I just don't understand circuits... at all...

[Ganymede12]

I am in my final year of A-levels, doing physics, and a topic I can't seem to get my head around is circuits. 

It's just that nothing seems to quite work the way it should. Ignoring the fact that conventional current is in the wrong direction, I have all these equations (which, fortunately, I don't have to remember for the exam) and  some just don't quite make sense with each other. 

So, there's P=(I^2)xR, which implies that Power and resistance are directly proportional to each other but then P=(V^2)/R implies that they are inversely proportional. Now, as I understand it, you use the first when the current is constant and the second if  P.D is constant. That's fine. But then there are some circuits in which neither is constant? and the Power varies? 

I've always thought that for equations like this to work, some element must be constant. In this past paper, everything is variable! 

There is a cell, ammeter and variable resistor connected in series with a voltmeter connected across he resistor. I have a graph showing me that p.d decreases as current increses/resistance is decresed and doing calculations shows that the power output from resistor is at its maximum at some point in the middle of the graph. 

I basically want to know which features have changed into reaction to what. When the resistance is changed, why does the voltage change? Why would he cell give a lower e.m.f for a circuit just because the circuit is easier to get through and how does it "know" to do so? 

Sorry for the long post. Any answers, explanations or links would be greatly appreciated. 

[MarkyD]

They all work out the same (for DC circuits)

Take 10V supply with a 20 Ohm load

P = (10^2)/R

P= 100/20 = 5 Watts

I = V/R

I = 10/20 = 0.5Amps

P = (0.5^2)x20 = 5 Watts

 

Bear in mind that resistance changes with temperature.  Conductors /metals increase resistance with temperature and Semiconductors decrease resistance with temperature

 

 

[Ganymede12]

1 minute ago, MarkyD said:

They all work out the same (for DC circuits)

Take 10V supply with a 20 Ohm load

P = (10^2)/R

P= 100/20 = 5 Watts

I = V/R

I = 10/20 = 0.5Amps

P = (0.5^2)x20 = 5 Watts

 

Bear in mind that resistance changes with temperature.  Conductors /metals increase resistance with temperature and Semiconductors decrease resistance with temperature

 

 

I can see that it does work, but I'm trying to figure out why it works. A lot of exam questions require explanationns rather than blind calculation, and personally I am much happier using  equations if I know where they come from. 

[MarkyD]

I'm sure Mr Ohms asked himself the same thing lying in bed  night after night wondering why the maths works like it does.

[Starwiz]

2 minutes ago, Ganymede12 said:

I can see that it does work, but I'm trying to figure out why it works. A lot of exam questions require explanationns rather than blind calculation, and personally I am much happier using  equations if I know where they come from. 

You can try to simplify it in your own mind.  I assume you know about the hole-electron theory, holes flow one way, electrons the other.  Imagine it as water flowing in a pipe.  If you increase the resistance by tightening a valve, the pressure builds up.  This is similar to voltage in an electric circuit.  The amount of water that flows is similar to the electric current.  And so it goes........  

John

[Ganymede12]

Okay... so what about the dependence thing?  If you decrease the resistance of a circuit then how can the cell "choose" to decrese as well? Shouldn't e.m.f be constant? 

[Davey-T]

Ah water running down the pipe, remember the coulombs from school but don't remember any equations, did end up doing a lot of plumbing 

Dave

[MarkyD]

Don't forget that a cell is basically a power source with a chemical resistor in series which probably has a lot to do with it.  I've never gone this deep as Mrs Blogs just wants her 13Amp socket changing and not the whys or wherefores

Sorry

[cloudsweeper]

The power max problem is standard theory.  The cell has internal resistance r, so there is always a volt drop across that. The load voltage is then less than the cell's EMF by those "lost volts". 

E = Ir + IR.............EMF = lost volts + load voltage.

 As far as power to the load is concerned, it can be shown by a bit of differentiation that the cell delivers max power to the load when internal resistance r is equal to load resistance R.

Doug.  

 

[Alien 13]

As mentioned above real world cells have internal resistance the wiring has resistance that changes with temperature but even after nearly 40 years in the electronics industry what always got me confused was the way diodes where marked ie anode non pointy end but positive and the stripe on the cathode

Alan

[Ganymede12]

Alright, so how is it that you can have circuits with variable resistors and changing the resistance changes everything else about the circuit? Why  doesn't, for example, just current change to compensate for the change in resistance? Why does P.D have to change as well?  

[SilverAstro]

1 hour ago, Ganymede12 said:

 

So, there's P=(I^2)xR, which implies that Power and resistance are directly proportional to each other

That would only be true if I^2 was a constant, it is not in this context,     

but then P=(V^2)/R implies that they are inversely proportional.

again, only true if V^2 was a constant, it is not.

But then there are some circuits in which neither is constant?

and thus you have given yourself the answer ! Would it help if I suggested you read up on Parametric Equations ?

 

[cloudsweeper]

42 minutes ago, Ganymede12 said:

Alright, so how is it that you can have circuits with variable resistors and changing the resistance changes everything else about the circuit? Why  doesn't, for example, just current change to compensate for the change in resistance? Why does P.D have to change as well?  

You vary the load resistance, so that changes the TOTAL circuit resistance, hence the current delivered will change. (EMF is fixed.)  Now, because you have changed the current, the lost volts Ir will change, and so the load voltage IR will also change.  (See earlier equation.)  So changing the load resistance changes the load voltage and  the current in the load (same as current in the whole circuit for a series circuit).

The key to understanding all this is the significant role of internal resistance in the source of EMF.

Doug.

 

Finally - look at it like this: if there were NO internal resistance in the source, changing external/load R would alter the current, but NOT the load voltage.  That would be constant, equal to source EMF, because there is NEVER any volt drop for zero internal/source resistance.

[wimvb]

Does this help?

voltage source.pdf

[Islander]

I^2 x R and V^2/R are derived from Ohm's law and the standard power equation.

Ohm's law says that the voltage across a resistance is a product of the resistance and the current in amps, i.e. V = I x R

Power in Watts is the product of the voltage across a load and the current through thatt load, i.e. P = I x V

So, if you know the current and the resistance and you want to work out the power then you have  V = I x R and P = I x V  and since we don't know the voltage across the load we can subsitiute the values for current and resistance from Ohm's law in place of the V in the current formula which gives us P = I x I x R or P = I^2 x R

Similarly, if we know the voltage and the resistance, since Ohm's law can solve current as I = V/R we can subsititure this in place of current on the power equation giving us P = V/R x V or P = V ^ 2/R

Voltage is the electrical pressure or 'push' that moves current around a circuit.  Current is the flow itself.  Resistance limits the amount of current that can flow for a given value of pressure or voltage.  Increase the resistance and less current is able to flow and similarly reducing the pressure or voltage reduces the current as well.  Conversely if you reduce the resistance or increase the pressure or voltage, then more current can flow.  The higher the current for a given value of load resistance across a circuit the greater the power dissipated in that load.  Increasing the power can be achieved by decreasing the load resistance or increasing the voltage.  Decreasing the power can be achieved by increasing the load resistance or decreasing the voltage.

If you have problems visualising this then think of it in terms of plumbing.  The voltage is the head or height of the water source.  The higher it is the greater the pressure available.  The current is the rate of flow of water and the resistance is the internal diameter of the pipework.  Varying these is directly analagous to varying voltage, current and resistance.

 

[cloudsweeper]

6 hours ago, Ganymede12 said:

Alright, so how is it that you can have circuits with variable resistors and changing the resistance changes everything else about the circuit? Why  doesn't, for example, just current change to compensate for the change in resistance? Why does P.D have to change as well?  

To continue my argument - in the simple case of r = 0, as R increases, I falls, so power falls - in other words P is inversely proportional to R - and this is backed up by using the

P = V^2/R "option" which now "works" since V IS CONSTANT as @SilverAstro pointed out.  Remember - this is only so for the "ideal" case of r = 0.  In the real world case where there IS internal resistance, P delivered to load follows a square law function, i.e. a curve with a max value for P , when r = R.

Internal resistance is the key to all of this.

Doug.

[Bukko]

Hello, Ganymede12,

Just stick with Islander's explanation, above.

This answers your original question nicely. All the stuff about internal resistance, etc. does not matter. In the real world, the V^2 component would be measured across the resistor.  For current, it would also change due to external factors, but  the current flow can be measured anywhere in the loop. If they want you to consider this, they will give you some internal resistance values. If not, then it is assumed to be an ideal case.

 

Good luck with your upcoming exams !!!

Gordon.

[cloudsweeper]

The original post says:

....and doing calculations shows that the power output from resistor is at its maximum at some point in the middle of the graph. 

This can only happen if there is internal resistance to consider.  Other posters recognised this.  Look at the text and graph at:

http://www.electronics-tutorials.ws/dccircuits/dcp_9.html

If there is a fiixed cell with no internal resistance, then load voltage is constant and power (V^2/R) is inversely proportional to R.  Another poster recognised this.

Take a look a graph showing inverse proportionality between P and R - it's asymptotic to both axes - it does not peak; there is no maximum.

Accepting the info in the original post, it has be said this is not a simple situation, but it is A-level Physics after all.

 

 

[Stub Mandrel]

The power equals the rate of work done pushing electrons past a resistance.

Power is proportional to how many electrons move AND the work done in moving each one through the circuit.

Current is proportional to the number of electrons - the rate.

The amount of work can be expressed in different ways:

1

 

[Macavity]

As a sometime Physicist, am I allowed to suggest that A-Level Physics can be a tad... Uninspiring?
In the late 60s, all Lead-Acid batteries, Rheostats and Ancient (albeit now venerable) Multimeters!

Quote

“If it squirms, it's biology. If it stinks, it's chemistry. If it doesn't work, it's physics..."

But no fault of Physics [IMO] or indeed the teaching thereof! A good thing to know the basics:
Ohm's Law" (Kirchoff's Law even). But, if you really want to understand "circuits" (for your own
satisfaction and interest), why not take a "trip" to e.g. MAPLINS and buy some components?

A Basic electronics "Breadboard", a pack of Resistors, a modern digital Multimeter etc. etc.
You too can explore such things in the comfort of your own home! Personally, I often found
it easier to learn at my own pace and outside of the classroom. LATER, I could bring the two
together and revisit the more "rigourous" formal stuff. Meanwhile I had built a REAL World
"circuit" that DID something... An Oscillator that made a noise? A basic Radio that worked? 

JUST an "off-beat" suggestion! There are a lot of KITS that allow you to learn such stuff...

[Davey-T]

Good idea but not much point knocking up a few components on a breadboard to flash a LED without some accompanying literature to explain what's occurring and why.

Dave

[George Jones]

In order to see what information the question gives and what the question wants, it would be useful to see the actual question. Can you (Ganymede12) scan in the question?

[Stub Mandrel]

I congratulate @Ganymede on NOT just giving the question (like so many students do) but asking for help understanding.

You have three quantities V, I, R all interrelated. If you change one and keep another the same than the third must change.

Start witha 1 ohm resistor with 1 volt across it, so that (I =V/R) 1 amp flows. Thiscreates a power of 1 watt P=IV P= 1x1  P=I^2 x R = 1x1x1  P= V^2/R = 1x1/1 = 1

 

If you double V to 2 volts:

if the resistance is kept the same, current doubles to 2 amps.   The power increases four times

P=I x V  = 2 x 2 = 4

P=I^2 x R = 2^2 x1  = 4

P=V^2/R =4

To keep the current at 1 amp, you must double R as well as V. The power doubles:

P=I*V = 1 x 2 = 2

P=I^2 x R = 1 x 1 x 2 = 2

P= V^2/R = 2 x 2 /2 = 2

 

 

The reason R can be at either the top or the bottom is because although power is proportional to the square of voltage and the square of current it is NOT directly related to the resistance.

If you have a constant voltage source and change the resistance, power increases as resistance reduces - but this is because more current flows.

If you have a constant current source and change the resistance, power reduces as resistance reduces -  because the voltage across it drops.

 

The best way to get your head around this is to imagine a fan heater, which is basically a big resistor. If you run it off the mains, say 240V, you will need about 4 amps to give about a kilowatt of power. If you run it off a car battery, 12V, you will need a huge 20 amps to get the same power. Now think what that means in terms of resistance - the car heater must be a puny 0.6 ohms, while the mains heater has to be a huge sixty ohms to limit the current to 4 amps. Work out the effect of plugging each heater into the wrong power outlet.

[Scott]

9 minutes ago, Stub Mandrel said:

I congratulate @Ganymede on NOT just giving the question (like so many students do) but asking for help understanding.

You have three quantities V, I, R all interrelated. If you change one and keep another the same than the third must change.

Start witha 1 ohm resistor with 1 volt across it, so that (I =VR) 1 amp flows. Thiscreates a power of 1 watt P=IV P= 1x1  P=I^2 x R = 1x1x1  P= V^2/R = 1x1/1 = 1

 

If you double V to 2 volts:

if the resistance is kept the same, current doubles to 2 amps.   The power increases four times

P=I x V  = 2 x 2 = 4

P=I^2 x R = 2^2 x1  = 4

P=V^2/R =4

To keep the current at 1 amp, you must double R as well as V. The power doubles:

P=I*V = 1 x 2 = 2

P=I^2 x R = 1 x 1 x 2 = 2

P= V^2/R = 2 x 2 /2 = 2

b

 

The reason R can be at either the top or the bottom is because although power is proportional to the square of voltage and the square of current it is NOT directly related to the resistance.

If you have a constant voltage source and change the resistance, power increases as resistance reduces - but this is because more current flows.

If you have a constant current source and change the resistance, power reduces as resistance reduces -  because the voltage across it drops.

 

The best way to get your head around this is to imagine a fan heater, which is basically a big resistor. If you run it off the mains, say 240V, you will need about 4 amps to give about a kilowatt of power. If you run it off a car battery, 12V, you will need a huge 20 amps to get the same power. Now think what that means in terms of resistance - the car heater must be a puny 0.6 ohms, while the mains heater has to be a huge sixty ohms to limit the current to 4 amps. Work out the effect of plugging each heater into the wrong power outlet.

Sorry to be a pedant but ohms law is E=IR, therefore I= E/R and R=E/I. therefore,if you work with 2 volts and 1ohm I=2/1 giving a current of 2A

Whilst this is the same as your answer, if you also increase the resistance to 2 then I=VR becomes 4A where as I=E/R becomes 1A

If this is wrong I'm truelly sorry and I'll post it in this post but I think my memory from college (40 yrs) are correct

A small but quite important difference

oh I should point out that E= voltage 

[Stub Mandrel]

No,

27 minutes ago, Scott said:

Whilst this is the same as your answer, if you also increase the resistance to 2 then I=VR becomes 4A where as I=E/R becomes 1A

 

No, 2V across 2R = 1A.