According to Ohm's law: $V=IR$. Therefore $V$ is directly proportional to $I$. Furthermore, $V=W/q$ and $q=IT$. Thus $V=W/IT$. Therefore $V$ is inversely proportional to $I$.

How can $V$ be both directly and inversely proportional to $I$? ($V$=Potential Difference, $I$=Current, $T$=Time, $q$=Charge, $W$=Work).

  • $\begingroup$ Let $y = x$. Now $y$ is proportional to $x$. If I define $k = x^2$, then $y = k/x$. Is $y$ suddenly inversely proportional to $x$ now? $\endgroup$ Oct 4, 2019 at 7:14
  • $\begingroup$ I get that but How should I relate the above question with it? $\endgroup$ Oct 4, 2019 at 7:25
  • $\begingroup$ What exactly is happening above? $\endgroup$ Oct 4, 2019 at 7:27
  • 1
    $\begingroup$ What's happening is that you are just cramming equations together without thinking about the physics, assumptions behind the equations, etc. It is a common mistake of beginner students. $\endgroup$ Oct 4, 2019 at 12:05

6 Answers 6


You are forgetting that $W$ is not an independent constant. In fact $W=I^2 R\ T$ so that $\frac{W}{I\ T}$ is indeed not inversely proportional to $I$.

  • $\begingroup$ How do you decide what is independent or dependent constant. I can write R=V/I and say R depends upon V and I so R is not independent. $\endgroup$
    – Ali
    Mar 27, 2020 at 10:10
  • $\begingroup$ @ALi In this case it is obvious (to me) that the OP is setting $R$, $I$, and $T$ as independent variables, and letting $V$ vary. In the end he wants to also use $W$ as an independent variable which he cannot do if $I$, $R$, and $T$ are independent. It is generally the user's choice as to what is independent and what is not, but that choice has to be consistent. $\endgroup$
    – Bill Watts
    Mar 27, 2020 at 17:27

If $V$ would be inveresly proportional to $I$, then there would be a constant $k$ such that

$$V = \frac k I$$

Sorry, in your formula

$$V=\frac W{I T} $$

which is the same as

$$V=\frac{\frac W T} I$$

the part $\frac W T$ is not constant.

$\left(\frac W T \right.$ is work by unit time, i. e. power, which is not a constant independent from $V$ and $I$.)


Ohms law gives the fundamental relationship between voltage and current for a resistor. V is proportional to I where the proportionality constant is R. The other equations are manipulations of various combinations of ohms law and power (P) or work (W).

When the various equations are manipulated, the constants of proportionality change and contain other hidden variables that change the relationship between voltage and current. For example, the relationship between voltage, current and power in a resistor can be written as


Now V is inversely related to I. But the proportionality constant is now P (power)

So which is it. Is voltage proportional to current as shown in Ohms law, or inversely proportional to current as shown here? The answer is both, depending on the proportionality constant.

Similarly, in the following power, $P$, is proportional to current, $I$


But if we substitute IR for V from Ohms law, we get


Is power proportional to $I$ or proportional to $I^2$. Both, depending on the proportionality constant being $R$ or $V$.

Hope this helps.

  • $\begingroup$ "But the proportionality constant is now P (power)" - but $P$ isn't constant with respect to $I$ for a resistor so I don't see how, for a resistor, $P$ can be (in any sense) a proportionality constant. Writing $V=\frac{P}{I}$ for a resistor does not imply that "V is inversely related to I" does it? $\endgroup$
    – Hal Hollis
    Oct 4, 2019 at 16:03
  • $\begingroup$ @HalHollis For a fixed power dissipation in a resistor, mathematically yes. Look, all I am trying to point out to the OP is that, with the exception of Ohms law, you shouldn't be using manipulations of related equations to determine the true relationship between voltage and current in a resistor. Nothing more, nothing less. $\endgroup$
    – Bob D
    Oct 4, 2019 at 16:21

In the relation V=W/IT potential I s not exactly inversely proportional to I because if we increase current I then the work done W will also increase by I^2 as W=I^2RT. So the W will also increase and the value of potential will also Increase.


Let for T and R be 1. Then

V=I^2 x 1 x 1/ I x 1


If we increase I from 1 to 2 then potential will also increase!

For I=1, V=1 For I=2, V=2 ..

This is Just because we cannot Increase current I but keep work done W be same or constant.


What exactly is happening above?

Note that the $R$ in

$$V = RI$$

is ordinarily understood to be a constant. We say something like

For the ideal resistor, the voltage $V$ is proportional to the current $I$ where the constant of proportionality is the resistance $R$

In your post, you wrote an equation involving the work $W$

$$V = \frac{W}{T}\frac{1}{I}$$

and claimed that this implies that the voltage is inversely proportional to the current.

Now, think about it. If that were true, then doubling the current would halve the voltage, correct? That is, if $\frac{W}{T}$ is a constant of proportionality, then doubling the current would not change the factor $\frac{W}{T}$, correct?

But the factor does change when the current is doubled. If $I' = 2I$, then it's easy to show that

$$\frac{W'}{T} = \frac{4W}{T}$$


$$V' = \frac{W'}{T}\frac{1}{I'} = \frac{4W}{T}\frac{1}{2I} = 2\frac{W}{T}\frac{1}{I} = 2V$$

So, in fact, doubling the current doubles the voltage, i.e., the voltage and current are proportional. The equation you wrote does not imply that the voltage is inversely proportional to the current.


Proportionality requires a constant.

  • $R$ is a constant (it doesn't change when $V$ or $I$ change).
  • $W$ is not a constant (it depends on $I$; higher current causes more work done).
  • $\begingroup$ How do you decide what is independent or dependent. I can write R=V/I and say R depends upon V and I so R is not independent. $\endgroup$
    – Ali
    Mar 27, 2020 at 10:16
  • $\begingroup$ @Ali We don't decide that. Nature does. We can only measure it to find out which parameter that is independent. Purely from the mathematical expression, we cannot know - we need to know about the physics of reality and the origin of a parameter to figure it out. $\endgroup$
    – Steeven
    Mar 27, 2020 at 10:19

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