Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

I am pretty lost working this textbook example (not homework).

Will someone verify if I did this correctly?

1) simplify the circuit

1a) \$20\text{k}\Omega + 40\text{k}\Omega = 60\text{k}\Omega\$

1b) \$60\text{k}\Omega \parallel 30\text{k}\Omega = (60\text{k}\Omega \times 30\text{k}\Omega) / 90\text{k}\Omega = 20\text{k}\Omega\$

1c) \$20\text{k}\Omega + 20\text{k}\Omega = 40\text{k}\Omega\$

Resulting in a collapsed circuit with just the 12V source and a \$40\text{k}\Omega\$ resistor.

Is this even the right start? It appears that with this simplified circuit \$V_o\$ would just be 12V. What did I do wrong? The answer is 2V.

share|improve this question
You were moving in the right direction but deviated away.Compute the potential difference across the equivalent resistance (20 k ohm) using the voltage divider law, that potential difference will be equal to the potential difference across the 30k ohm resistor.After getting that information, use the voltage divider law again to get the value of Vo. – KillaKem Jun 12 '14 at 1:09
I referenced vdiv in my text and I see it now. Thanks! – asdf Jun 12 '14 at 1:39
up vote 4 down vote accepted

Try re-drawing the circuit like so:


simulate this circuit – Schematic created using CircuitLab

(It is much easier to discuss a circuit when you use reference designators (R1, R2...) on the components.)

You correctly determined that R3 and R4 are in series, and are equivalent to a 60K resistor. You also found that R3 and R4 in series, are in parallel with R2, to make an equivalent 20K resistor. You then added that combination to R1, and got lost.

R1 in series with the combination of the other three resistors make a voltage divider, giving a voltage (to be determined by the student) at V1.

R3 and R4 make another voltage divider, multiplying the voltage at V1 by 1/3, giving the desired voltage at Vo.

You can easily solve this problem by looking at the resistor values - there's no need to calculate currents (although doing such calculations may be useful practice...)

share|improve this answer
I understand now, thank you! – asdf Jun 12 '14 at 1:43

This is definitely a good start.

The "mistake" is that the terminal Vo didn't collapse to the top of the equivalent resistance, it is buried inside that 20 kOhm resistance. The final part of the solution is "unraveling" the lumped resistor now that you can identify all of the parameters of that one resistance (voltage, current, resistance).

Start picking apart the resistor network by finding the currents through each equivalent resistor. The thing to remember is that an equivalent resistor represents the entire network, and if the variable you need to find is inside that lumped resistor, you're not done solving the problem.

The 0.6 mA current flowing through the whole resistor network is correct. Knowing this current, you can look at the first 20k resistor and identify the voltage across that resistor. With that new voltage, you can solve for more currents, and then more voltages.

share|improve this answer
Thank you. I am going to try this piece by piece approach as well as the voltage divider approach. I'll post back in a few minutes. – asdf Jun 12 '14 at 1:34
I'm not sure I see how to find Vo using 0.6mA, but you helped me understand that since Vo is buried in the lumped resistor, my original approach was incorrect. I ended up using voltage division and got the right answer. So thanks! – asdf Jun 12 '14 at 1:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.