And why do we need to calculate a value for RL to calculate the value of Rs?

Hi,

Well for one thing it is often a good idea to have two ways to calculate something, if we can find two ways that is.

However, what is your logic for calculating Rs?

 

Hi,

Well for one thing it is often a good idea to have two ways to calculate something, if we can find two ways that is.

However, what is your logic for calculating Rs?

Read Post #4 or Post #6 and do... the... math.

 

Read Post #4 or Post #6 and do... the... math.

Hi,

The math is what comes into question.

Ok, but from post #6 once you calculate the voltage you seem to jump to the conclusion, but in reality you have to do a subtraction, then once you do the subtraction, you then have to do a division. It may still be shorter than having to calculate the value of RL, but it seems kind of sparse not knowing what RL is ever. Knowing RL gives us a chance to check the results which might be a good idea.

I think the OP has gone south so we could probably talk more freely now

 

Without having to calculate anything, if the voltage at the load has to be 5 V, then Rs must be zero.
The value of Cout is irrelevant.
I = 200 mA is also irrelevant.

 

Without having to calculate anything, if the voltage at the load has to be 5 V, then Rs must be zero.
The value of Cout is irrelevant.

Hi,

Yes I agree, but really the reason we know that Rs must be zero is because we did a subtraction in our heads and avoided the units. If the voltage of the load must be 5v and the output is already 5v then 5-5=0, but note that the '0' there is in the units of volts, not Ohms. To be more succinct, we have to divide by the current, which then gives us the resistance in Ohms.

I realize that for people who have done this a lot in the past may not have to go through all the steps, but for a learner they may have to have it spelled out.
Also interesting is if we wrote a program for this, we would have to include all the steps.

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