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((1/12)+(1/12))^-1 , Sorry, the only reason I posted that was b/c ive been second guessing myself. But the Req of the two 12 ohm is 6. So now we have a 3 ohm resistor in series with Vs and a 6 ohm Req in parallel with Vs
((1/12)+(1/12))^-1 , Sorry, the only reason I posted that was b/c ive been second guessing myself. But the Req of the two 12 ohm is 6. So now we have a 3 ohm resistor in series with Vs and a 6 ohm Req in parallel with Vs
there in series, its just R1+R2 so 3+6=9, and using ohms laws to calculate the current through this Req is V/R=I so Vs/9=I.
Cant we backtrack now since we know what (I) is, and since current is the same for resistors in series(the 3 and 6) it'd be (I),
then if we calculate the voltage across these two resistors(3and6) and then expand the 6 ohm back to the 12 ohm resistors in parallel.
Since we now have a voltage, and voltage is the same for resistors in parallel we can calculate I1...
Yes that's right. Here is a drawing to show this process. I used current division rather than voltage division, but obviously voltage division works too. You should know both ways.
This kind of topology change is common when trying to solve circuits like this. First we have to combine resistances when possible (what i like to call "folding") and then put them back once we find out what we need to know ("unfolding").
This drawing shows this for the first circuit but the second circuit will need this kind of logic even more. We'll have to fold up the circuit quite a bit and then later unfold a little at a time.
there in series, its just R1+R2 so 3+6=9, and using ohms laws to calculate the current through this Req is V/R=I so Vs/9=I.
Cant we backtrack now since we know what (I) is, and since current is the same for resistors in series(the 3 and 6) it'd be (I),
then if we calculate the voltage across these two resistors(3and6) and then expand the 6 ohm back to the 12 ohm resistors in parallel.
Since we now have a voltage, and voltage is the same for resistors in parallel we can calculate I1...
So now we know that the current through each of Vs, the 3 ohm resistor is and the combination of the two 12 ohm resistors is the same because they are in series with each other. Now, what is the current I1, through just one of the parallel combination of 12 ohm resistors?
After finding the voltage that goes through the 3 ohm and 6 ohm, i'll only show the 6 ohm voltage,which is ((2/3)*Vs) .Since we now have a voltage for the 6 ohm we can expand the 6 ohm back into the two 12 ohm resistors and find the current through one of them, which is ((1/18)*Vs)=I1 I got this answer from ohms law which is I1=V/R and after pluging in is I1=((2/3)*Vs/12)
After finding the voltage that goes through the 3 ohm and 6 ohm, i'll only show the 6 ohm voltage,which is ((2/3)*Vs) .Since we now have a voltage for the 6 ohm we can expand the 6 ohm back into the two 12 ohm resistors and find the current through one of them, which is ((1/18)*Vs)=I1 I got this answer from ohms law which is I1=V/R and after pluging in is I1=((2/3)*Vs/12)
Your answer is correct, but convoluted. Voltage does not go "through" resistors, current does. Voltage is applied across the resistors. You don't have to expand or contract any resistors. You already figured out that the current through Vs and the 3 ohm resistor was Vs/9. Following that, the current divides equally between the 12 ohm resistors. That makes it I1=Vs/18 through each of the 12 ohm resistors. Why work yourself into a lather figuring out the voltage across each resistor? I never asked you for those voltages.
Now that you know I1, you can equate it to the I1 that you previous figured out for the right side. Then using Vout=9*Vs, you can calculate "A". Can you do it?
A=81, looking back for some reason when I tried reducing the left side of the circuit, I marked the current I1 and didn't realize it like an idiot. But I get what your saying,I think I was just excited I got it, so I calculated the others lol.
A=81, looking back for some reason when I tried reducing the left side of the circuit, I marked the current I1 and didn't realize it like an idiot. But I get what your saying,I think I was just excited I got it, so I calculated the others lol.
I sense that you don't really have a feel for what is happening in an electrical circuit. You seem to operate mechanically without a method or strategy for working on a problem. You did not seem to know what a series or parallel circuit was, had trouble with current division, got fixated on voltage when it was not necessary, and were easily distracted by other aspects of the circuit that had nothing to do with the solution.
I think you know you should review your course material from ground zero until you understand the basics well enough so you don't discombobulate yourself while performing the simplest analysis. Otherwise, you will have a rough, tough, bumpy ride doing these problems.
I agree 100%, but as for the other problem. Since were given Vout and the 2 and 1 ohm resistors are dividing the voltage across them. Using the equation .2=(1/1+2)*Vo we can calculate Vo=.6, now that we have Vo we can calculate the voltage across the 2 ohm by using voltage division again and substituting in what we have so far, I'll call the voltage across the 2 ohm V2, V2=(2/3)*.6=.4 now that we have both voltages that were being divided cant we just add them together to get .6 V ? For the total voltage across them and then use ohms law to find the current and then use current division?
Voltage division is used when we have two or more resistors in series and we know the voltage at the top of the UPPER resistor.
For example, if we only had the source Vo and the 8 and 4 ohm resistors just to the right of that, we could use voltage division to calculate the voltage across the 4 ohm resistor. Calling the voltage at the top of the 4 ohm resistor v4 we would then have:
v4=Vo*4/(8+4)
and that would be all we needed.
But since in the full circuit we are given Vout=0.2v and that is ALREADY across the lower resistor, we have to do it differently. To use voltage division we would have to rearrange. Calling the voltage across the lower 2 ohm resistor v2 we have:
Vout=v2*(1/(1+2))=v2*1/3=v2/3
and rearranging we get:
3*Vout=v2
or
v2=3*Vout
Note already that we have not yet calculated Vo, we have just calculated v2. To calculate Vo we would first have to calculate v4 using this same method, then calculate Vo after that.
A simpler method to get v2 however is to just note that the output voltage Vout is given and that the current through the 1 ohm resistor must be:
i1=0.2/1=0.2 amps
and since current in s series circuit is the same everywhere, we know that 0.2 amsp flows through the upper 2 ohm resistor also. That means the upper 2 ohm resistor is dropping 2*0.2=0.4 volts. When we add that 0.4 to the 0.2 output, we get 0.6 volts again. We could then repeat this to get v4, then repeat once more to get Vo.
Note again we are calling the voltage across the lower 2 ohm resistor v2, and the voltage across the lower 4 ohm resistor v4. It's better when the resistors are marked R1, R2, etc.
That second is a lot simpler, thanks. Referring back to the way I used,I didn't know what to call the voltage source so I just called it Vo in the equation when I tried substituting and algebraic manipulation, when I calculated the voltage across the 2ohm,I knew it wasn't the actual source Vo.But since it isn't the main source Vo what would be the technical term for it, just voltage across it Vs. So V2=(R1/R1+R2)Vs
Well, try to remember that when you type out a formula in text you have to watch where the paren's are going, because when we have a denominator we have to enclose it in paren's or else the next guy who reads it may get lost not knowing if it is a denominator or not. So the formula should be typed:
v2=R1/(R1+R2)*Vs
or
v2=Vs*R1/(R1+R2)
and here you see there is no way to understand it incorrectly because R1 and R2 are in the denominator and so they are enclosed in paren's.
The technical term for this voltage if the component was called "R9" for example would simply be:
"The voltage across R9".
Sometimes this is written as "vR9" where the 'v' is lower case, but this convention varies a bit from author to author.
Using the same idea, the current through R9 would be written as "iR9".
If the other end of R9 is connected to ground we might also call it a 'node' voltage.
For example, we might call "Vout" a node voltage too.
The reason why we are using "vR9" is because if we call it by the resistor value itself and there is more than one resistor with the same value then we have to make sure we mention which resistor we are really talking about. That's why it is better to label the resistors R1, R2, R3, etc.
Since theres .6 volts as the total voltage for the 2 and 1 ohm resistors,and since voltage is the same for resistors in parallel there has to be .6 volts across the 2 ohm resistor. The resistor under the 4ohm.Correct.
Since this has to do with the last topic I figured I'd post it here instead of flooding the forum.While going over extra problems on the internet I stumbled across this and got stumped.I just don't know were to begin,the majority of them seem to share nodes to be in parallel but also seem to be in series.Req inbetween the 1 and 18 ohm resistor is needed,forgot to draw the arrow and the Req symbol.But the arrow points inward,inbetween the 1 and 18 ohm resistor
Yes this new circuit has resistors that are connected in a manner that makes it more difficult to combine resistances in an effort to simplify the network.
When this happens one way to handle it is to look at it like a real life problem instead of a theoretical one. What would you do if someone handed you this network on a breadboard and told you to find the total resistance at the two input nodes, given you also have test equipment like a power supply and voltage and current meters? You would connect the power supply to the input nodes, turn up the voltage a little, then measure the voltage across the nodes and the current through the nodes, then use Ohm's Law as R=E/I to get the total resistance across the two input nodes.
We can also do this in theory by placing a 1v source across the two input nodes and then using say nodal analysis. Once we calculate the node voltage of the node just to the left of the 1 ohm resistor, we then know the input current and input voltage so we can use Ohm's Law to get the total resistance.
There are also delta/star transformations we can use but that might sill be a little difficult so the 1 volt test voltage method is probably the simplest way. That uses a method (nodal analysis) that you should be familiar with anyway or should learn if not.
[LATER]
After looking again, it appears that transforming the delta on the far left to star might be enough as then it may be possible to combine the other resistances to make a simpler network. We could try this too if you like.
By delta to star(im guessing you mean Y) I was able to solve it,but it'd be nice for a double check. Solving labeling the delta as Rc=18,Rb=6,Ra=6 and solving for R1=b*c/(a+b+c) 18*6/30=3.6 R2=c*a/(a+b+c) 18*6/30=3.6 and R3= a*b/(a+b+c) 6*6/30= 1.2 , I posted a picture of what I believe the circuit would look like,please let me know if this is correct so far. And i'll continue
Hey that's really quite good, except you left out the other 6 ohm resistor. See that?
After you include that missing resistor you can see how the resistors can be combined.
Also, make sure to get the other input node included too.
The "Wye" configuration is also sometimes called a "Star". This becomes even more apparent in 6 phase systems.
To double check, set a voltage of 1v at the two input nodes and calculate the current through the 1 ohm resistor, then use R=E/I to see if you get the same total resistance.
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