how we can solve it ..by use parallel and series..(indctor)...??

 

how we can solve it ..by use parallel and series..(indctor)...??

Inductance is treated the same as resistance.

Simplify the circuit in steps...
- Add series inductors together
- calculate parallel inductance in each sub-circuit
- repeat until all branches (sub-circuits) are consolidated into one inductance value.

 

It is always a good idea to estimate the answer before you get started so that you have a good sanity check on whatever final answer you get.

In this case, you can get a quick min/max on the final answer by noting that you always have to go through the 6mH and the 10mH inductors even if the mess to the right turns out to be virtually nothing. So you have a minimum of 16mH. Then, even if the mess to the right of the 8mH inductor were completely removed you would have 24mH and the when the mess is put back in it can only make it go down from there. So you have a max of 24 mH. So you know that the answer is (20±4)mH. If you get anything outside of that, you know your answer is wrong. Plus, in many situations, this estimate is good enough and you can just call it 20mH and move on. If you are taking an exam and are pressed for time, putting this as a temporary answer, along with an indication that is IS just an estimate, might be enough to get most of the points even if you run out of time to come back to it.

 

It is always a good idea to estimate the answer before you get started so that you have a good sanity check on whatever final answer you get.

In this case, you can get a quick min/max on the final answer by noting that you always have to go through the 6mH and the 10mH inductors even if the mess to the right turns out to be virtually nothing. So you have a minimum of 16mH. Then, even if the mess to the right of the 8mH inductor were completely removed you would have 24mH and the when the mess is put back in it can only make it go down from there. So you have a max of 24 mH. So you know that the answer is (20±4)mH. If you get anything outside of that, you know your answer is wrong. Plus, in many situations, this estimate is good enough and you can just call it 20mH and move on. If you are taking an exam and are pressed for time, putting this as a temporary answer, along with an indication that is IS just an estimate, might be enough to get most of the points even if you run out of time to come back to it.

here it mean will be like this..the first step parallel..?

 

here it mean will be like this..the first step parallel..?

No, no.

I said take care of any series first.
For example, the upper right triangle. Add the 8 and 12 to make a "20". Then do the parallel of the "20" with the 5.

On the bottom triangle, add the 8 and 4 to make a 12, then do the parallel calculation with the 6

At that point you will have your two answers and they will end up making a new pair in series with each other.

Keep re-drawing the circuit as you go to find the new situation. Keep grouping and solving until you have one final value.

 

here it mean will be like this..the first step parallel..?

For two elements to be in parallel, they must have the same voltage across them at all times. The test for this is to put your fingers on the two ends of one element and then see if you can move them to opposite ends of the other element without lifting either finger and while staying on the same two nodes (i.e., without crossing any other elements).

Do the devices you are trying to claim are in parallel meet that test?