Do you "lose" electricity when you course it through subpar conductors?

Question

Imagine I had a basic circuit - say the classic 9V battery on one end, a couple of wires, and a little light bulb on the other.

Of course, in a real world example those wires would probably be insulated copper, but say I had an example of this circuit where the wires were actually made of a much less conductive metal.

If I were to compare the copper wires to the less conductive wires, what would the difference be? Would the less conductive wires actually deliver less electricity to the light bulb, would some amount of electricity be "lost" in the process, would it ultimately deliver the same amount of electricity to the light bulb but take longer? What would happen?

Answer 1

Assuming your circuit is a simple loop from the battery to the bulb and back to the battery, then the current will be the same at all points in your circuit. So no electricity is "lost" in the circuit.

However, as the electrons (which make up the electric current) pass through a wire with a resistance that is greater than zero, their energy is reduced. This reduction in energy shows up as a drop in voltage from one end of the wire to the other.

If a current $I$ flows through a wire with a voltage drop of $V$ then the reduction in electrical energy per unit time along the wire is $IV$. If the wire is simply carrying the current from one place to another (rather than being part of an electromagnet or an electric motor, for example) then all of this energy is transformed into heat energy. This is why a wire that is either carrying a large current or has a high resistance (or both) gets hot and can even melt.

In an incandescent bulb, some of the electrical energy is transformed into light, which is what we want. However, quite a lot is still transformed into heat, which is why incandescent bulbs become hot.

If you compare a light bulb circuit with high resistance wires with one with low resistance wires, and the voltage supplied by the battery is the same in both cases, then the circuit with high resistance wires has a grater voltage drop across the wires. The sum of the voltage drops from one side of the battery, round the circuit to the other side of the battery must equal the battery's voltage. This means that the voltage drop across the light bulb is smaller in the circuit with higher resistance wires, and the light bulb will be lit less brightly.

Answer 2

Terminology

You're asking about amounts of "electricity", but that's an ill-defined term. Physics knows measurable quantities like:

• Power, expressed in Watt units (W), telling how much work (e.g. heat) can be delivered per time unit. For a light bulb, it roughly corresponds to its brightness. Power can be computed as the product of current and voltage.
• Current, expressed in Ampere (A), telling how many electrons flow through a given point / wire / ...
• Voltage, expressed in Volt (V), not so easy to explain in everyday language. You can roughly think of it like a pressure or force that drives the electrons through the wires.
• Resistance (e.g. of a wire), expressed in Ohm (Ω), telling you how much voltage you need (and therefore lose) to make a given amount of current flow through the wire. The voltage needed is the product of resistance and current (Ohm's Law).

Example

Let's say, your bulb is meant for 9V, 0.1A, and the length of wire you use has 1Ω resistance.

As electrons don't accumulate in places (at least not in simple circuits like yours), all electrons entering a point also exit it (at the opposite side), meaning that the same current is observed in every place in the conductors as well as through the bulb.

What about the voltages? You start with 9V at the battery, and this has to be divided between the voltage needed to "push" the electrons through the resistive wire, and the voltage arriving at the bulb.

With 1Ω resistance and a 0.1A current (1 - see below), the voltage loss is 0.1V, meaning that only 8.9V arrive at the bulb, meaning that it will be a little dimmer than expected.

The battery provides a power of 0.9W (9V * 0.1A), and this gets split:

• 0.01W (0.1V * 0.1A) is lost in the wires (heating them up),
• 0.89W (8.9V * 0.1A) are consumed by the bulb (producing light and heat).

The greater the wire resistance, the more of the precious battery power gets wasted in the wire.

Summary

All the current the battery provides also arrives at the bulb.

At the bulb, the voltage is lower than the voltage provided by the battery, the difference being lost over the wires.

Part of the electrical power that the battery delivers is lost over the wires, only the remaining amount can do useful work. The light is dimmer.

If instead of a light bulb you want to drive a heater (not very practical from a small 9V battery), it will be slower, meaning to take longer to reach the same temperature.

(1) To be exact, a 9V, 0.1A bulb will draw a bit less than 0.1A when supplied with only 8.9V instead of 9V, but let's ignore that here...

Answer 3

If I were to compare the copper wires to the less conductive wires, what would the difference be?

The difference would be a greater drop in voltage across the less conductive wires than the copper wires, resulting in a lower voltage available to operate the light bulb.

Would the less conductive wires actually deliver less electricity to the light bulb

The less conductive wires will cause an increase in resistance and less current for a fixed battery voltage, as $I=V/R$ per Ohm's law. The reduction in the current delivered to the circuit means less power dissipated in the wires and bulb, as the power dissipated varies as the square of the current and linearly with resistance, or $P=I^{2}R$. For an incandescent bulb, that means it will operate more dimly.

would some amount of electricity be "lost" in the process, would it ultimately deliver the same amount of electricity to the light bulb but take longer? What would happen?

The term electricity is vague, but electrical energy is never lost, just converted to different forms (light, heat, mechanical energy, etc.). But as I already indicated, the voltage across the bulb will be less meaning it will consume less power. The time it takes to deliver the power is virtually unchanged since the electric field is established in the circuit near the speed of light, regardless of the resistance in the circuit.

Hope this helps.

Answer 4

It would just take longer (the electric current would be less) if you apply the same voltage at the ends of the conductor. But the electricity is not lost, it is just slowed down at the same force that pulls the charges through the wire.

You can somehow compare a worse conductor with a thinner pipe in hydraulics. If you try to pull your cola through a thinner straw, you need more pull from your tongue. But the cola is not lost.