# Kirchoff's laws: How do I find R1?

Discussion in 'General Electronics Chat' started by EnJoneer, Oct 11, 2016.

1. ### EnJoneer Thread Starter New Member

Oct 11, 2016
1
0
I know the answer is somewhere around 3.2 x 10^-6 Amps because I have used Thevinin's also, but despite using various combinations of equations and methods, I can only seem to get 3.2 x 10^-3... and no this isn't a mA value.

I have also redrawn the circuit diagram to make calculation a little easier and still can't get there, I must be making a silly little mistake?

Any help will be much appreciated!

2. ### wayneh Expert

Sep 9, 2010
12,388
3,244
You might start with the simplification that since R1<< R2 or R3, ignore it. Solve, then add R1 back in.

Or you could write down the known equations. As you know, you need enough equations to match the number of unknowns.

3. ### Dodgydave AAC Fanatic!

Jun 22, 2012
5,154
772
At a guess, you have 21V and 25k in series, thats 840uAmps, which gives 8.4V drop across the R2 10k resistor, giving 17.4V across R1...

4. ### drc_567 Senior Member

Dec 29, 2008
138
17
• A Thevenin approach here is not simple to visualize. If you use two separate current loops, you will have two independent equations and two unknowns ... downhill from there.

5. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
So you are asking us to find the "silly mistake" in your work but yet you don't show any work in which we can even begin to look for your error.

Also, your title says you are trying to find R1. But R1 is given as 160 kΩ. Then your first sentence states that you know the answer is a current. Which current? You have three branches. Don't make us guess. Engineering is not about guessing. Are you looking for the current in R1? If so, then the answer is 3.61 μA left to right, which is in reasonable agreement with what you know the answer to be.

My guess (and it is only a guess since you didn't show your work) is that your result IS in mA and this isn't obvious to you because you didn't track your units properly and so you just tacked on the units that you wanted the answer to have.

6. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Why is it not simple to visualize?

For the open circuit voltage you have to voltage sources in series totaling 21 V going through two resistors in series totaling 25 Ω, so you have a current that is just a bit under 1 mA (0.84 mA). You can walk though which ever source turns your fancy and you will get an open circuit voltage of 0.6 V.

Or you can get the short circuit current almost by inspection. You have 0.9mA flowing one way and 0.8 mA flowing in the other for a net of 0.1 mA.

The effective resistance is just R2 and R3 in parallel, which is simply 6 kΩ.

From there, finding the current in R1 is trivial.

7. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Guess again.

You might want to consider the direction of that current before concluding that there is 17.4 V across R1 (easy mistake to make, particularly if you don't actually draw the current direction on the diagram and just work with a mental image).

8. ### drc_567 Senior Member

Dec 29, 2008
138
17
... most instructive.