# Why should you need to know the electric current in a circuit?

1. Jul 28, 2009

### user111_23

Okay, I know what current is. It's Q/t. But why should I know what the current is? What makes current so important besides the fact that it's a factor in voltage, power, resistance, etc?

Sorry for the plethora of electricity topics I have created. :tongue:

2. Jul 28, 2009

### negitron

You've got a 12-amp fuse connecting a 120 V source to a 1500 W load. Will the fuse blow?

3. Jul 28, 2009

### user111_23

I'm guessing yes? I don't really know.

4. Jul 28, 2009

### mgb_phys

For a given voltage, and in most power applications voltage is fixed, current tells you how much power you have - both to the load and heating/destroying the cables

5. Jul 28, 2009

### negitron

And that's why knowing the magnitude of the current is important.

6. Jul 28, 2009

### MATLABdude

Is this slow blow, or fast blow? Is it 1500W continuous, peak, or average?

This further illustrates that not only is current (and understanding of it) an important, fundamental concept, but also that it's just the beginning of the picture...

EDIT: Oh wait, you're posting in General Physics... (Which is probably why you gave the mathematical formalism of current) Still, it's an important concept to get.

7. Jul 28, 2009

### pixel01

In fact the current is the one that makes things done in a circuit. You may have experienced static shock in the winter, it's thousands of volts, but the current is tiny so you are safe.

8. Jul 29, 2009

### user111_23

Okay, knowing the current for safety reasons is there. But what does current have to do with the performance of electrical devices?

Knowing the voltage and resistance can help me calculate the current. What is the significance of this electric current provided that the circuit is already running?

9. Jul 29, 2009

### user111_23

How does current tell me how much power (in watts) is there?

By the way, why should I care about power anyway?

10. Jul 29, 2009

### maverick_starstrider

? Current, voltage and resistance are all related. Just look at how a transistor operates (or an op-amp). An understanding of current is the most important consideration when designing a circuit since, as has been already pointed out, the voltage is usually fixed by the power source. Remember, for example, the basis of the magnetic force is the CURRENT in a wire, not the voltage.

11. Jul 29, 2009

### user111_23

Why is it important to know that the voltage is fixed?

I'm asking a lot of questions, I know. :tongue2:

12. Jul 29, 2009

### Bob S

If you have a reactive load, like a refrigererator or vacuum cleaner or power tool with an electric motor, the current is out of phase with the voltage, so the volt-amps exceeds the watts, sometimes by a large factor.

13. Jul 29, 2009

### maverick_starstrider

... Because without knowing the voltage you can't determine the current?.... Lol, what are you rockysheaphear 2.0 (or 0.5)

14. Jul 29, 2009

### user111_23

All I'm asking is: What does current have that other factors in electricity don't, and why is this quality important?

15. Jul 29, 2009

### turin

negitron basically nailed it in the first response. When it comes down to a physical circuit, there are devices in there with fixed electrical properties (i.e. resistance) and fixed thermal properties (i.e. at what temp it will melt, or worse, catch fire). Current is the direct external measure of electricity that flows through a given device that determines how hot it gets. So, current is the most direct measure of electricity that determines how efficient a circuit will be and how dangerous a circuit will be.

A fuse is one example. Perhaps a more profound example is the wiring in your house, supposing that there were no circuit breakers. Suppose you connect a bunch of power tools to a single circuit in your house. Power tools have an amperage rating, which basically tells you what? - the current that the power tool draws. Why is that important? If you have, say 100 amps worth of power tools connected to a single household circuit, then the wiring in the walls is gonna get hot, real hot, and possibly start a fire. This is the case even though they all operate at 110 volts.

16. Jul 29, 2009

### user111_23

Thanks, you helped a lot!

I have a few more questions; 1; is resistance a factor in the heating of wires? 2; how does current determine how efficient a circuit is?

17. Jul 30, 2009

### turin

I noticed that pixel and maverick also aluded to electronic devices, which is a different issue for which current is also important, so I just wanted bring that back into the conversation as well. Modern electronics use electricity in a sophisticated way. That is, electric and magnetic fields are important design considerations, as well as the motion/mobility of charge carries. For instance, the operation of a BJT (a still quite common type of transistor) is most straightforwardly modeled in terms of the current that flows into and out of the connections.

Yes, mathematically. The simple version of electric heating is RI2. But it is the current that brings the energy into the wires; the resistance is just a parameter. As has been pointed out, this can also be expressed as VI for resistive/ohmic devices, since V, I and R are related. However, if the device is highly reactive, then only part of that voltage goes into the heating, and RI2 is the more direct calculation, where R is the resistive part of the impedance of the device. (I hope that I am not confusing you more. Look up "power factor".)

Consider the power lines that run from the power station to your house. These lines transport a lot of electrical power; they need to supply your house with all of the needed electricity, as well as hundreds of other houses.

I will simplify and approximate the situation a bit for the sake of clarity:

In order for you to get, say 1 kW at 100 V, that would require 1 kW / 100 V = 10 A to be transmitted down the power line. If the electric company actually did that over, say 10 km, and suppose the resistance of the power line was 1 Ω/km, then the power loss in the line would be

( 1 Ω/km * 10 km ) * ( 10 A )2 = 1 kW.

That is, the 10 A distribution would result in a 50% efficiency. However, the electric company does not do this. Instead, they use a transformer to transfer the power at a much lower current, say 1 A for this direct residential application. Then, the power loss in the line would be

( 1 Ω/km * 10 km ) * ( 1 A )2 = 10 W.

That is, the 1 A distribution would result in a 99% efficiency. I have ignored other loss mechanisms, such as the loss in the transformers, but the basic idea is that lower current makes the transmission more efficient.