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

• Draconifors
In summary, Homework Equations state that Ohm's law is V=I*R, EMF equations state that ε=I*r+I*R, and the Attempt at a Solution states that 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). I added up the two voltages (=18V), and solved for R using V=I*R
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!

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

Is it 4A? Because 12V/3Ω?

Draconifors said:
Is it 4A? Because 12V/3Ω?
Hello Draconifors. Welcome to PF !

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

SammyS said:
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.

Draconifors said:
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?

TSny said:
Yikes!
Yikes is pretty much my own reaction, honestly.

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

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

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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.

Draconifors said:
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.

TSny said:
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.

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?

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

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.

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?

Draconifors said:
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?

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TSny said:
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?

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.

#### Attachments

• circuit 3.png
2.2 KB · Views: 656

## 1. What is EMF?

EMF stands for electromotive force, which refers to the potential energy difference between two points in an electrical circuit. It is typically measured in volts (V).

## 2. How do you calculate EMF?

EMF can be calculated by multiplying the current (I) in the circuit by the resistance (R) in the circuit. The unit for EMF is volts (V), so the formula is EMF = I x R.

## 3. What is the current in this circuit?

The current in this circuit is 3.4 amps (A). This is the amount of electrical charge passing through a point in the circuit per second.

## 4. What is the voltage in this circuit?

The voltage in this circuit is 6 volts (V). This is the potential energy difference between two points in the circuit.

## 5. How do you find the resistance in this circuit?

The resistance in this circuit can be found by dividing the EMF (6 V) by the current (3.4 A). So, R = 6 V / 3.4 A = 1.76 ohms. Resistance is measured in ohms (Ω).

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