Electircal circuit sum query

[Guest kenshi64]

On the attached image of a circuit I was given a sum in the exam as follows, but I can't make sense of it:

The circuit has an EMF of 12V and an internal resistance of 5ohms.

Q1) Total resistance of the circuit

A) 50(45+5) I don't get why they add 5 since that's the internal resistance!

Q2) Current in the internal resistance

A) I= (12/50) = 0.24. Why is 50 used? the internal resistance is 5 not 50! and EMF is provided too, please explain!

Q3)Potential difference between points X and Y

A)across 30ohm voltage drops by 3.60 V (so potential at X is 3.60 V);

across 60ohm voltage drops by 7.20 V (so potential at Y is 7.20 V);

so potential difference between X and Y is (negative) 3.6 V;

---Here I am completely lost as to how they got these values, I'm infinitely grateful for feedback, this sum has been eating away at me for long!

[Drake Glau]

Total resistance will be the internal+everything else.

For resistors in circuit you just add them and in parallel you add their inverses.

So x=90ohms and y=90 ohms

For the whole circuit (minus the internal) it would be (1/90)+(1/90)=1/r

So thats 2/90, or 1/45, so the total is 45. Now add your five because its wants the TOTAL, and you get 50ohms.

They use 50 because the overall resistance is going to dictate the current running through the cell. Think about how the electrons leave the negative side of the cell, travel through the whole circuit, then finally make it back and through the cell. By the time it got back to cell, they had gone through the whole circuit, so it uses the whole resistance.

Yeaaa Q3 is stumping me too, been a while. I'm pretty sure it has something to do with Kichhoff's Laws (the voltage and current laws) as well as some circuitry math, I'm just unable to do it at the moment I guess =/

[matgcauthon]

For Q3, you have to think about a voltmeter. It measures the difference in voltage between two points. So if it is connected in parralel with a light bulb,it will measure the difference in voltage before and after the light bulb. So here, it is simply measuring the difference in voltage between point X and point Y, that being 7.2-3.6=3.6V. So whenever you have that sort of question, just work out the voltage at each point, then subtract. and it is always positive, so don't worry bout that.

Hope that helps!!!

[Guest kenshi64]

For Q3, you have to think about a voltmeter. It measures the difference in voltage between two points. So if it is connected in parralel with a light bulb,it will measure the difference in voltage before and after the light bulb. So here, it is simply measuring the difference in voltage between point X and point Y, that being 7.2-3.6=3.6V. So whenever you have that sort of question, just work out the voltage at each point, then subtract. and it is always positive, so don't worry bout that.

Hope that helps!!!

Okay that's a start, but how do I get the current in the external circuit to begin with!? Thanks!