Shock Value B/Circuit Lab C
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Re: Shock Value B/Circuit Lab C
What do you do in the lab part of Shock Value (Div.B)?
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Re: Shock Value B/Circuit Lab C
This is my first time doing a study event, so I'm not really sure what needs to go into the binder. Are you guys putting every little thing that you find about circuit lab (every article, all notes, etc) into the binder, or do you use it more as reference material, and just put the stuff you would have to look up (resistor color bands, etc)?
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Re: Shock Value B/Circuit Lab C
The lab portion kinda depends on your event supervisor, there might be cases where you don't even have a lab portion on the test. But usually it's some basic electric material. The most generic is where they'd give you some light bulbs, some wires, a resistor, and a switch and ask you to configure it until the light bulb lights up. (mind you, you should be prepared to be able to handle or know how to use the materials they give you. snap circuits, alligator clip wires etc..)PicturePerfect wrote:What do you do in the lab part of Shock Value (Div.B)?
Another thing they might ask you is the use of multimeters: you should try to find out or know how to find the resistance, voltage and current.
I'd start with putting the reference material in the binder, because it is there as a supplement. You should really have everything in your head. Besides flipping pages in the binder could cost you some valuable time. Especially since it's an event like Shock Value. (in many cases you definitely won't have the full time to do the testing portion. sometimes they they split up the test in to stations like Forestry or Rocks and Minerals.) But if you want to put the little details, sure that's fine. There might be occasions where you need to look up some details. Like what's the Ohm's first name, or who built the incandescent light bulb. The important part is that you can find stuff easily in your binder.thatrobotguy wrote:This is my first time doing a study event, so I'm not really sure what needs to go into the binder. Are you guys putting every little thing that you find about circuit lab (every article, all notes, etc) into the binder, or do you use it more as reference material, and just put the stuff you would have to look up (resistor color bands, etc)?
Edited:
WindPower, Bridges.
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Re: Shock Value B/Circuit Lab C
Hey guys, I'm stressing a bit and need help with a question:
How would I solve the following problem: (From the test exchange)

The voltage across A is 3 volts. The resistance of B is 1 ohm. The resistance of C is 3
ohms. The resistance of D is 4 ohms.
7. If the negative terminal of the battery is ground, what is the relative voltage at
point E? (In other words, what is the voltage between the negative terminal of the battery and point E?)
How would I solve the following problem: (From the test exchange)

The voltage across A is 3 volts. The resistance of B is 1 ohm. The resistance of C is 3
ohms. The resistance of D is 4 ohms.
7. If the negative terminal of the battery is ground, what is the relative voltage at
point E? (In other words, what is the voltage between the negative terminal of the battery and point E?)
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Re: Shock Value B/Circuit Lab C
A=3Vgeminicross wrote:Hey guys, I'm stressing a bit and need help with a question:
How would I solve the following problem: (From the test exchange)
The voltage across A is 3 volts. The resistance of B is 1 ohm. The resistance of C is 3
ohms. The resistance of D is 4 ohms.
7. If the negative terminal of the battery is ground, what is the relative voltage at
point E? (In other words, what is the voltage between the negative terminal of the battery and point E?)
Resistance of B = 1 Ohms
C= 3 ohms
D= 4 ohms
So first, find equivalent resistance of the circuit. 1/R1 + 1/R2 = 1/RT for the two in parallel.
So 1/3 +1/4 =7/12 which to find RT is the reciprocal, so 12/7.
Now, the now 12/7 ohm resistor is now in series with B, so 12/7 + 1 = 19/7
I=V/R
I=3/(19/7) = ~1.105 amps
Now, to find the voltage through point E, you need to find the voltage through resistor B. V=IR
Since B is 1 ohms, then the the voltage is 1.105V.
Which is the voltage at point E.
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Re: Shock Value B/Circuit Lab C
ThanksBozongle wrote:A=3Vgeminicross wrote:Hey guys, I'm stressing a bit and need help with a question:
How would I solve the following problem: (From the test exchange)
The voltage across A is 3 volts. The resistance of B is 1 ohm. The resistance of C is 3
ohms. The resistance of D is 4 ohms.
7. If the negative terminal of the battery is ground, what is the relative voltage at
point E? (In other words, what is the voltage between the negative terminal of the battery and point E?)
Resistance of B = 1 Ohms
C= 3 ohms
D= 4 ohms
So first, find equivalent resistance of the circuit. 1/R1 + 1/R2 = 1/RT for the two in parallel.
So 1/3 +1/4 =7/12 which to find RT is the reciprocal, so 12/7.
Now, the now 12/7 ohm resistor is now in series with B, so 12/7 + 1 = 19/7
I=V/R
I=3/(19/7) = ~1.105 amps
Now, to find the voltage through point E, you need to find the voltage through resistor B. V=IR
Since B is 1 ohms, then the the voltage is 1.105V.
Which is the voltage at point E.

That's honestly the best explanation I've heard from anywhere/one
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Re: Shock Value B/Circuit Lab C
No problemgeminicross wrote:ThanksBozongle wrote:A=3Vgeminicross wrote:Hey guys, I'm stressing a bit and need help with a question:
How would I solve the following problem: (From the test exchange)
The voltage across A is 3 volts. The resistance of B is 1 ohm. The resistance of C is 3
ohms. The resistance of D is 4 ohms.
7. If the negative terminal of the battery is ground, what is the relative voltage at
point E? (In other words, what is the voltage between the negative terminal of the battery and point E?)
Resistance of B = 1 Ohms
C= 3 ohms
D= 4 ohms
So first, find equivalent resistance of the circuit. 1/R1 + 1/R2 = 1/RT for the two in parallel.
So 1/3 +1/4 =7/12 which to find RT is the reciprocal, so 12/7.
Now, the now 12/7 ohm resistor is now in series with B, so 12/7 + 1 = 19/7
I=V/R
I=3/(19/7) = ~1.105 amps
Now, to find the voltage through point E, you need to find the voltage through resistor B. V=IR
Since B is 1 ohms, then the the voltage is 1.105V.
Which is the voltage at point E.![]()
That's honestly the best explanation I've heard from anywhere/one

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Re: Shock Value B/Circuit Lab C
People who knows lot about multimeters: Do you know what might be happening that causes my values from readingsto constantly change? It seems I can never get an accurate reading, it jumps from 4 to 5 to 2, etc
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Re: Shock Value B/Circuit Lab C
What is it that you are measuring?Bozongle wrote:People who knows lot about multimeters: Do you know what might be happening that causes my values from readingsto constantly change? It seems I can never get an accurate reading, it jumps from 4 to 5 to 2, etc
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Re: Shock Value B/Circuit Lab C
Current in ampsSchrodingerscat wrote:What is it that you are measuring?Bozongle wrote:People who knows lot about multimeters: Do you know what might be happening that causes my values from readingsto constantly change? It seems I can never get an accurate reading, it jumps from 4 to 5 to 2, etc
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