Shock Value B/Circuit Lab C Question Marathon

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Shock Value B/Circuit Lab C Question Marathon

Postby Jim_R » August 28th, 2013, 8:13 pm

Question Marathon for Shock Value B/Circuit Lab C.
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby ScienceOlympian » August 28th, 2013, 8:15 pm

Okay. Let's start out with a SUPER easy basic question.
Let's say that a circuit has one resistor (lightbulb) and one battery.
The battery produces 10V, and the resistor has 2 Ohms.
How many amperes are running through the circuit?
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby sercle » August 29th, 2013, 9:31 pm

ScienceOlympian wrote:Okay. Let's start out with a SUPER easy basic question.
Let's say that a circuit has one resistor (lightbulb) and one battery.
The battery produces 10V, and the resistor has 2 Ohms.
How many amperes are running through the circuit?


Classic Ohm's Law question!
answer
5A


Edit: Oops! It appears I forgot to create a question; thanks to the below poster for continuing.
Last edited by sercle on September 1st, 2013, 8:58 pm, edited 1 time in total.
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby TwelveSquared » September 1st, 2013, 5:01 pm

Technically, I believe the previous poster was supposed to create another question, but since they didn't, I will do so, with a somewhat more complex question.

The circuit is as follows:
Bt1 has an unknown voltage. Bt1 is wired to R1, which has a resistance of 5Ω. R1 is wired in parallel with R2(4Ω) and R3(4Ω), which are in series with each other.

if the total current through the circuit is 2 ohms, what is the voltage of Bt1?
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby iwonder » September 1st, 2013, 6:47 pm

TwelveSquared wrote:if the total current through the circuit is 2 ohms


Wait hold up what?

If you meant 2 amps...
spoiler alert
R2 and R3 can be represented as an 8 ohm resistor. Put 8 ohms in parallel with a 5 ohm resistor, and it can be represented as a 3.077 ohm resistor (nothing is reasonably this precise...). 2 amps flowing through a 3.077ohm resistor is 6.154 volts. If it were on a test I'd put down 6 volts for the battery voltage.


Try this one on for size...

What's the Norton equivalent current for the following circuit?
A 5 volt battery is attached to a 1kohm resistor, which is then attached to a 470ohm resistor, and finally through a 330ohm resistor back to the other terminal of the battery. The point for simplification of the circuit is across the 1 kohm resistor.
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby TwelveSquared » September 2nd, 2013, 7:57 am

iwonder wrote:
TwelveSquared wrote:if the total current through the circuit is 2 ohms


Wait hold up what?

If you meant 2 amps...

Yes, I did. I thought I later edited that, it must not have gone through on account of my terrible internet connection.
Anyways, I'll take a stab at your problem.
Spoilerz contained here
Only just entering Div C, I'm not very familiar with Norton's Theorem, but here's what i think you're looking for:
Total current: 15V \ (1 + .470 + .330) = 8.33...
Current through load: 1 / (.470 + .330) * 8.33 =~ 10.96
Equivalent Resistance: .08/1 = .08
So, the Equivalent circuit is a 10.96 mA current source in parallel with a .8kΩ(or 800Ω) resistance.

EDIT: Made the same mistake i just corrected above. Your question is...
Transform a delta-circuit where Ra = 2Ω, Rb = 3Ω, and Rc = 4Ω into a Y-circuit containing resistors R1, R2, R3.
(round answers to 4 significant figures)
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby iwonder » September 2nd, 2013, 8:24 am

TwelveSquared wrote:Yes, I did. I thought I later edited that, it must not have gone through on account of my terrible internet connection.
Anyways, I'll take a stab at your problem.
Spoilerz contained here
Only just entering Div C, I'm not very familiar with Norton's Theorem, but here's what i think you're looking for:
Total current: 15V \ (1 + .470 + .330) = 8.33...
Current through load: 1 / (.470 + .330) * 8.33 =~ 10.96
Equivalent Resistance: .08/1 = .08
So, the Equivalent circuit is a 10.96 mA current source in parallel with a .8kΩ(or 800Ω) resistance.



Anwser to Norton Question
Pretty close actually! You got the correct resistance, just not the right current :P

When you short out the two 'terminals' in the circuit you're left with a 5v battery and an 800ohm resistor, so the current is 6.25mA for the equivalent. Since it's all in series the total current is your current through the load (that may not have been clear), and it's a 5v battery not a 15v one :P
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby TwelveSquared » September 5th, 2013, 5:44 pm

TwelveSquared wrote:
iwonder wrote:
TwelveSquared wrote:if the total current through the circuit is 2 ohms

Transform a delta-circuit where Ra = 2Ω, Rb = 3Ω, and Rc = 4Ω into a Y-circuit containing resistors R1, R2, R3.
(round answers to 4 significant figures)

... I'm fairly certain we're not the only ones in this event... That's the question, if the edit was somehow missed.
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby iwonder » September 5th, 2013, 8:16 pm

I was trying to give someone else a chance, but I guess no one took it :D

Here you go
The wye circuit (Y is the abbreviation, like delta and the triangle I can't type) R1 = 1.333 ohms, R2 = .888 ohms, R3 = .666 ohms, keep in mind that's all mental math, so it might not be quiet correct
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby TwelveSquared » September 9th, 2013, 3:22 pm

iwonder wrote:I was trying to give someone else a chance, but I guess no one took it :D

Here you go
The wye circuit (Y is the abbreviation, like delta and the triangle I can't type) R1 = 1.333 ohms, R2 = .888 ohms, R3 = .666 ohms, keep in mind that's all mental math, so it might not be quiet correct

Image
Correct.
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby iwonder » September 9th, 2013, 3:51 pm

Nice pic.

Soo... Let's see if we can get someone else into this.

3 3 ohm resistors are in parallel and connected to a 5v battery. A fuse is in the battery that is rated for 4A. Will the fuse blow?
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby FawnOnyx » September 9th, 2013, 5:25 pm

Alright fine I'll answer this one xD
Three 3 ohm resistors in parallel is 1/3 + 1/3 + 1/3 = 1/Req, so the equivalent resistance is 1 ohm. Ohm's law gives I = V/R = 5v/1ohm = 5 amps. Therefore the fuse will blow.
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby iwonder » September 9th, 2013, 5:56 pm

Awesome :P your turn!
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby FawnOnyx » September 9th, 2013, 11:08 pm

Wanted to do some logic gates since they sound interesting:

Image

1. What will the output be if A is true and B,C, D are false?
2. What will the output be if A and B are false and C and D are true?
3. What will the output be if A,B,C, and D are all false?

Bonus: write a boolean algebra expression for the output in terms of A, B, C, and D
Not worth your time: make a truth table for this gate system

Disclaimer: I just looked up logic gates for the first time so I'm not sure if these questions are 100% legit.
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Re: Shock Value B/Circuit Lab C Question Marathon

Postby TwelveSquared » September 11th, 2013, 7:23 pm

Ok, I waited since I virtually alternated answering every other question so far, but I'll go ahead.
I'm a programmer, so I can hopefully handle logic...
(although, i will note that i think the rules only cover DC, not digital logic. But it's still good practice anyway.)
These comparisons are what run your computer...
Note: Since this is electronics, "true" is usually represented digitally as a 1 (or a high voltage, usually 5V), and false is 0(low voltage, usually 0V), which is what I'll use for the rest of this.
Problem 1:
Since A is high, the OR gate outputs high(1). Since both inputs in the NAND are low(0), it also outputs 1. These two bits are put into a NOR gate, resulting in 0.
Problem 2:
A and B are false, OR gate outputs 0. same output for the NAND. Since both inputs are 0, the NOR outputs 1, the total output.
Problem 3:
OR gate has same situation, outputs 0. the NAND outputs 1, when put through the NOR results in 0.
Boolean expression:
(I had to look this up...)
_____________
``````````__
(A + B) + (CD)
EDIT: Ignore all the ````. The site removed the whitespace when i tried to do it with spaces. The two lines represent NOT(inversion).
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