# Voltage-current relationship in the wire

I'm reading Art of Electronics by Paul Horowitz and Winfield Hill. Although it is a well-written guide to electronics, I'm stuck with the first topic.

I would appreciate any answer that can help me to understand relationship between voltage and current as well as between current and wires.

Given the voltage (V) is drop (or potential difference) in charge between two points, the current (I) is how much electrons (charge) moving through some point during unit time, and the power (P) which concept I cannot understand in a nutshell (mostly by intuition.)

We're going to make power constant. Now, following the formula P=V*I, if we increase V, then I should drop and if we decrease V, the I will rise. At this moment my head goes boom, because if we lower potential difference between two charges significantly, the rate of flow of the charge will drastically increase. I cannot fully understand this

The second problem: If with the constant voltage and power (in formula) I was about to change width of the wire, should the current rise or drop? A wider wire should give us less resistance and therefore more space for charge to flow.

I read alot on this topic and watched tons of videos, but still, this easy concept is hard for me to comprehend.

• V=I x R and P=V x I, so if you want to keep P constant while raising voltage you have to add resistance (make wire narrower). To keep P constant while lowering voltage you would lower resistance (fatter wire). Dec 10, 2022 at 14:33
• @user1850479 , makes sense. Thank you Dec 10, 2022 at 14:39
• When you're looking at power and wires, you need to be careful about whether you're looking at the power delivered by the wired to the load they're connected to or the power lost in the wires due to heating caused by the wires' resistance. This is something which trips up many noobs to the subject. Dec 16, 2022 at 4:24

Given the voltage (V) is drop (or potential difference) in charge

This may be the source of your confusion. Current, as your first learn of it, is the movement of charge. But voltage drop is not a "drop in charge" or a "difference in charge". What is briefly called potential difference, is the difference in electrical potential between two points. Electric potential can be said to exist everywhere, including places where there is no electric charge. It is a field. Likewise, potential differences can exist between two points where neither the points themselves, nor the points in between contain any charge.

At this moment my head goes boom, because if we lower potential difference between two charges significantly, the rate of flow of the charge will drastically increase. I cannot fully understand this

Potential difference and current, although completely distinct on a conceptual level, are nevertheless related on a physical level -- through Ohm's law. In many cases, the current through a component (or wire) is fairly exactly proportional to the voltage drop across the component, and this holds true for a wide range of voltage drops and currents. In that case we say that the component has a particular resistance (within a certain range of voltages and currents). Further, we say that this resistance is the ratio between the potential difference and the current.

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

Or, rearranged

$$V = IR$$

However, if you keep in mind that this relationship is really a relationship between two conceptually distinct things, and not a "drop in charge" vs a flow of charges, perhaps your head will not go "boom".

If with the constant voltage and power (in formula) I was about to change width of the wire, should the current rise or drop? A wider wire should give us less resistance and therefore more space for charge to flow

If you keep voltage and power constant, then by Watt's Law, $$\P = VI\$$, the current must also remain constant. Watt's Law, unlike Ohm's Law is valid regardless of the type of component, or the magnitude of the voltage or current.

[Later, you will learn that some components can store energy, and then release it later. You will also learn that there is a kind of current called displacement current, that does not require a movement of charges. However, these are refinements of the concepts, not alterations in the exactness of the relationship given by Watt's Law. That the ratio between current and voltage is fairly constant is an empirical law valid for a range of materials and magnitudes of voltage and current. That the power transformed in an electric circuit is the product of the voltage and current we take to be more axiomatic. Knowledgable people may wonder whether in a particular component the ratio between voltage and current remains constant if the voltage changes. However, knowledgeable people do not wonder whether $$\P = VI\$$, unless they are questioning some model or theory of physics.]

Returning from the aside, if your voltage is constant, and the only resistance in your circuit happened to be a wire, if the wire were made twice as wide, the resistance would be half as much, so the current would be twice as much. Thus, the power would be twice as much. The power, in this case being the unchanged voltage times the doubled current.

The information you're missing is simple:

You can't keep power constant while changing voltage, without changing something else in the circuit as well.

In a simple resistive circuit, V = IR. The voltage across a resistor is directly proportional to the current through it, which means that power is uniquely determined by just the voltage (or equivalently, just the current). If you change the resistance as well as the voltage, then you can keep power constant while changing the voltage.

There are active circuits that can "change the resistance" automatically to keep power constant; switching converters powering a fixed load generally act like this, for instance. (Thinking of them as having a variable resistance glosses over so many details that I would say it's wrong to think of them this way, but it's a serviceable lie-to-children until you start actually working with them.)

Suppose you think of the wire as a resistor, and for simplicity suppose it has a fixed resistance.

If you apply a voltage across the resistor (or wire) then a current will flow. Once you have decided the voltage you have no control over the current. It will be I = E/R. Similarly, if you decide to put a current through the wire, the voltage across it cannot be independently controlled, it will be E = I * R.

And, in either case, the power dissipated by the wire will be P = E * I. You really can only only control one thing independently (voltage, current or power), the others just follow.

If someone goes and changes the wire on you, say they double the length, then you'll have to change the voltage (to double) to maintain the same current. The power would then be doubled since each length has the same voltage and the same current, and there are two of them.

If you want to maintain the same power then you would need to increase the voltage by $$\\sqrt{2}\$$ - that would give you less ($$\\frac{\sqrt{2}}{2} = 1/\sqrt{2}\$$) current, but the power would be the same. That falls out from P = E^2/R or P= I^2*R.

It may not be obvious, but the same thing applies to things that don't have a fixed linear resistance such as LEDs. You can change the current, and the voltage and power will change, but you don't get to independently specify more than one of those three things.