# Potential difference and current

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).

• 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? – Marius Ladegård Meyer Oct 4 '19 at 7:14
• I get that but How should I relate the above question with it? – Shreya Soni Oct 4 '19 at 7:25
• What exactly is happening above? – Shreya Soni Oct 4 '19 at 7:27
• 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. – Aaron Stevens Oct 4 '19 at 12:05

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 $$I$$.

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

$$V=\frac{P}{I}$$

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, $$I$$

$$P=IV$$

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

$$P=I^{2}R$$

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

Hope this helps.

• "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? – Hal Hollis Oct 4 '19 at 16:03
• @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. – Bob D Oct 4 '19 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.

V=W/IT

Let for T and R be 1. Then

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

V=I^2/I

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

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

Thus

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

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).

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 $$I$$.)