I will be really grateful if someone can explain this to me. A battery connected in series with a resistor R of resistance 5.0 Ω. The electromotive force (e.m.f.) of the battery is 9.0 V and the internal resistance is r. The potential difference (p.d.) across the battery terminals is 6.9 V. Calculate the current in the circuit.

- anonymous

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- anonymous

Now the formula I used was E=V+Vr where V= terminal p.d and Vr=p.d across battery 9=6.9+I(5) I= 0.42A But apparently the answer is incorrect. Why? @experimentX @ganeshie8

- sidsiddhartha

but it seems correct

- sidsiddhartha

E-ir=IR

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## More answers

- anonymous

No, the answer is 1.38 A.

- anonymous

I know. That's what is confusing me. The formula is right, then why is the answer incorrect?

- sidsiddhartha

|dw:1398454706860:dw| that's the ckt is'nt it

- anonymous

Here's the diagram that's given with the question.

##### 1 Attachment

- anonymous

Yes. O.o

- anonymous

They got the answer by simply putting it this way. V/R=6.9/5=1.38 A. I want to know why do it this way? :O

- inkyvoyd

another way to look at it is that kirchoff's voltage law dictates that the sum of all voltages in a closed loop must equal zero. But because the voltage across the battery equals -6.9 (when measured in the direction of conventional current flow), the voltage across the resistor must similarly be 6.9. Thus the current in the resistor must equal 6.9/5 or 1.38

- theEric

Just a side note, since it wasn't explicit, @inkyvoyd 's reasoning explains why the answer was as simple as 'V/R."

- anonymous

I still don't get why the formula I used cannot be applied in this case. :/

- theEric

Oh! I'm not sure what your variables mean in \(E=V+V_r\), despite the short explanations. What I can tell from your equation, 9=6.9+I(5) is that you are saying \(9\) volts is going to be the sum of the \(6.9\) and \(I(5)\) voltages. Kirchhoff's law states that the sum of voltages as you trace around the circuit will be zero. That is the sum of all. So, you can say \(\sum\)all voltages=0 \(\Downarrow\) \(\sum\)forward voltages-\(\sum\)backward voltages = 0 Right? My notation might not be proper, but let me know if you agree.

- theEric

Forward versus backwards is just a choice.

- anonymous

Right.

- Vincent-Lyon.Fr

If 6.9V is the potential difference across the battery, then it is also the potential difference across the resistor. Hence current I = 6.9V/ 5Ω = 1.38 A There is no problem with that.

- anonymous

Okay, so isn't it that the total emf of the battery is equal to the p.d across the resistor and the p.d across the battery?

- theEric

\(\sum\)forward voltages-\(\sum\)backward voltages = 0 \(\Downarrow\) \(\sum\)forward voltages=\(\sum\)backward voltages Does that make sense, too? If there's a 1V forward, then there must be 1V backward. Otherwise, there's some voltage unaccounted for. That's silly. So! The given forward voltages are \(6.9\rm V\) from the battery (which is 9V forward and something internal backwards) The given backward voltages are \(I(5\Omega)\) from the resistor. So \(\rm 6.9V=I(5\Omega)\) ------------------------------- What you were thinking was that the forward voltage is \(9\rm V\) the backward voltages are \(\rm6.9V\) \(I(5\Omega)\) The part that is incorrect is that you suggest that \(\rm6.9V\) is the backwards voltage. That's what it looks like to me. Really, it is given as the forward voltage. The backwards part of it is the difference.

- theEric

That's my analysis of your equation! So it was just a slight lapse. You can say that \(\rm9V=(9-6.9)V+I(5)\), and I think that's how your mind wants you to think.

- anonymous

I thought 6.9V was the voltage drop in the internal resistance of the battery.

- anonymous

I think my concepts are all flawed.

- theEric

Oh! Okay! Well that was it, then! It says that the battery (chemistry stuff) produces a \(\rm9V\) potential difference. And then it says that that the voltage across the battery is \(\rm6.9V\). So, that \(\rm6.9V\) is the voltage the battery actually contributes to the circuit when it's hooked up with that \(5\Omega\) resistor.

- anonymous

Oh okay. So that means 2.1V is lost in the battery itself, right?

- theEric

Don't worry! You actually had the right idea!! You just misinterpreted the meaning of the \(6.9\rm V\)! If you were told that the internal resistance caused a \(6.9\rm V\) drop, then you can apply what you thought. In fact, that would lead to the Kirchhoff's law equation you made exactly. \(\rm 9V=6.9V+I(5\Omega)\) \(\huge\color{lime}{:)}\)

- theEric

Right! 2.1V is what is lost in the battery.

- anonymous

Okay. Can you please explain what exactly is terminal p.d in context of such questions?

- theEric

So, since we both understand that, we can make another equation, this one correct, being \(\rm9V=2.1V+I(5\Omega)\)

- theEric

The potential difference across the terminals would be what you could actually measure with a voltmeter on the outside of the battery, if that helps.

- anonymous

Yes, I get it now. xD Thank you so much.

- theEric

What is the potential difference from one terminal to the other. Cool! You're welcome :)

- anonymous

Okay, so here in this question is the terminal p.d 2.1V or 6.9V?

- theEric

In this question, the potential difference between the terminals is \(\rm6.9V\). The EMF is the potential difference made by the chemical reaction. That is \(\rm9V\).

- anonymous

Okay, thank you. I'm sorry, I took so much of your time.

- theEric

Why do I come here if not to do this? Hehe, I like this. I'm glad I can help. Don't apologize and you're welcome! :)

- theEric

I have to go. Good timing :) Take care! :)

- anonymous

Goodbye. You too. ^_^

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