# Circuits confusing

1. Jun 25, 2013

ThIS IS NOT A HOMEWORK! I need to know the difference which is why i used simple setups to help ask a question. This helps me in doing more complicated circuits after understanding where a potential drop occurs and when it doesnt

2. Jun 25, 2013

### gary32

Hey, Do you understand how to draw circuit diagrams? If so then you're best to draw the circuits (doesnt have to be a piece of art or anything) just so you can see whats going on, then draw in the values you know.

3. Jun 25, 2013

### gary32

I have attached a drawing of your first circuit (a),

No electricity can flow, because there is a break in the circuit,
Lets say your battery is 9v,
The potential difference between point 1 and point 2 would be 9v,
There is technically no potential difference between point 1 and point 3, if you got a voltmeter and put 1 probe on point 1 and the other probe at point 3, the reading would be 0v as there is no potential difference
There is also no potential difference between point 2 and point 3.

If I closed the switch, there would be a potential difference across the bulb BUT no current flowing as there is a broken bulb, no current can flow through it, but the potential is there. If the bulb was okay, then the current would flow through the bulb.

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4. Jun 25, 2013

But i dont know when there's a voltage drop and when the potential is the same :( especially for the later parts about broken bulbs.
I do know how to draw circuit diagrams
But i dont know what happens at these points- tried using different colours too
Most importantly, i need to know part g and h , and whether an open switch affects the p.d across a bulb- zero or still equals emf? (Assuming no internal resistance of cell)

5. Jun 25, 2013

### Staff: Mentor

If a circuit is broken in some way, either by one gap or by many, no current will flow so that any component that gets its potential difference thanks to Ohm's Law (V = I*R) will not exhibit a potential difference across it. So now you need to know between what points you will measure a potential difference given that the circuit is open so that no current is flowing.

To find that, first follow the path from the voltage source's + terminal until you reach the first gap in the circuit (open switch, broken bulb, cut wire,...). Everything along that path or connected to it will be at the same potential, that of the + terminal of the source. Next follow the path from the - terminal of the voltage source until you reach the first gap in that direction. Everything along that path will have zero potential. If you placed a voltmeter's ground lead anywhere along the "-" path, and its positive lead anywhere along the "+" path, you will read the potential across the voltage source.

As for the rest of the circuit that may lie between those two gaps, it might as well be erased from the diagram; it has no connection to either pole of the battery and is thus "floating". If either test lead is connected anywhere on that sub-circuit and the other is connected to anywhere on the source-connected portions, you will read no potential.

All of this ignores the very, very, tiny capacitance that is bound to exist across the open gaps and could theoretically at least allow some potential differences to exist across the "floating" circuitry. But the charges involved will be so small that you couldn't measure them with practical test gear.

6. Jun 25, 2013

Okay so lets say theres 5 bulbs, bulbs 2 and 4 broken. Counting From LHS. Emf is 12v.
Left hand side of bulb1: 12v
RHS of bulb 1: ( 12- 12/3 ) v ?

LHS of bulb 2 : (12- 12/3) v
RHS of bulb 2 : same
...
RHS of bulb 5: 0v

I read in a book p.d is zero across EVERY bulb (when 2 or more broken bulbs out of 10 in a series closed circuit) which i dont understand.

I acknowledge p.d across the LHS of bulb 1 and RHS of last bulb,bulb 5 in my case and bulb 10 in the example's case is indeed =emf (when cell has no internal resistance)

7. Jun 25, 2013

### gary32

If there is 1 broken bulb in a series circuit then NONE of the bulbs will light as the current cannot flow because it is stopped at the break, there is NO potential difference across ANY of the bulbs for that reason.

See what I attached, that is how a bulb works (a filament one anyway), so basically when the filament breaks, its just like cutting a wire which of course stops ANY current flow

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8. Jun 25, 2013

I think i got it from the 2nd last reply and the first reply to this topic :P But Except this case?Answer is C and i got it based on the 2nd last reply Hmph.

Many millions of thanks to you 2 who helped! Hugs anyone? Hahaha.

9. Jun 25, 2013

### gary32

Yes, There is a potential difference over X, the reading would be 240v.
There is no potential difference over Y, the reading would be 0v

10. Jun 25, 2013

Btw, Why would potential at Q decrease compared to original? Its earthed before that so potential at Q is pd across the Resistor R0 (answer is D)
Also, is it because R parallel is halved only or is it because BOTH R parallel being halved and split(meaning current decreases) of current that potential at P decreases?

11. Jun 25, 2013

### gary32

I'm not sure why you would ground a circuit where it is grounded but,

The switch switches another resistor in place so the total resistance of the parallel decreases, when resistance decreases, so does the potential,
I'm not sure why the potential at Q decreases, however, I assume it's because the current has decreased going through R0

12. Jun 25, 2013

I think for potential at P it decreases cuz
Both R and I falls- R becomes R/2, I becomes I/2 is it?
For Q i think overall I has increased cuz of the decrease in total resistance with the switch from series to parallel for the left part---so the potential DROP after the earthing part at resistor R0 is larger (since l falls and Resistance of fixed resistor R0 is constant) resulting in more negative potential hence the decrease than before.
Think you didnt take notice of the direction of current flow. Just thought of these :P correct me if im wrong!

13. Jun 25, 2013

### gary32

I noted the current flow, there is no or little actual flow down to the ground terminal as shown in the drawing,
Between positive and ground there should be your full 12v (or whatever your voltage is), and between ground and the negative terminal on battery should be minus something?

When the switch is open, the current through R1 is the same as current through R0 I would have thought, when the switch is closed, that current is decreased, which also decreases the current through R0 thus reducing the potential

What are your resistor values and voltage?

14. Jun 25, 2013

The resistors have the same resistance. It is a general question without values but i think we got it. Except that im still unsure of the part about potential decrease for P- due to fall in R, I or both. I think now, that its quite ambiguous cuz fall in Total resistance actually increases I supplied by cell. Then its hard to tell whether potential at P changes! :0!! :9(and i cant say larger current causes larger potential drop thru internal resistance hence reducing terminal pd more cuz internal resistance is negligible):0 HOw
Picked this from a book this time and thought it was quite near to understanding of these open circuits n potential things :P we could discuss more on this when i find more related questions :) which i happen to have a lot of reference to churn out questions by myself based on those.

Last edited: Jun 25, 2013
15. Jun 25, 2013

### gary32

I have thrown some values in for an example,
Hopefully that may clear things for you

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16. Jun 25, 2013

### gary32

I also just created a simulation of what is happening,
The voltage drop (potential decrease) occurs when the switch is closed,
The simulation shows me opening the switch at 4secs, closing it at ~6secs, opening it ~8secs etc

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17. Jun 25, 2013

It did prove the potential at P fell since (increase in I supplied by cell due to fall in R total) is LESS than (fall in R total due to series to parallel change) which meant the v fell since IR fell, but well if no values were given how do i tell? Hmph

18. Jun 25, 2013

### gary32

Indeed, You can tell by making up the values or even better knowing the equations inside out,
No matter what size resistors they are in parallel, the total resistance across will ALWAYS be less than the smallest resistor, that means you know if you add another resistor in, you know the resistance is going to decrease so the current is going to increase so the potential is going to decrease. Thats the explanation without numbers.

I like to make up numbers to confirm things, especially when transposing more complex formulas,
I will use ohms law though as an example of this
Say i know V=IR
and i transposed for I which is, I = V/R as you know,
But i want to make sure that that's right,
To check, I would make up a value for I and a value for R so lets say I=5 and R=10
5x10 = 50 so V is 50

Now implement that into your transposed formula,
So, if you had that 50/R which is 10 = 5, what you was looking for, so that tested that that transposition is 100% correct, then you have no doubts what-so-ever.

Its just a little tip I like to use to make sure my transposition is 100% correct in exams, if your transposition is wrong then your answer is wrong, by checking transposition, you've atleast got a chance then haha,

Take this formula for example as attached, I wouldn't risk loosing marks over some stupid transposition errors so i would check it if i had time left (got a bit off topic now haha)

EDIT: attached came up horribly on screen view, it is formula used in RLC circuits

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19. Jun 25, 2013

Too high order for me- that formula P
I guess thats the only way - to sub in values :(( which takes a whole lot of time for me and theres time limit for exams :(

20. Jun 25, 2013

### gary32

You don't need to sub in values to transpose, I just use it to test my transposition if i have time,

To transpose, you just need to practice ;)