The Student Room Group
well, whatever manipulation u do, ur still going to end up with E so im a bit confused lol
or.. would u write P/I = V and equalise this to E = Ir + IR ??? eeek! sorry, cant help u on this
Reply 3
P can also be written as P= I^2R if that helps..
Reply 4
sorry I phrased the question wrong.
I have two formulas
p = IV
and IR+ Ir= E
i need to condense it into a formula in the form mx+c= y
and it must only include the letters
E , P, R and r
oh lol, make I the subject of both equations and the equalise them to eachother, then rearrange
or maybe not...coz uve still got V....hmmm
Reply 7
I'd make I= P/v and subsitute it into E= IR + Ir.

Whoops: Nevermind
but then uve still got v??:confused:
Reply 9
xx-footy-fan-xx
oh lol, make I the subject of both equations and the equalise them to eachother, then rearrange

but I can't seem to get it as a linear equation I keep getting loads of squared values
Reply 10
Ya so you can use the equation P=I2/RP=I^2/R

then when i do that I get
Unparseable latex formula:

RE^2/(R+r)^2 {\frac}

you can yes. but u said that we can only use P = IV and E = Ir + IR ???
Reply 12
How about:

P= IV But V= IR so P= I^2R

Therefore I = Squareroot of P/R

Then sub that into E= IR + Ir?

You'll get something akward but eh..it still works, I think!
p = IV IR+ Ir= E

P/V = I I(R+r) = E ==> I=E/(R+r)

P/V = E/ (R+r)

P(R+r)/V = E
(PR + Pr) / V = E
PR + Pr = EV then i dont know... im lost lol
Reply 14
xx-footy-fan-xx
you can yes. but u said that we can only use P = IV and E = Ir + IR ???

no i meant those were the two i am given at the start
i dont know. the only thing i can think about is: IR = V and therefore E = V + Ir
then rearrange to get V = -Ir + E to get it in the form u want.
Reply 16
Uh, is that a no on my solution then?
lol, im convinced that this question is impossible
Gathering
Ya so you can use the equation P=I2/RP=I^2/R

then when i do that I get
Unparseable latex formula:

RE^2/(R+r)^2 {\frac}


yes that's the method, but it's:

P=I2RP=I^2R not divide by R.

Though it depends whether your P is defined as the power from the battery or P is the power dissipated by the resistance R