Scioly.org

You couldn't - not directly anyway.

With a multimeter, just set it to ohms and measure it.

If you wanted to use a voltmeter, you'd need to set the resistor up with a power source, then measure the current it draws (with an ammeter) and the voltage drop across it (with the voltmeter). Then perform Ohm's Law.

just wondering, what resources are people using to study this event? the offical page on soinc doesnt seem to have much...

Ohm's Law states: E = I x R

Where:
R = Resistance in Ohms
E = Electromotive Force in Volts
I = Current in Amperes

If R increases, E must also increase in order for I to remain the same.

In a simple circuit with only 1 EMF source, 1 resistance and 1 current path, increasing the resistance will cause current ( I ) to decrease. In circuits with more than a single resistance in series, each resistance uses (drops) only a portion of the total (applied) voltage. Increasing one of the resistances causes the voltage it drops to increase. Since the applied voltage remains the same, voltage dropped by the other series resistors must decrease.

For Series Circuits:

E(total) = E(R1) + E(R2).... + E(Rn)
R(total) = R1 + R2.... + Rn
I(total) = I(R1) = I(R2).... = I(Rn)

For Parallel Circuits:

E(total) = E(R1) = E(R2).... = E(Rn)
R(total) = 1/[(1/R1) + (1/R2).... + (1/Rn)]
I(total) = I(R1) + I(R2).... + I(Rn)

Edit: Corrected typo in equation for Ohm's Law

Impressive! I'm guessing it's copy and paste, but did you get this from the soinc website?

I joined Shock Value because I loved to build circuits. I get very confused with all of the terms in the Shock Value meets, but it's very fun and I'm sure that I'll learn all of the terms at some point.

So total resistance in parallel can be expressed as

Rt=1/[(1/R1)+(1/R2)+...(1/Rn)]

So if we had two 2 ohm resistors wired in parallel the total resistance would be

Rt=1/[(1/2)+(1/2)]
Rt=1/[(1)]
Rt=1 ohm

Now if we make it four 2 ohm resistors the total resistance would be

Rt=1/[(1/2)+(1/2)+(1/2)+(1/2)]
Rt=1/[(2)]
Rt=0.5 ohms

How does increasing resistance cause less resistance?

That's correct.

You're not increasing resistance, you're decreasing it.

Adding resistors in series increases resistance. Adding resistance in parallel decreases resistance.

The best way to describe it might be to use an analogy to a water pipe. Say you have a 1/2" diameter water pipe - you can only fit so much water through it at a time. But if you have two 1/2" water pipes, you can fit twice the amount of water through at a time - thus your "resistance" is halved.

If you have only two resistors, the formula can be simplified to . Or, if you have multiple resistors of the same value in parallel, the total resistance is equal to the value of the resistors divided by the number of resistors.

You're not increasing resistance at all. You're providing another path for the current to flow through - thus you're decreasing total resistance.

Do you need a switch in a circuit? The judges at Invitationals seem to think so.

You don't have tonier a switch to complete a circuit
however If they give you one you should use it

scifipi wrote:Impressive! I'm guessing it's copy and paste, but did you get this from the soinc website?

Actually I created it using the gray matter found in the intercranial spaces between the left and right auditory canals!

Blue cobra wrote:How could you find the resistance of a resistor with a voltmeter?

As stated previously you can't do it with just a voltmeter. One way would be to place it in series with another, known, resistance and pass a current through them. By measuring the voltage drop across each of the resistors, you can find the unknown resistance.

Where:

R(1) = Known Resistance
R(2) = Unknown Resistance
I(1) = Current Through R(1)
I(2) = Current Through R(2)
E(1) = Voltage Dropped by R(1)
E(2) = Voltage Dropped by R(2)



Ohm's Law states:

E = I x R

or for a given resistance: R(1)

E(1) = I(1) x R(1)

Solving for I:

I(1)=E(1)/R(1)

Since the two resistances are in series, the current is the same for both.

I(2) = I(1) = E(1)/R(1)

Using Ohm's law again and solving for R, the equation for the unknown resistance, R(2) is:

R(2) = E(2)/I(2)

Substituting for I(2) produces:

R(2) = E(2)/[E(1)/R(1)]

or

R(2) = [E(2) x R(1)]/E(2)

In conclusion: All you need to know is the resistance of the unknown resistor. By measuring the voltage drops across both resistors, you can calculate the value of the unknown resistor.

I need one more event and was looking at a couple of the open spots for my team and came upon shock value could you tell me what it is about and what you need to study/know?