Solving for the Voltages and Currents in a (simple?) Ciruit

1. Sep 16, 2008

runawayshoes

1. The problem statement, all variables and given/known data

For the three (simple?) circuits, I'm asked to solve for the voltages at A and B, and also the current I. For figure 1, I assume that is just a simple voltage divider, with the voltage at A to be 5*(1/21) correct?

I'm a little more confused with the circuit in figure 2. What I attempted to use was superposition. So I assumed the right source is turned off, which means that there is no current flowing through the 5k resistor, and the bottom 10k and 1k are then connected in parallel correct. Then I just use voltage divider again, then I do the same thing for the other side and then just sum the voltages. The answer I got was 3.15V. If someone could verify this that would be SO great. Now as far as solving for the current I, I'm completely stumped. I'm used to solving for the current through a component so when it is asking for the current through just the wire I'm confused.

As for figure 2, he states that B is floating until a probe is connected. I'm not sure how that affects the problem. So I just assumed that no current flows through the right 1k and just used voltage divider again. I would really appreciate it if someone can help me with these problems.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 16, 2008

Defennder

One missing step. That gives you only the voltage drop across the (top) 10K resistor. You need to subtract this from 5V to find the voltage at A.

You're complicating it unnecessarily. Don't think you have it right, though. One simple way to approach this would be to find the equivalent resistances between 5V and A, and A and ground. Then apply voltage divider principle. Then the potential at A is what you want.

What is a "probe" here? And anyway, considering that B is an open circuit node, your approach is correct. I don't know what "probe" means though.

3. Sep 19, 2008

bjersey

Ok, the first one is wrong. The voltage at point A for figure one should be around 2.62V because this is what I actually measured in the lab. I dont get it, this should just be a simple voltage rule right? How come when I apply it I don't get the right voltage using 5*(1/21) ?

4. Sep 19, 2008

Defennder

I already told you what's wrong with your approach for 1. You omitted a final step.

5. Sep 19, 2008

Sakha

For the problem 1, you are doing 5*(1/21), instead of 5*(11/21), which will give 2.619.
Remember the formula is $$Vout=Vin(\frac{R2}{R1+R2})$$.

Last edited: Sep 19, 2008
6. Sep 20, 2008

CEL

For figure 2, both the upper resistors are connected to +5V, so you can connect their upper terminals together. The same with the two lower resistors, that are connected to ground.
So, you have two sets of two parallel resistors and can solve the problem with a voltage divider.