddy Current]

Something doesn't add up here. Combining all the resistors gives 523 ohms. At 50V, that's 95mA thru the 420R resistor. Voltage across the parallel pair is 9.8V, so you would have 40mA, not 22mA on the 242R resistor. The 9.8V across the 180R resistor would give 55mA, so Kirchhoff is happy.

If you ignore the 50V source and 420R resistor, and just consider the parallel combination of a 242R and 180R resistor, you get 5.3V across the pair and 29mA thru the 180R resistor.

/mike

That was one of the solutions i tried but 29mA is not one of the multiple choices.

 

[don_resqcapt19]

That was one of the solutions i tried but 29mA is not one of the multiple choices.

Is one of the choices 73mA?

 

[gar]

130720-0916 EDT

Eddy Current:

From your first post:

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?

Your are told you have two resistors in parallel, a 242 ohm and a 180 ohm, and that 22 mA is flowing thru the 242 ohm resistor. The question is how much current is flowing thru the 180 ohm resistor? Anything else in that paragraph can be ignored. Thus, you have a simple parallel circuit problem.

Both resistors have the same applied voltage. Thus, calculate that voltage from the given information. The result rounded to a reasonable value is 0.022*242 = 5.324 V. Then calculate the current thru the 180 ohm resistor. 5.324/180 = 0.0296 A.

If 29.6 or 30 mA is not in the answer list, then the author of the question made a mistake somewhere. At the 1% tolerance level neither of the resistance values are standard. At the 5% tolerance 180 is standard, 242 is not, 240 is, but not at the 10% level. So is 242 a typo?

If you provided us with all of the multiple choice answers, then we might piece together what might have been other resistance values for the 242 ohm resistor.

Another way to calculate the current is 0.022*242/180 = 0.0296 or 29.6 mA.

I have said nothing different than most of the other replies, just a slightly different perspective. Do provide us with the available answers.

If 242 is changed to 560, then the three resistor network comes close to 50 V with 22 mA in the 560. Also 420 is not a standard nominal value.

.

 

[kwired]

Your are told you have two resistors in parallel, a 242 ohm and a 180 ohm, and that 22 mA is flowing thru the 242 ohm resistor. The question is how much current is flowing thru the 180 ohm resistor? Anything else in that paragraph can be ignored. Thus, you have a simple parallel circuit problem.

Both resistors have the same applied voltage. Thus, calculate that voltage from the given information. The result rounded to a reasonable value is 0.022*242 = 5.324 V. Then calculate the current thru the 180 ohm resistor. 5.324/180 = 0.0296 A.

I didn't do any calculations, but that was my thoughts. Use Ohm's law and Kirchoff's laws to find the missing elements and ignore any information provided that doesn't apply to what you are trying to find. If you have two of the three elements in Ohm's law you can calculate the third. Resistors in parallel will have the same voltage across them. Resistors in series will have the same current flowing through them.

 

[Eddy Current]

Here are the choices 36.4, 59.4, 60.3, 73.6

Apparently the answer is 36.4 according to someone who marked in the book before me but i wouldn't trust that some of the other answers they had were wrong.

 

[gar]

130720-2118 EDT

Eddy Current:

Clearly the values given as answers do not belong to the problem as stated. They are not even close.

Suppose 180 ohms is correct, then
36.4 mA = 6.552 V
59.4 mA = 10.692 V
60.3 mA = 10.854 V
73.6 mA = 13.248 V

For 22 mA and these voltages we get resistances of:
297.8 ohms
486 ohms
493.4 ohms
602.2 ohms

180 ohms and 300 ohms are valid 5% tolerance values. 297.8 is close to 300, but it is hard to see 242 to 300 as a typo. 486 is close to 487 a 2% tolerance. 493.4 is close to 493 a 1% or better tolerance. 602.2 is close to 604 a 2% tolerance.

22 + 73.6 = 95.6 and that produces too much voltage in the 420 resistor and the source would have to be greater than 50 V.

Suppose the currents are the total thru the 420 ohm resistor. Then the 180 ohm values would be :
36.4-22 = 14.4, and 2.592 V, and 117.8 ohms
59.4-22 = 37.4, and 6.732 V, and 306 ohms, 305 is a 1% value
60.3-22 = 38.3, and 6.894 V, and 313.4 ohms. 312 is a 1% value
73.6-22 = 51.6, and 9.288 V, and 422.2 ohms, 422 is a 2% value

Conclusion: the answers do not belong to the stated question.

Who is the problem author?

Someone check my math.

.

 

[hurk27]

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???