Question

power of AC circuit
p=du/dt=d(qv)/dt=q dv/dt+v dq/dt.
Why do we use in AC p = IV!! and cancelling the other term.

Answer 1

Do u know
dk/dt=0
If k is a constant

Answer 2

dv/dt=0
If potential difference v=constant

Answer 3

U know
Current i=q/t
Or, i=dq/dt

Answer 4

Response plz!!!

Answer 5

for AC voltage is constant really !!

Answer 6

I think it is r.m.s value of ac voltage

Answer 7

And rms value of ac voltage is constant

Answer 8

But when proved RMS by making p =I*V

Answer 9

Rms=root mean square

Answer 10

Anyway we almost always use rms value of voltage

Answer 11

I know .If you look at the proof of RMS is just by making p=integral of v^2/r and then you will get rms for v

Answer 12

So u got it. Right?

Answer 13

NO
we assumed that p = I*V then got R.M.S!
I am asking about why p = I*V it should be =i*v+q dv/dt.

Answer 14

U know
p=I*v+q*dv/dt
p=I*v+q*0
p=?

Answer 15

Here
dv/dt=0
Because v=rms value of voltage=constant

Answer 16

dv/dt isn't 0
v isn't the rms why are you saying that?

Answer 17

Then u will cancell it

Answer 18

why should I cancel it?

Answer 19

Why drop the dV/dt term? Look at when you do work, ( and this is key ) in moving a charge through a potential difference. If the charge is not moved or transported by the potential diff no work is done and no power required. So even when the Potential diff. changes with time, it can only contribute to work and power is a current is produced. So in the term q*dV/dt even though dV/dt is non zero q is fixed in time, no charge is transported. That's my story and I'm sticking to it. (at least until someone can come up with a better idea)

Answer 20

There is a drawback when measuring power in AC circuits using P=IV due to a reactance which will create a phase difference between voltage and current. This phase angle, if present, has to be considered in the computation. P(AC) = V I Cos(phase angle). Normally RMS values apply to the voltage and current.

Answer 21

I still don't know why would we put dv/dt =0

Answer 22

@Catch.me i think u r frm usa. Right? So u use ac current of 110 volt. This 110 volt is rms value of ur ac current and this is constant. Right? Derivation of any constant is zero?

Answer 23

I mean v=110volt
So
dv/dt=d(110)/dt=?

Answer 24

AS I know RMS appear when we make p = V^2/R. which assumed that p=I*v only, so this is a cyclic reasoning to put rms in account!!

Answer 25

.

Answer 26

@Catch.me

in the posts you have made above, what do you think dq/dt means?

Answer 27

dq/dt = i :P
time rate of flowing charge(changing current)
\[i = n*A*v _{d}*e\]
what do you mean?!

Answer 28

yes, i think you do mean that dq/dt = I

but, you started with this: p=du/dt=d(qv)/dt

which is based on U = qv? is that correct?

if that is the case, then dq/dt is not current......

go back to the fundmental equation (Coulumb) and you will see what i am trying to explain.

Answer 29

0_o! I can't get your point coulomb is for statics.

Answer 30

i would argue that coulumb is for everything. as in everything can be taken back to coulumb and, there, dq/dt means an actual change in q with time.

i know there's a really ugly fudge wherein we calc a drift current and dq/dt re-appears in an equation as "rate of transfer of charge", I but that has a different meaning. because, back to basics, if U = QV, then indeed U' = Q'V + QV', but meaning Q is getting bigger or smaller with time.

if the charge is actually physically moving you are looking at something very different. in the context of a voltage drop across a resistor and current I flowing through that resistance, it is very simple:

\[P =\text{ work done per unit time}= \frac {VQ}{t} = VI \,\]

if V = V(t) and this I = I(t), this becomes:
\[P = V(t)I(t)\]

of course, apologies if i am missing the point you are trying to make.

Answer 31

my confusion is why don't we take power for AC circuits as U' = Q'V + QV'

Answer 32

the power used by an AC circuit at any instant of time is that instantaneous current times the instantaneous voltage (p=iv),

with v =|V|cos(wt+a)

and i =|I|cost(wt+b).

If you average this over a cycle you get net power P= |V|*|I|cos (a-b) .

Also note that for work to be done you need a potential difference and a charge flow at the same moment.

Answer 33

**power for AC circuits as U' = Q'V + QV'**

i will exagerate this purely to make the point i have been trying to make. why not take **power for AC circuits as F = ma** or why not take **power for AC circuits as E = mc^2**

i am genuinely not wishing to be facetious or rude. i just think that, whilst the mathematical outcome of d/dt (QV) is as you say it is, it is important to know what that really means in the physical world. U = QV if there is a charge Q in a field at a point where potential = V. if the charge or the field change with time, the d/dt (QV) explain how potential energy of that charge/field arrangement changes with time. so that's a charge in a field.

in order to derive the formula for electrical power in a wire, you need to build a mathematical model that reflects that reality, including electrons moving in a uniform electric field and obeying Ohm's Law.

IOW, U = QV is not the correct start point. IMHO.

Answer 34

Most? books start by saying the energy required to move charge Q under the influence of a potential difference V is VQ. The power required is determined by the rate at which you move the charge or the current produced by the potential difference. Therefore the power is VI. This is of course born out by experimental verification.

Answer 35

"Most? books start by saying the energy required to move charge Q under the influence of a potential difference V is VQ. "

that's completely by definition. great.

"The power required is determined by ***the rate at which you move the charge*** or the current produced by the potential difference."

agreed, but that's not dQ/dt, is it?

dQ/dt is the rate at which the charge changes wrt time. doesn't happen. the electrons don't change theor coulumbs, nor does the body of electrons that flood past a point undergo a charge change.

i disagree with @Catch.me on the main thrust of this thread, for reasons that seem very clear to me; but i confess i am struggling to find a way to see beyond the fudge that is dQ/dt (change in amount of coulumbs of a charge per unit time) = I (amount of charged particles that pass a fixed point per unit time).

Answer 36

Current is the flow of charge, the rate at which the charge passes a given point in coulombs per sec. . Just like liters per sec is rate of flow of a fluid. I amp is one coulomb (i.e., 6.25 E^18 electrons) per second.

Answer 37

i know that, and it doesn't help, unfortunately.

dQ/dt measures the rate of change of Q wrt time. Q is not changing. dQ in that mathematical sense = 0. the charge's position is changing. that's totally different. and that's the point.

of course, I know that I = Coulombs/ s : measured past a point in a circuit.

the OP was ripping through the axiomatic U = QV (from Coulomb) and trying to re-write the Power equation. my problem is that I struggle to find a way to relate power to I and V without making simplistic assumptions about what dq/dt means. and i struggle to find any kind of first principles derivation online. forget about AC, that is irrelevant.

same applies to a simple DC circuit -- i think i may be over-complicating the issue. but dQ/dt as a mathematical proposition **IS** change in Q over change in t.

the definition is : U = QV, thus dU = dQ V + dV Q

in a circuit, the story is different.

Answer 38

@IrishBoy123

You are missing the point dq/dt is the rate at which charge is being moved not created or destroyed across a potential difference. That is what the term work is all abou,t moving something under the action of a force.

Answer 39

i assure you I am NOT missing the point, in fact this is the point and i am delighted that we have reached this far!!

in the physical world, and in our modelling equations, q is the charge and t is time. we have an equation that relates the magnitude of that charge and its potential to its potential energy.

mathematically, something is missing that relates q to its velocity. of course you can just stop using dq/dt for what it means -- rate of charge of quantity of charge with time -- and use it instead as current and everything is completely hunky dory -- i see that clearly, but my point is that that is incomplete.

U = QV
dU = dQ V + dV Q
dQ = 0, electrons are not vanishing
dU/dt = Q dV/dt
it is V that is changing with time ie it is the potential of that charge that changes with time -- forget about AC, confine to DC for now --as the charge moves in the field. if it moves at a drift velocity v, then there is a way to reconcile this all with notions of current and Coulumbs per second.

that is what i am looking for.

btw, really appreciate your input and consideration. honestly, i am not nit picking. there is a point to this even if it might seem that i am being pedantic.

if there is no point to this, then the OP is right, yet i am sure that he is not.

Answer 40

@Catch.me

I think this is the reasoning is you are calculating instantaneous power

So its true that the potential difference is changing.
But we are calculating at every instant, (for a given potential difference) what is the power dissipated.

So p = VI = the instantaneous power.
Only then we average it out, and then introduce RMS value and all that


I also like Gleem's answer, q dv/dt although mathematically correct, I don't think it denotes anything physical. If there is no transfer of charge, and only the potential changes with time, there isn't any work done what so ever!. (or may be that creates a magnetic field?)

Answer 41

@IrishBoy123

Wow.. I never ever thought about it. You are absolutely right.
dq/dt literally means (rate of change of an entity named as charge with respect to time)

For example
when a capacitor is charging, we can think about dq/dt because the charge on the capacitor plate is increasing. (Or discharging, when it is decreasing)

But when we have a current, we define it as the amount of charge flowing through a cross section per second.
So if dq is a small amount of charge that flows through a cross section in time dt, then
current
I = dq/dt.

So dq/dt has two completely different meanings. (one is the rate at which charge is growing or decreasing, and another the rate at which charge is flowing)

But I think ultimately they both will have the same value.
Conservation of charge.

Imagine charging of capacitor.
dq/dt = rate at which charge on the capacitor is built. lets say it is 5C/sec
What that means is every second charge on teh capacitor is increasing by 5C.
But if the charge on the cap is increasing by 5C, then there must be a flow of 5C of charge every second through any cross section of the wire. Thus the current must also be the same.

Do you think that makes sense? :P

Answer 42

@Mashy

thank you.