kzfaq.info/get/bejne/j9p5ZZxlyNCvfZc.html

kzfaq.info/get/bejne/j9p5ZZxlyNCvfZc.html

that is how we calculate the total resistance in parallel. 1/R(p.total) =1/R1 + 1/R2. @ENGRTUTOR , may you please confirm if l am right.

Technically the fraction IS the more exact value, but the main reason is to show how the math occurred. Recall that a resistor pair in parallel is R1*R2/(R1+R2), so for example if R1 is 10 and R2 is 3, you get: 30/13. 30/13 is an exact number, which you can approximate as 2.3077 but that is not the exact number, nor is it clear where 2.3077 comes from without showing the fraction first. You also could have done it using the 1/Rt = 1/R1 + 1/R2... formula, but that also gives you the same fraction, it just happens to work for more than 2 resistors at once. Tip - any parallel combination of resistors will always have LESS resistance than even the smallest resistor. In this case, 10 and 3 were combined to give roughly 2.3. If you get a number larger than the smallest resistor, you did it wrong, and your most likely error is you forgot to flip the number at the end, that is 1/x.

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no . because the equivalent resistance is even less than 2 ohms and we all know how the current likes the path of least resistance

He is using nodal Pointing v1,V2 in junction

There is no 10A, it is a voltage source 10V. Current sources are opened to remove their influcence (ie. set current to 0A) and Voltage sources act as a short when their influcence is removed (ie. voltage between two pts is 0).

Yeah, at 6:30 I think it's supposed to be ((1/2)/((1/2)+(1/8.33)))*1 since it's current division (which is I1 = (R2/(R1+R2))*Isource)

Yes that's correct but why did he use 1/R instead of just R --> 1/2 instead of 2 and 1/8.33 instead of 8.33 ??

+Titan Xelon Idk why he's doing that but if you just follow the formula you'll get the answer he got. Just remember that according to the formula the numerator is the OPPOSITE resistor from the resistor we are interested in. I1 = [r2/(r1+r2)] * Is. There's no reason to do reciprocals like him. The formula works

Good question, that's because they're in parallel and he didn't wanna get the voltage then divide by the resistor. Piece of cake

The RECIPROCAL always works. The simple formula you list is a special case where there are only 2 resistors.

looks like the problem is very complex

@@Shafixy nice

There are plenty of ways to solve this problem. I am showing ONE.

Some students for assignments/homework are forced to demonstrate the use of this method. Some problems are harder to solve without it.

You can use KCL and KVL and set up a system of equations, and solve through algebra. The superposition method allows you to solve more directly, so you don't have to do nearly as much algebra. It's a simple matter of drawing each version of the circuit with the other source nullified, combining resistors, and using voltage dividers and current dividers to isolate the either voltage or current that applies to the resistor of interest.

Help!

yes

Yes it is current division. That is why he does (1/8.33)/(1/2 + 1/8.33) * 1A. Notice that he does not do R1/(R1+R2). Instead he does (1/R1)/(1/R1 + 1/R2). Look at SLIDE 3 of this PDF note (tuttle.merc.iastate.edu/ee201/topics/analysis_techniques/dividers.pdf)

Yes, you can do it that way as well

Why did he put 1/2 in the resistors when he can just use the formula I1 = 2 / (8.33+2) * 1A ?????

NOP

Actually... He used a strange formula for it, but it's the correct answer.

Yeah, I was thrown off by that as well.

yeah,....that's method is not eassy

2:50 “Lets use voltage divider.” Then places a huge ass equation without explaining any of it.

kzfaq.info/get/bejne/j9p5ZZxlyNCvfZc.html

how did he get 15/50 ohm 5:24

@@jianyanchoong5762 He added the two resistor using parallel addition.

kzfaq.info/get/bejne/j9p5ZZxlyNCvfZc.html

I had the same doubt, I just plugged in some numbers in the calc and realized that (1/r1)/(1/r1+1/r2) = (r2/r1+r2)

kzfaq.info/get/bejne/j9p5ZZxlyNCvfZc.html

It is not a Voltmeter or Ammeter -- the problem shows a VOLTAGE SOURCE and a CURRENT SOURCE. Setting Current source to 0 means no current (aka broken wire)... Setting Voltage source to 0 means your are measuring voltage across the same node (aka same wire).

this means you're just dumb to begin with. if foreigners can understand heavily accented english; in this case non-indians understanding indian accented english, then why can't you?