Shock Value B/Circuit Lab C Question Marathon

[Jim_R]

Question Marathon for Shock Value B/Circuit Lab C.

[ScienceOlympian]

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?

[sercle]

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 (Click to Show Hidden Text)

5A

Edit: Oops! It appears I forgot to create a question; thanks to the below poster for continuing.

[TwelveSquared]

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?

[iwonder]

if the total current through the circuit is 2 ohms

Wait hold up what?

If you meant 2 amps...

spoiler alert (Click to Show Hidden Text)

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.

[TwelveSquared]

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 (Click to Show Hidden Text)

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)

[iwonder]

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 (Click to Show Hidden Text)

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 (Click to Show Hidden Text)

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

[TwelveSquared]

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.

[iwonder]

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

Here you go (Click to Show Hidden Text)

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

[TwelveSquared]

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

Here you go (Click to Show Hidden Text)

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

Correct.