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A series of easy (but I can't do) physics questions

OP

Sidhe

you have ohms and volts right of the first resistor in x

Calculate amps

I=V/R

You now have amps and resistance.

V=IR

For the second resistor.

repeat for Y

Power(Watts)=VI

do x
do y

With your value for amps and volts for each.

Compare values. Answer the question, breathe a sigh of relief, go have a cup of tea or a coke you earned it :smile:

OMG, why can't someone do this for me. Ok according Sidhe,

For X,

I = V/R = 1.5/0.5 = 3 A

V = IR = 3 * 3 = 9 V (wtf - that's greater than 1.5V - just great)

For Y,

I = 1.5/2 = 0.75 A

V = IR = 0.75 * 3 = 2.25 V

Power = IV

For X, P = 3 * 9 = 27 W

For Y, P = 0.75 * 2.25 = 1.7 W

Is this right? Well it does work out for the correct choice A, but not sure whether values are correct - I have a feeling it's definitely wrong.

I HATE CIRCUITS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! I'd rather deal with chemicals -_-

And no I did not earn any tea/coke - I am so stupid and lazy (for physics anyways).

Reply 21

plmokn

Welcome to the joy of ambiguous symbols!

I = V/R = 1.5/0.5 = 3 A

The R here is the total resistance of the circuit i.e. the two values of resistance added together not just 0.5.

V = IR = 3 * 3 = 9 V

This R you've got right (it's the voltage across the load resistor you want to find).

Reply 22

Sidhe

aye it's 1.5/3.5

For the first one.

Gives you a nice manageable voltage. :smile:

there's your mistake the rest is fine. You just have the wrong numbers is all.

And the reason people don't do it for you is you wont learn anything that way, plus I don't know what the rules are here, but generally homework helper sections frown upon people doing someones homework for them, without any input from the homework doer type person.

Reply 23

Eau

OP

15

plmokn

Welcome to the joy of ambiguous symbols!

The R here is the total resistance of the circuit i.e. the two values of resistance added together not just 0.5.

This R you've got right (it's the voltage across the load resistor you want to find).

LOL OF COURSE!!! Thanks, you confirmed by doubts about my answer. :biggrin:

Reply 24

Theorist

10

plmokn

P=Rload*I^2
Write an expression for I, plug it into the expression above for power, and use calculus to find the turning point (remember you're varying the load resistor and keeping the internal resistance constant).

The actual differentiating should be simple if you simplify the fraction as much as possible first and think about what needs to be maximised/ minimised.

As it happens this is quite important if you like loud music.

I'm stuck getting the expression to differentiate.

[indent]Do I give the internal resistance an arbitrary value? What about the e.m.f?
Am I right in saying that we wish to differentiate

P

with respect to

R

?
Am I also right in thinking that I substitute

I = \frac{\epsilon}{r + R}

into the original equation?[/indent]

Please help me get the expression. I'll slog from there. :wink:

Reply 25

Theorist

10

Sidhe

I don't know what the rules are here, but generally homework helper sections frown upon people doing someones homework for them, without any input from the homework doer type person.

You should the see the Mathematics forum. It's a race to give the answers, seriously!

Reply 26

plmokn

Am I right in saying that we wish to differentiate P with respect to R?

Am I also right in thinking that I substitute ... into the original equation?

Exactly, keep everything as symbols.

Disclaimer: the result you get isn't going to massively change your perception of the world!

I've put a couple of hints below in white (highlight to view) that will make the differentiation easier.
take a factor V/r , where r is the internal resistance, out the front of the fraction, also divide the top and bottom of the fraction by R. Then define a variable say s=R/r, and differentiate P wrt s. Since you should have a fraction of the form 1/something you actually only need to differentiate the bottom and set the result=0

Reply 27

Sidhe

Dharma

You should the see the Mathematics forum. It's a race to give the answers, seriously!

That's silly people won't learn anything if you just do all there work for them, I have observed the maths section though, and I don't think people just spoon feed, I think they might give pointers as well more than just doing the whole thing from scratch. Mind you with maths you always know the answer because you can just get it from a calculus helper on line, or some software package. :smile:

Reply 28

Theorist

10

plmokn

Disclaimer: the result you get isn't going to massively change your perception of the world!

Still. It's about the process involved. :smile:

plmokn

I've put a couple of hints below in white (highlight to view) that will make the differentiation easier.

take a factor V/r , where r is the internal resistance, out the front of the fraction, also divide the top and bottom of the fraction by R. Then define a variable say s=R/r, and differentiate P wrt s. Since you should have a fraction of the form 1/something you actually only need to differentiate the bottom and set the result=0

I officially give up. I can't do it. :frown:

I don't see how you can take out a factor of

\frac{V}{r}

when not all the terms of

(r+R)^{2}

have a common factor.

Could you please tell me how to do it?

Reply 29

Eau

OP

15

Lol, yes when you post something in the maths forum, answers arrive in matter of seconds!

double post

We've got:

P=RI^2

P=V^2 \times \frac{R}{R^2 + r^2 + 2rR}

P=\frac{V^2}{r} \times \frac{1}{R/r + r/R +2}

P=\frac{V^2}{r} \times \frac{1}{s+1/s +2}

Then to find a turning point we only need to differentiate the bottom of the second fraction wrt s and set=0 to find the value of s and therefore the value of R.
Differentiating and rearranging gives

s^2=1

which means R=r at the maximum.

The result is simple to extend to complex impedances (once you know about them) which take into account capacitors and inductors. It's important to note that this doesn't imply the maximum efficiency is 50%, you can get almost all of the energy supplied by the emf dissipated in the load resistor, but not at the same time as maximising the power output.

Reply 32

Theorist

10

plmokn

P=V^2 \times \frac{R}{R^2 + r^2 + 2rR}

P=\frac{V^2}{r} \times \frac{1}{R/r + r/R +2}

How did you get from that step to that step? Are you not missing anything? :confused:

plmokn

P=\frac{V^2}{r} \times \frac{1}{s+1/s +2}

Then to find a turning point we only need to differentiate the bottom of the second fraction wrt s and set=0 to find the value of s and therefore the value of R.
Differentiating and rearranging gives

s^2=1

which means R=r at the maximum.

How did you differentiate the expression, and what is the identity that relates

R

to

s

?

Reply 33

plmokn

Take out a factor r from everything on the bottom, then divide the top and the bottom of the fraction by R.

I defined s=R/r, you don't really need, you could differentiate directly wrt R but it makes it slightly simpler.

To see why its valid to just differentiate the bottom of the fraction when looking for turning points, consider a function f(x)=1/g(x) differentiate both sides, then f'(x)=-g'(x)/(g(x)^2), so if g'(x)=0, f'(x)=0, where ' means the derivative of.

Edit: I'm not sure what level of maths you've studied, so if you've not met the chain rule in differentiation (I think it's C3/4?) you'll probably be completly lost by what I've done above.

A series of easy (but I can't do) physics questions

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