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Classic Ohm's Law question!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?
5A
Wait hold up what?TwelveSquared wrote:if the total current through the circuit is 2 ohms
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.
Yes, I did. I thought I later edited that, it must not have gone through on account of my terrible internet connection.iwonder wrote:Wait hold up what?TwelveSquared wrote:if the total current through the circuit is 2 ohms
If you meant 2 amps...
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.
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.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.
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
... I'm fairly certain we're not the only ones in this event... That's the question, if the edit was somehow missed.TwelveSquared wrote:iwonder wrote:Transform a delta-circuit where Ra = 2Ω, Rb = 3Ω, and Rc = 4Ω into a Y-circuit containing resistors R1, R2, R3.TwelveSquared wrote:if the total current through the circuit is 2 ohms
(round answers to 4 significant figures)
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
iwonder wrote:I was trying to give someone else a chance, but I guess no one took it
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
