# Electricity, finding voltage and resistance

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1. Sep 17, 2016

1. The problem statement, all variables and given/known data
Design a voltage divider circuit to provide an output voltage v0 = 5V from a 12V source. The current taken from the 12V source is to be 100mA.
a)Find the values of R1 and R2.
b)Now suppose a load resistance of 200 ohms is connected across the output terminals (in parallel with R2). Find the value of v0.

2. Relevant equations

V = IR
V1 = VR1/(R1+R2)

3. The attempt at a solution
I would assume that I would use the above equation V1 = VR1/(R1+R2), but I am stuck on what to do next.

2. Sep 17, 2016

The picture above was taken from online, but the voltages written in the question are the correct voltages to be used. Disregard the voltage given in the picture

3. Sep 17, 2016

### Staff: Mentor

If the current taken from the 12 V source is 100 mA, how much current flows through $R_1$?

4. Sep 17, 2016

Is it 100 mA because it is in series, so current is constant?

5. Sep 17, 2016

### Staff: Mentor

Yes. How many volts does $R_1$ have to drop from 12 V in order for there to be 5 V at the output?

6. Sep 17, 2016

That would be 7V, so R1 = V/I = 7V/.1A = 70 ohms?

7. Sep 17, 2016

### Staff: Mentor

That's right. Handle $R_2$ in the same manner.

8. Sep 17, 2016

So for R2, V = 5V since voltage needs to add up when in series. Therefore, R2 = V/I = 5V/.1A = 50 ohms?

9. Sep 17, 2016

### Staff: Mentor

Sure. Check your results using the Relevant equation that you gave in your first post. Make sure you use the right resistance values in the right places in that equation!

10. Sep 17, 2016

Ok, so I am not entirely sure if this is correct for part B, but I did, v0 = VR3/(R2+R3) = (5V)(200 ohms)/(200 ohms + 50 ohms) = 4V

11. Sep 17, 2016

50 ohms being from R2

12. Sep 17, 2016

### Staff: Mentor

No, you should combine the new resistance with the $R_2$ first, then apply the voltage divider equation to the new circuit.

Adding the additional load to the original voltage divider circuit will change the amount of current being drawn from the 12 V source, so the drop across $R_1$ must change; The 5 V you calculated previously for the output of the divider no longer holds.

13. Sep 17, 2016

So what voltage would I use in the voltage divider equation? Would it be the original 12 V?

14. Sep 17, 2016

### Staff: Mentor

Yes. The source hasn't changed.

Edit: You're new circuit looks like this:

15. Sep 17, 2016

Using that then, I do V = (12V)(40 ohms) / (70 ohms + 40 ohms) = 4.36 V and that doesn't appear right

16. Sep 17, 2016

### Staff: Mentor

And yet it is!

If you are interested in playing around with these sorts of problems and checking your results, you might consider getting a copy of a circuit simulation software where you can "wire up" the circuit and see how it behaves, check voltages and currents and so on.

There are free software packages that you can download. For example, LTSpice is used by many people.