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I am confused on this.

We know that P = IV and V = IR. That means P = V^2/R by substitution.

I understand that by that second equation we can say "power is proportional to voltage squared" but what if we had gone with the first one, P = IV? In this equation power is proportional to voltage by itself. What if I had not known about Ohm's Law? Would it have been wrong to say power is linearly proportional to voltage based on P = IV?

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    \$\begingroup\$ Yes, because I is a function of V, as long as we're talking about resistors. Power is linearly proportional to voltage, though, if you're talking about a constant current device. \$\endgroup\$ – Hearth Jan 20 at 3:30
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    \$\begingroup\$ @Hearth So power is proportional to V^2 if resistance is constant? \$\endgroup\$ – user525966 Jan 20 at 3:43
  • \$\begingroup\$ Yes, like Spehro and Tony said. \$\endgroup\$ – Hearth Jan 20 at 3:47
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    \$\begingroup\$ P=IV is fundamental (always true, instantaneously, in any circuit). V=IR is only true for resistors. \$\endgroup\$ – mkeith Jan 20 at 8:05
  • \$\begingroup\$ I think this question arose due to something I left unsaid in a comment to your previous question. I apologize. \$\endgroup\$ – DKNguyen Jan 20 at 14:15
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It depends on the circumstances. Without knowing anything else, we don't know whether power is proportional to \$V\$, \$V^2\$, or neither.

If a variable voltage source is connected to a resistor, then the power is proportional to \$V^2\$.

If a variable voltage source is connected to a constant-current load (something which admits the same amount of current regardless of the voltage across it), then the power is proportional to \$V\$.

If a variable voltage source is connected to a constant-power load (one which admits more current when the voltage is less, and less current when the voltage is more, so as to consume a constant amount of power, and thereby acting as a negative resistance), then the power is independent of \$V\$. A switched-mode power supply may behave in a manner similar to this. It would be pretty straightforward to design a device which consumes \$1\ \mathrm{W}\$ of power when connected to any voltage between \$5\ \mathrm{V}\$ and \$24\ \mathrm{V}\$, for example.

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  • \$\begingroup\$ Might be worth expanding to explain that if you have constant-current circuit, you must have a variable resistor so that the resistance goes up with the voltage. If that happens, then the P=V^2/R equation has to take account of both V and R increasing, so reduces to the simple proportionality. \$\endgroup\$ – Oscar Bravo Jan 21 at 12:55
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    \$\begingroup\$ @OscarBravo: The concept of resistance is only meaningful if the ratio between voltage and current remains reasonably constant when those values change. If R=E1/I1 and R=E2/I2, then E1=I1R aand E2=I2R. If one knows E1, I1, and I2, and wants to know E2, one can thus derive E2 as E2=I2(E1/I1), i.e. E2=E1(I2/I1). If R isn't constant, then E1=I1(E1/I1) would be vacuously true for any non-zero current I1, but the usefulness of R as a concept stems from its ability to relate (E1/I1) and (E2/I2). \$\endgroup\$ – supercat Jan 21 at 21:07
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    \$\begingroup\$ Constant power is proportional to \$V^0\$ ;-) \$\endgroup\$ – Nayuki Jan 22 at 20:25
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If (iff) R is fixed the power is indeed proportional to \$V^2 ~ and~ I^2\$.

You may easily prove the latter by similar substitution.

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If the resistance is constant, then increasing the voltage increases the current proportionally, exactly proportionally in the ideal case.

Power is not proportional to voltage squared in cases where the load is not a constant resistance. For example, with a rectifier diode it will increase considerably faster than voltage squared. If you measure power into a current regulator diode, it will increase more like proportional to voltage since current will be more-or-less constant over a range of voltages.

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In the general case: Variable1 = a constant of proportionality * Variable2 and so Variable1 is proportional to Variable2.

In the specific case of P = (1/R)(V^2) 1/R is a constant of proportionality. For a fixed resistance value, R does not change and so P is proportional to V^2.

In the specific case of P = I * V I is not a constant of proportionality because, for a fixed resistance value, as V increases so does I in proportion so P is still proportional to V^2.

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The power by definition* is the product of the current and the voltage across a one-port (a two-terminal device). The power is only proportional to the square of the voltage if the \$I(V)\$ relationship is linear. If it is constant, i.e. \$I=I_0 \rightarrow P \approx const \cdot V\$. The \$I(V)\$ characterisic could be anything else, but for a common case where \$V=I\cdot R\$, \$\rightarrow P= V^2/R = I^2 \cdot R\$. If you have a diode, or transistor, or a voltage dependent resistor, there will be no physical resistance value to use, and power is calculated as \$P(t)=I(t)\cdot V(t)\$.

* let's assume it is the definition. It can be derived from more basic principles.

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Instantaneous power is the instantaneous product of voltage and current. But current is almost always somehow dependent on voltage (though not always linearly). So to determine power in most circuits you need to develop a formula for current as a function of voltage. For a "purely resistive" load that formula is V/R, so power is (V^2)/R.

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*I am only trying to portrait the answer through mathematics

There exists a law used widely in electrical engineering, as I'm sure you have heard of it, the Ohm's Law, which states at a given R and I, the voltage could be calculated as V=IR. you need to understand that this law is not a linear equation, it's rather an instantons fact that stands, meaning at every differential time step, the product of resistance and current would give you the Voltage.

Now, the definition of power in physics is the rate at which work is done. in mathematical terms, that would be:

$$d(work)/dt = Power$$

So, lets now look into the primal defnition of voltage,work required per unit of electric charge, so it is essentially work/charge. If you look at this in differential terms, then it would become $$V = d(work)/d(charge)$$

thus $$d(work) = V*d(charge)$$ so if you differentiate both sides with respect to time: $$d(work)/dt = V*d(charge)/dt$$ (btw mathematicians would go berserk if they see you move differential segments around so don't do it around them!). So now we have the 2 tems we know in this equation: $$Power = d(work)/dt$$ $$Current = I = d(charge)/dt$$ giving us the final format: $$P=V*I$$ thus proving it represents something that's instantaneous. Now, think of voltage as pressure applied across a tube, and current representing how much water goes through, as you can see one is dependent on the other, so you cannot isolate one variable to say that the multiplication of both would rely on (meaning that I*V represents two variables not a constant and a variable). Now imagine these 2 equations representing the relationship between voltage and current: $$V=0.5*I---------------------------(1)$$ $$V=0.75*I^2-------------------------(2)$$ so for (1), if you calculate power, it would indeed be proportional to I^2, however for (2), it will have a different relationship. Also look at Ohm's law like this that: $$R(1) = 0.5$$ $$R(2) = 0.75*I$$ so for (1) R is constant and for (2), R actually depends on an external factor (the current itself in this case). This is a pure mathematical example and ofc does not hold in an experimental setup (as there are different factors influencing how the relationship would change)

So as a final line of answer, If the relationship between V and I is linear, then yes power is proportional to V^2, but this is not a generalized rule

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For a one-port, power is proportional to V times I. What you are missing is the constitutive equation of the one port, i.e. the relation that links the voltage across it to the current flowing through it

$$ V = f(I) \; or \; I = g(V) $$

Without knowing that relation you cannot say if the power that goes into (or even out of) your port is proportional to V, V^2, V Exp[V] and so on.

In general, if you want to focus on the voltage, you have

$$ P = V I = V g(V) $$

and unless you specify further your V, you really cannot say. If the one-port is a resistance following Ohm's law, the relation is linear: \$g(V) = 1/R \; V\$, so

$$ P = V * 1/R * V = 1/R * V^2 $$

For a diode you will get a power that depends on the product of the voltage with an exponential function of the voltage.
I cannot imagine a passive device that dissipated a nonzero power whose power is only proportional to V: it would imply that the current does not depend on the voltage applied.

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