Hello sir, in example 2 when we found i1 it is in fraction if we change it to decimal it is 0.66A so in i0 when we replace i1 to find it it will be -1.32A hence, isn't Rth Vo/io = 1/-1.33 = -0.76 ohms. so only because i changed the fraction to a decimal it changed the answer totally?? if so then when should i just keep it as a fraction and when should i change it to decimal?
Usually during the solution process, use fraction, and then you convert the final answer to decimal, but if you want to be changing to decimal every now and then, leave your decimal to 4dp.
Hi! May I know why are there two value of Ix? What is the difference between the Ix obtained in solving the resistance vs the Ix obtained in solving the voltage?
The idea is. To find the rth we need to solve for Ix, that's the first Ix. For the second Ix helps you to find Vth. The two are independent. So in another question it can be a different ball game altogether
Current flows from the 6v battery. You can make 6v the reference source to do the current distribution, it is correct, but my reference was the 1.5ix current source that is why I have the 0.5ix moving towards 6v. You can equally do a current distribution using the 6v as reference and get the same answer
What was the purpose of the polarity over the 5 ohm for vx .Shoudnt the voltage across ir be -5Ix.Since in the direction assumed there will be a negative drop?
This is contradicting to content I read from Basic engineering circuit analysis by David Irwin 11th edition page 192, I suggest u check that page and see where u go wrong if at all u are wrong
@@SkanCityAcademy_SirJohn The part where you apply an independent current source, the book says you only do this if the entire circuit is made up of only dependent circuits, yet here yo dealing with one that has both independent and dependent
@@SkanCityAcademy_SirJohnI rechecked and both methods work, the first time I replied I had not yet tried both, but now am sure they both work, thanks for the great content
How is it possible that the voltage drop across 4 ohm resistor is 5.33V? That means 5 ohm and 3 ohm resistor only dropped 0.67V from the 6V battery. Can you please explain that?
The values for Ix are different because you are doing two different things: 1. You want to find RTh hence you need to excite the circuit because mind you, the independent sources are removed and if independent sources are removed, dependent sources have no purpose because they are controlled by these independent sources. So ones you excite the circuit with a 1v or 1A source you get a value for Ix. Also when finding Ix or any current value, do not consider the loop with a dependent current source. 2. After finding RTh, you need to find Vth, and this time you leave the terminal open, hence the value for Ix will surely be different. For the last part. You can use the other loop from vth to the far left. Ignore the loop with the current source.
Okay, Do current distribution. That is Current in 4ohms is (ix + 1), in 3 ohms is ix and in 5ohms is 0.5ix. Apply kvl to the loop containing resistors, 3, 4, 5. Ix = -8/9 Vo = 4(-8/9 + 1)v Rth = vo/io = 4/9/1 = 0.444 ohms
