Shock Value B

[robotman]

Power is the rate at which work is done.(edited so that it was right )
Power is converted into different forms for different things. light energy for light bulbs, heat energy for a heater or sound energy for a speaker.
There are a few things that i think they could ask about power (this is from a worksheet)
An electric space heater draws 15.0 A on a 120 V line. How much power does this heater use?

The full worksheet plus some others are now on the wiki
Shock Value Wiki

[blue cobra]

Thanks, I think I'm starting to get it.

When reading a circuit diagram, are we supposed to assume conventional flow or electron flow? But since there are no semiconductors in this event will it even matter?

[Helix_Nebula_Freak]

I actually think you need to ask the event coordiantior for the type of flow to be assumed...

[robotman]

Thanks, I think I'm starting to get it.

When reading a circuit diagram, are we supposed to assume conventional flow or electron flow? But since there are no semiconductors in this event will it even matter?

For most times I would assume that it was conventional flow because most circuit diagrams are asking you what is being powered or what is the resistance.

just to be safe i suggest asking the event coordinator

[andrewwski]

Power is not the amount of work a circuit can do. That would naturally be work.

Power is the rate at which work is done.

Power is measured in watts. Work is usually measured in joules, but also in watt-hours or kilowatt-hours. 1 kwh equals one kilowatt being drawn for one hour - or power * time.

Conventional vs. electron flow shouldn't really matter when looking at circuit diagrams - the positive and negative terminals of a component (or schematic symbol) don't change regardless of how we think of the electrons as flowing.

But assume conventional flow unless otherwise stated - it's pretty standard. The electron flow model is not used as commonly, although as you know it's actually correct.

[blue cobra]

What would the voltage be in a circuit if you had a 9v battery and a 1.5v battery wired in parallel?

[andrewwski]

BAD IDEA.

The 9V is going to be feeding current into the 1.5 V until their voltages equalize. But the 1.5 V battery is not designed to have that much of a charge, so it will likely get very warm, or possibly catch on fire. It would be very dangerous.

The total voltage would be somewhere in between 1.5 V and 9 V, but it would depend on the capacity of the batteries themselves. Remember a 9 V is actually 6 small 1.5 V cells in series.

But don't ever connect them that way. It would be very dangerous.

[Hamtown009]

How do you calculate voltage drop? I've looked at some things but I can't seem to get the grasp on it.

[blue cobra]

Voltage drop was a difficult concept for me to grasp as well. But as I understand, it is as follows:

I like to think of calculating voltage drop as applying ohm's law to one resistor. Ohm's law tells us that voltage=current x resistance. So suppose you have a resistor circuit with a 9v battery and two 100 ohm resistors. Use ohm's law to find the current in the whole circuit.
I=E/R
I=9/200
I=0.045 amps
Current is the same everywhere in the circuit, but voltage is not (which is why we can have voltage drops). So now that we know the voltage before the resistor, the current at the resistor, and the resistance of the resistor, we can calculate voltage drop, using Ohm's law.
E=IR
E=0.045 x 100
E=4.5v (That's the voltage drop!)
Since both of our resistors are the same- 100 ohms- we'll find their voltage drops are the same.
E=IR
E=0.045 x 100
E=4.5v
Something else to notice is that all of the voltage drops in a circuit add up to the voltage source. 4.5+4.5=9.

Some of the more knowledgeable people on here may be more helpful than me, but if you haven't already, check out Episode V of the Wiki. It's about voltage drops, and it links to another sight about it as well.

[Hamtown009]

Ohhhhh. ok so you take your total resistance and use ohm's law to figure out the current. then you use the current and multiply it by one of the resistors to get the voltage drop?