A Physics question - circuits - E=V+Ir

[Lord Percy]

I know there are a fair few people who are quite clued up on all things to do with electronics. For me, sadly, I've always utterly despised it for some reason.

So anyway, in my efforts to self teach A-level Physics, I've hit a brick wall here. It's a question regarding general circuit analysis, when there are two power sources (such as a car; alternator + battery), and also includes the factor of internal resistance: E=V+Ir

If anyone can explain the answers to these questions, taking as many words as required to make it childishly simple for me, I'll be very happy indeed.

I can do questions a) and b)i), but from there I'm unashamedly stumped. I think once it's been explained, it should click quite easily. I've been fine with everything else so far

Cheers

Also. Never use OCR books to self-teach Physics

https://www.thestudentroom.co.uk/attachment.php?attachmentid=201112&d=1362417876

[HJM]

Hey can't help with the physics as such, but a key source i used for revision (maths and chem) was Khan Academy, the guy does a load of subjects which i'm sure are relevant to yourself, i can only hope he has covered this topic, if not maybe there are others which can help with other topics.

https://www.khanacademy.org/science/physics
____________________
"90% of all statistics are made up"

CBT-18/04/2013

[smegballs]

So are we to assume there is a 9V battery between the two wires on the left? Would make it clearer if it was shown.

[Lord Percy]

[smegballs]

https://i.imgur.com/j1BcnQQ.png

Okay so we are only interested in the loop at the bottom.

And kirchoff says that all the emfs in a loop must be equal to zero.

We know the currents and resistances, so we can find the voltage drops across the resistors.

200*0.026 = 5.2
700*0.006 = 4.2

So going clockwise around the loop we can write the following.

x - 5.2 (negative as against current flow) + 4.2 (positive with current flow) = 0

x - 1 = x
x= 1


You could also write as

x - R1I1 + R2I2 = 0

and then sub in the values for R and I

x - (200*0.026) + (700*0.006) = 0
etc

[smegballs]

Okay for part c (i),

When there is zero current flowing in the 700 resistor, where must be equal voltage on either side of it, therefore the node is sitting at 1V.

If the node is sitting at 1V then the current through the 200 ohm resistor will be:

I = V/R
I = 1/200
I = 0.005 = 5mA

to do c (ii) we need the voltage divider equation.

https://forum.allaboutcircuits.com/image_cache/httpinteractive.usc.edumembersmedianpashenkovvoltage_divider.jpg

To find R1 we re-arrange to get:

R1 = (R2(Vin - Vout)/Vout

Sub in the values.....

R1 = (200 ( 9-1)) /1

R1 = 200 *8 = 1600 ohms

[smegballs]

Dunno how good your algebra is so here's the re-arrangement for you.


Vout = Vin * R2 / (R1 +R2)

Vout (R1 + R2) = VinR2

R1Vout + R2Vout = R2Vin

R1Vout = R2Vin-R2Vout

R1Vout = R2 (Vin - Vout)

R1 = (R2 (Vin - Vout)) / Vout

[Lord Percy]

Absolutely brilliant

I'm totally fine with algebra so that's all ok. The only thing I've been stumped with is the concept of electricity flowing in all these various directions. I think I have a psychological aversion to it just because I decided I hated it at school. Slowly it's all beginning to click.



Also it may be because I'm rushing through it very fast indeed

[smegballs]

Keep at it man!

My electric/electronics knowledge is all self taught, mainly from the perspective of building radio equipment. As it turns out, RF gear is quite a bit more complicated than most simple stuff so if you can get your head around designing a radio, then LED drivers and whatnot is pretty easy.

[BenZXR]

R1 = 1000cc, all you need to know
____________________
CBT 03/10/12 | Theory 11/01/13 | Mod 1 16/01/13 | Mod 2 18/01/13
Current: Suzuki GSXR600 00'.
Old: Peugeot Speedfight 56', Kawasaki ZXR400 51'.

[cimbian]

Read the question again and keep in mind that under Kirchoff's law all current must be accounted for.

The question clearly states that the circuit is powered by a 9V battery of assumed internal resistance of 0 Ohm and the current is shown as 32mA (assume this to be off to the left of the diagram and is a constant voltage).

Now, you also have a cell of a stated 1V and as this has a circuit with current flow its current must be accounted for WITHIN that circuit and the ONLY current flowing in that circuit is 6mA.

Temperature changes in the thermistor will alter its resistance and therefore the current flowing through it; however, the 200 Ohm resistor is a fixed value which means a change in current will lead to a change in Voltage across it.

If the cell is to remain at 1V but the potential (Voltage) between the two fixed resistors changes then the Voltage dropped across the 700 Ohm resistor must also change, which must result in a change of current.

Keep in mind that in this circuit you are dealing with constant Voltages and fixed value resistors. A combination of Ohms Law and understanding the need to account for ALL current in a loop under Kirchoff's Law should tell you what you need.
____________________
22PlusY
Current: Moto Guzzi 1100 Breva. Previous: Honda XL650V TransAlp
Bearded, Balding, Born again Buddhist Biker

[Lord Percy]

Right, so...

- Voltage remains constant.
- Resistors impede current.
- Energy per coulomb stays constant, but rate of flow of charge drops when resisted. i.e. the electrons passing through the resistor have the same energy as when they went in, but they're bottlenecked to a slower flow.
- Current conserved within the main loop as a whole
- Current splits accordingly when circuits go into parallel
- Current conserved within separate smaller loops (like that one on the side)

The bit that confuses me is the part to do with emf=pd or something.

So the starting source is 9V, but then the loop on the side is 1V. And then something about it being conserved?

I'm convinced it's all gonna suddenly click like magic. I've breezed through everything else in this course so far! Electricity sucks! So convoluted

[smegballs]

Strictly speaking current is conserved at nodes as well as loops.

https://i.imgur.com/Ef2e4Tx.png

Inputs + Outputs = 0

I've had questions before that couldn't be solved using loop methods, instead you had to assume a constant current and use node theory to work it all out.

[Lord Percy]

Sure, that makes sense. I think the addition of this cell on the right is what's confusing me. How does it change the circuit? In the diagram, what would the values be if the cell wasn't there?

[smegballs]

[cimbian]

Not sure I see how you are allowing for the PD over R2 being 1V less than over R3.

EDIT: I see, you are answering the last question from OP... cell is not there.
____________________
22PlusY
Current: Moto Guzzi 1100 Breva. Previous: Honda XL650V TransAlp
Bearded, Balding, Born again Buddhist Biker

[Lord Percy]

Guys you're all heroes

Spent a good 5 hours at the library today and powered through question after question after question, and it's all finally clicked. The points I have now grasped (which generally lead me to understand it all) are:

- Current always conserved, therefore splits accordingly at junctions, and takes path of least resistance, hence the divide in current is a ratio of the resistances along each path..
- In series, e.m.f is divided accordingly, depending on the resistance of each component.
- In parallel, e.m.f stays maximum within each loop, and is subdivided among component in series along each loop

Basically Kirchoff's 1st and 2nd laws

Onwards and upwards! Thank ferk that's all over with. Only need to get through waves, particles, astrophysics and quantum shizzle now, which I find drastically easier