Calc EMF: Find ε & R in Circuit w/ 3.4 A & 6 V

[Draconifors]

Homework Statement

In the circuit shown in the figure above, the ammeter reads 3.4 A and the voltmeter reads 6 V. Find the emf ɛ and the resistance R.

Homework Equations

Ohm's law; V= I*R
EMF equations: ε=I*r+I*R
ε=I*r+V

The Attempt at a Solution

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I got the resistance of R right: It's 5.29Ω. To do that, I found the current in the 2Ω resistor (6V/2Ω=3A).
Then, I know that since the 2Ω and the 4Ω resistors in are in series, the current in the two of them are equal. (4Ω*3A=12V).
Then, I add up the two voltages (=18V), and solve for R using V=I*R, which gives me 18/3.4 = 5.2941Ω.

However, when it comes to finding the emf, none of my answers are correct. I've tried "reducing" the circuit so it becomes 3 resistors in parallel (4Ω,2Ω and R on one side, 1Ω and the battery in the middle, and the 2 3Ω ones on the other side). However, this doesn't seem to give me the right answer. At first I thought it might be 6 volts since circuits in series are supposed to have the same voltage everywhere, but then I tried rationalizing it and it didn't work.

Any push in the right direction would be greatly appreciated!

 

[TSny]

What is the potential difference across each of the 3Ω resistors?

 

[Draconifors]

Is it 4A? Because 12V/3Ω?

 

[SammyS]

Is it 4A? Because 12V/3Ω?

Hello Draconifors. Welcome to PF !

Potential difference can't be 4A . That has units for a current.

 

[Draconifors]

Hello Draconifors. Welcome to PF !

Potential difference can't be 4A . That has units for a current.

Thank you!

And ohh. That's true. The only thing I can think of is just ignoring my A and saying it's 4V but I wouldn't know why.

 

[TSny]

The only thing I can think of is just ignoring my A and saying it's 4V

Yikes!

How does the potential difference across one of the 3Ω resistors compare to the potential difference across the resistor R?

 

[Draconifors]

Yikes!

Yikes is pretty much my own reaction, honestly.

And they're all in parallel, so it has to be the same, right?

 

[TSny]

Yes, the potential difference across each parallel branch is the same.

 

[Draconifors]

Ok so if each of the 3Ω resistors has 12V going through it, then I should be able to find all the current running through the circuit and then using ε=I*r+V, no?
I find that I = 14.4, r=1 and V=12, but the answer is not 26.4.

 

[TSny]

Ok so if each of the 3Ω resistors has 12V going through it, ...

No, not 12 V. What did you use for the potential difference across the resistor R when you calculated R?

Nit-picky side note: "voltage of a resistor" represents potential difference between a point on one side of the resistor and a point on the other side. "Voltage" does not "go through" something.

 

[Draconifors]

What did you use for the potential difference across the resistor R when you calculated R?

I used 18V, as that was the sum of the current in my 4Ω and 2Ω + the voltmeter.

 

[TSny]

OK. If the resistor R is in parallel with one of the 3 Ω resistors, what is the potential difference across the 3 Ω resistor?

 

[Draconifors]

It's also 18V, as potential differences are the same throughout a circuit with resistors in parallel.

 

[TSny]

Yes, each 3 Ω resistor will have a potential difference that's the same as the potential difference across R.
When two branches of a circuit are in parallel, then the potential difference across one branch will equal the potential difference across the other branch.

 

[Draconifors]

Ok so there's the same potential difference going through the branch with the 1Ω resistor and the battery, right? Does this mean I need to calculate the I of the whole circuit and then use ε=I*r+V, with 1Ω=r and V=18V?

 

[TSny]

Ok so there's the same potential difference going through the branch with the 1Ω resistor and the battery, right?

Yes, although I cringed a bit again when I read your phrase "potential difference going through the branch". Charge flows through the branch, but potential difference (or voltage) isn't going anywhere.

Does this mean I need to calculate the I of the whole circuit and then use ε=I*r+V, with 1Ω=r and V=18V?

Not sure what you mean here. What do you mean by "I of the whole circuit"?
Is the I in ε=I*r+V the same as the I of the whole circuit?

 

[Draconifors]

Is the I in ε=I*r+V the same as the I of the whole circuit?

That was what I understood of my teacher's explanations of the formula, yes. Is it only the I going through each branch?

 

[TSny]

There are different ways you could approach the problem. For example, you could consider the junction point J shown in the figure on the left below. Use the "junction rule" that the total current leaving a junction must equal the total current entering the junction.

Or, you could approach the problem by using the rules for combining resistors to get a simpler circuit as shown on the right.