The Value of a Missing Resistor?

[Reldar]

I'm sure this is extremely basic for you chaps, but I'm a beginner and I just can't seem to figure out how to find the missing value.

Not looking for a direct answer, but any guidance would be greatly appreciated.


The question:

If R1 = 6 ohms
R2 = 4 ohms
When an electromotive force of 10 volts is applied to the circuit a current of 1 ampere flows through R1.
What is the value of resistance Rx?


Thank you.

 

[(*steve*)]

Use ohms law to determine the resistance of R1 and R2.

Use ohms law to determine the resistance which would drop 10V at 1A.

Note that Rx is not shown in your diagram, so the next trick is to determine from what you have calculated above, the value and placement of Rx.

 

[Laplace]

I would not worry about the resistance of R1 and R2. If you know the current through R1 then you can calculate the voltage across R1. This will be the same voltage across R2, from which you calculate the current through R2. Add these currents together to get the current through Rx. Subtract the voltage across R1 from 10 volts to get the voltage across Rx. So now you have the current and voltage for Rx, use Ohm's law to determine Rx. (This is assuming a certain location for Rx.)

 

[(*steve*)]

OK, I assumed a total of 1A, not 1A through the top resistor. That probably makes more sense given the location of the arrow.

 

[GonzoEngineer]

Did a teacher give you that diagram?

If so, I would find a better teacher....BWAHAHAHAHAHA!

 

[Gordy]

This is a trick question that should never be asked. It does nothing to help students learn electronics theory. So here is the help you need. RX will be lower than the value of R2. The formula for resistors in parallel using recipicle 1/x. So RX will be about 2.816 Ohms. The total amount of current through the circuit is about 3.55 Amps. Remember the confusing question is only asking for the value of RX and nothing else.

Study parallel circuits often because these kind if questions will most often be asked on tests you take.

I hope this helps you.

 

[(*steve*)]

Did a teacher give you that diagram?

If so, I would find a better teacher....BWAHAHAHAHAHA!

The power of ambiguous diagrams!

 

[Gordy]

The power of ambiguous diagrams!

I remember when I was in college the instructore played tricks like that to get us to think. Some students were upset and some dropped out of the class. I for one knowing what was happening, understood what the indtructor was trying to do. I helped several students that were very confused and later was told to keep my mouth shut by the instructor. Later that semester, he had hardly no respect from the students. Hi way of teaching we found out was not in sync with the books we read from. It was horrible!

 

[gorgon]

I've no idea what 'trick question' you are talking about here. If 1A flows through R1, you'll have 6V over it. The current through R2 will be 1.5A, a total of 2.5A through in total. Rx need to bleed 4V at 2.5A to get a total of 10V, and should be fairly easy to compute with Ohms law.

 

[(*steve*)]

I've no idea what 'trick question' you are talking about here.

The diagram is not as clear as it could be.

I presumed that the 1A was flowing through the circuit, not just one of the resistors. It would have been clearer to show it above the resistor, not between them.

The position of Rx is not clearly shown, although the logical place is in series with the rest of the circuit.

 

[john monks]

With 10V across R2 you get 10V/4ohm or 2.5A. This current with the 1A going through R1 totals 3.5A. Now you take 10V/3.5A to get Rx. You get 2.857ohms.

 

[Laplace]

With 10V across R2 ..... take 10V/3.5A to get Rx.

How can you get 10V across R2 and also 10V across Rx?

 

[(*steve*)]

With 10V across R2 you get 10V/4ohm or 2.5A. This current with the 1A going through R1 totals 3.5A. Now you take 10V/3.5A to get Rx. You get 2.857ohms.

Notice that we don't give answers here. Just help on achieving them.

Because your answer is wrong, there's no point in removing it.

 

[Sid723]

My answer is correct depending on how the original problem is supposed to be interpreted.

Notice, I did not show how the answer was derived. If the problem is interpreted the way I say, then the answer I gave will be derived.

In any case, the problem is flawed in how it was written, so it doesn't matter if the answer was given for a flawed problem or not.

 

[(*steve*)]

I think we can agree that the original problem is this:

From this, you can calculate the voltage across R1 and therefore the voltage across R2 and Rx.

From the voltage across R2 you can calculate the current through R2.

Using this information you can calculate the current through Rx. Since you also have the voltage across it, you can calculate the resistance.

 

[Sid723]

@Steve:
That is how I interpreted the problem. Although the RX symbol is not shown in the original circuit and so it gives the impression that 10 volts is across R1 and R2. That is how the original circuit confuses everyone.

@Reldar:
So, is Steve's new circuit diagram what the original circuit should have been? You may want to recheck your source for the original circuit diagram to see if it matches.

 

[Sid723]

Now that I slept on it, this is what I believe this problem was originally intended to say:

Find what the "Internal" resistance (Rx) of the power source would be if 10 volts was applied to the circuit from that source.

Make sense now?