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

In summary, the conversation discusses solving for voltages and current in three circuits, with one circuit using the voltage divider principle and the other two using superposition. There is confusion about finding the voltage at point A in figure one and the effect of a floating node in figure two. The summary provides clarification on these points and summarizes the correct approach for solving the problems.
  • #1
runawayshoes
7
0

Homework Statement



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.

Thanks in advance.

circuitsbl_09082008.jpg

 
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  • #2
runawayshoes said:
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?
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.

runawayshoes said:
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.
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.

runawayshoes said:
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.
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
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 don't 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
I already told you what's wrong with your approach for 1. You omitted a final step.
 
  • #5
For the problem 1, you are doing 5*(1/21), instead of 5*(11/21), which will give 2.619.
Remember the formula is [tex]Vout=Vin(\frac{R2}{R1+R2})[/tex].
 
Last edited:
  • #6
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.
 

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

1. What is a circuit?

A circuit is a closed loop through which an electric current can flow. It consists of various components such as resistors, capacitors, and inductors, connected by conductive wires.

2. How do I solve for the voltages and currents in a circuit?

To solve for the voltages and currents in a circuit, you will need to use Ohm's Law and Kirchhoff's Laws. Ohm's Law states that the current through a conductor is directly proportional to the voltage and inversely proportional to the resistance. Kirchhoff's Laws state that the sum of currents entering and leaving a junction must be equal, and the sum of voltage drops in a closed loop must be equal to the voltage source.

3. What is a simple circuit?

A simple circuit is a circuit that has a single voltage source and one or more resistors connected in series or parallel. It is often used to illustrate basic concepts in circuit analysis.

4. What are the steps for solving a simple circuit?

The steps for solving a simple circuit are as follows:1. Identify the components in the circuit and their values.2. Determine the direction of the current flow.3. Apply Ohm's Law and Kirchhoff's Laws to calculate the voltages and currents at different points in the circuit.4. Use the calculated values to create a circuit diagram with labeled voltages and currents.5. Check your calculations by applying Kirchhoff's Laws again.

5. What are some common mistakes when solving circuits?

Some common mistakes when solving circuits include:- Forgetting to take into account the direction of current flow- Using incorrect equations or values for components- Not considering the effects of parallel and series connections- Mixing up the calculations for voltage and current drops- Not checking the final circuit diagram for accuracy

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