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P=vI need some explanation

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nirVaan

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hello there

im finding it hard to undestand the equation p=vi .i know its definition. but when i think a bout it more it does not make sense to me. why is power equals to volatage into current what does this mean what is power in terms of electronic

if V is 4 and R is 2 then current is 2
if V is 8 R is 2 and then current is 4

in both these is the work done by the voltage is same ?
and why does power increase exponentially when v and i increase linearly

im all confused can you help me to figure this out.

Thanks
 

Hi nirvaan,

that's really an easy question. If you take the definition of P = VI, now substitute I = V/R. It becomes P = (V)Λ2/R.

That's why power increases exponentially with the voltage. Now, here is one important thing, if you have a circuit and a voltage source in the circuit. suppose 12 V source, and you have a resistance of 2 ohms. Now, you want to measure the voltage drop across that resistance, how do you measure it? You need to how much is the current value in ampere. You need to measure that current using ameter and then you can multiply that current value to resistance to get the voltage dropped across the resistance. and power is the dissipated energy in the resistance. i.e., P = V^2/R

Thanks
 

    N

    nirVaan

    Points: 2
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thankz

can you tell me how voltage into current actually equate to energy dissipated in the resistance. i hope you guys understand what im trying to say.

i mean if you take V=IR its the relation ship is apparent there is a pressure there is a flow of electrons and there is a resistance that opposes the flow how this equate to V=IR is understandable.

but for me understanding how voltage multiplied by current equals to power is really not.

and thankz for the reply
 

nirVaan,
The volt is a unit of electromotive force. The coulomb is an electrical charge equivalent to 6.24150962915265E18 electrons. The electromotive force acting on the electrons causes the electrons to move. This is force through distance, or work. An Ampere, by definition is 1 coulomb per second. The result being work per unit time, or power.
Regards,
Kral
 

    N

    nirVaan

    Points: 2
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thankz !!!!!!!

it make sense now. but can you guys tell me

f V is 4 and R is 2 then current is 2
if V is 8 R is 2 and then current is 4

in both these is the work done by the voltage is same ?

thank you
 

No... you've doubled the potential energy, or force. So you've moved twice as many electrons (doubled the current). So you've done twice the work. Power has increased fourfold.
 

    N

    nirVaan

    Points: 2
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okay thanks guys i have one more small little problem.

A: f V is 4 and R is 2 then current is 2
B: if V is 8 R is 2 and then current is 2

which one is more efficient im guessing its A(because less potential to move same number of chargers).if so what if current is 4 in B is the same now in terms of efficiency?

and does power have any relationship with efficiency in a circuit like if more power less efficient ?

sorry if these question seems stupid or if it feels im going on circles.this is what came to mind mind while reading the reply's

thankz
 

nirVaan said:
B: if V is 8 R is 2 and then current is 2
Click to expand...


Thats impossible unless u use a current limitter or a voltage divider like shown in the figure.

Obviously A is better since it consumes less, uses less components so less expensive.
 

    N

    nirVaan

    Points: 2
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yes im sorry in B: it should be V is 8 R is 4 and then current is 2

but i get the point now

thankz
 

Remember that volt (V) is a derived unit of SI and can be expressed in this way:

1 V = 1 (kg . m^2)/(A . s^3) #### (note that gram, meter, ampere and second are the basic units)
or
1V = 1 (N . m)/(A . s) #### (1 newton = 1 kg . m / s^2)
or
1V = J / C #### ( one volt is equal a one joule of energy per coulomb of charge)

You know that J = W . s and C = A . s

Replacing:

1 V = (W . s) / (A . s) =>>> 1V = 1W/1A (law of Joule)
 

    N

    nirVaan

    Points: 2
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rkodaira said:
Remember that volt (V) is a derived unit of SI and can be expressed in this way:

1 V = 1 (kg . m^2)/(A . s^3) #### (note that gram, meter, ampere and second are the basic units)
or
1V = 1 (N . m)/(A . s) #### (1 newton = 1 kg . m / s^2)
or
1V = J / C #### ( one volt is equal a one joule of energy per coulomb of charge)

You know that J = W . s and C = A . s

Replacing:

1 V = (W . s) / (A . s) =>>> 1V = 1W/1A (law of Joule)
Click to expand...

Thanks for this theory rkodaira,

Have you a material detailing the MKSA unities ?
I have a doupt regarding the Power difinition.
Using a cad tool (let say Design compiler, Eldo/spectre) after synthesis/simulation of a circuits the tool says that the designed chip consumes a total power of , let say 400 mW.
When we talk about bulbs of ligth we have 75 W bulbs, 100W etc.

Is that amount of consumed power is per second ?
That mean, for exemple the Chip, the chip consumes 400 mW per second ?
 

First, thank you for the points.

1) I do not have any material about MKS (meter, kilogram, second ?) system, but I think a quick search in the Google will solve it.

2) You are making some confusion about power definition. We do not say 400mW per second, it is wrong. Because POWER is the rate of energy (or work) consumed/generated/dissipated per time unit, so 1W = 1 joule of energy per second.
POWER = ENERGY (or WORK) divided per TIME
Does not exist such unit POWER divided per time.

So your IC has 400mW of dissipation, or it has a rate of 400mJ per second of heat dissipation, or in other words, it dissipates 400 mili joules per second (uses electrical current to make some work and generates heat).

Do not worry about this confusion, a lot of people does the same.
 

Nirvaan,
the effeciency is defined as η=(Output Power)/(Input Power),
Where : η is the efficiency of the system.
so the system can be considered more efficient when the output power increases,
so in the first case : V=4V, R=2Ω , this yields to I=V/R=4/2=2A
and the output power is equal to : P=V^2/R=16/2=8 W.
In second case : V=8V , R=2Ω , this yields to I=4A
and the output power is equal to : P=V^2/R=64/2=32 W.
For the constant value of input power in both cases, we see that :
η for the first case = 8/(Input Power)
η for the second case = 32/(Input Power),
The second case is more efficient than first case because it has a greater value,
I hope my answer be correct and explained the idea to calculate the power.
 

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