Someone help with simple circuit problem?

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In summary, you have a series circuit with a total resistance of 14 Ohms, and you find that the voltage drop across R2 is 6.43 volts.
  • #1
ViEternal83
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A series circuit with
R1=5 Ohms
R2=9 Ohms
V=10V

What is voltage drop across R2?

I got 11.11 Watts.

If someone can tell me if I'm wrong or not, it would be much appreciated.

Is this some sort of super smart people's forum where this kind of question is laughed at?
 
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  • #2
You shouldn't get watts for a voltage drop - watts is power. So your answer is automatically wrong because its incorrect units, and I don't get where you got that number but its wrong.

Few ways to solve this problem, here's one that should be easy to visualize. You know in a series circuit that the current will be identical across the components, so let's find that current and work with that using ohm's law.

Add the series resistance to get the total resistance. You know the voltage, find the current across this total resistance. Now you know the current you can find each voltage drop and both should add to equal your starting voltage.

If you really are looking for power dissapated by R2, its 4.59 watts. I'm not going to simply tell you the voltage drop, show your work and I'm sure someone will help you to learn how to do this yourself, very easy once you get the idea of using ohm's law and this is VERY valuable knowledge.

voltage = current * resistance

Cliff
 
  • #3
Oh, haha.
I had no idea what I was doing apparently.
But thanks for telling me I was wrong and the information.
I'll try again.
 
  • #4
Req = R1+R2 = 5 + 9 = 14 Ohms
I = V/Req = 10/14 = 5/7

Voltage drop across R2 = V = I*R2 = 5/7 * 9 = 6.43 Volts
Is this correct?

Thanks for the help.
What a nice forum.
 
Last edited:
  • #5
Yup, you got it now. Simple algerbra and you can just about figure it all out from ohm's law until you get to AC and non-linear behavior, but that's another story for a different day! :)

Cliff
 

1. How do I solve a simple circuit problem?

To solve a simple circuit problem, you need to follow the basic steps of circuit analysis. First, identify all the components in the circuit and their values. Then, use Ohm's law and Kirchhoff's laws to calculate the voltage, current, and resistance in each part of the circuit. Finally, use these values to determine the overall behavior of the circuit.

2. What is the difference between series and parallel circuits?

In a series circuit, the components are connected one after the other, creating a single pathway for the current to flow through. In a parallel circuit, the components are connected side by side, creating multiple pathways for the current to flow through. This results in different behaviors, such as the total resistance being greater in a series circuit and the total current being divided among the components in a parallel circuit.

3. How do I calculate the total resistance in a series or parallel circuit?

In a series circuit, the total resistance is equal to the sum of all the individual resistances. In a parallel circuit, the total resistance is calculated using the formula 1/Rt = 1/R1 + 1/R2 + ... + 1/Rn, where Rt is the total resistance and R1, R2, etc. are the individual resistances.

4. What is a short circuit and how does it affect a circuit?

A short circuit occurs when there is a direct connection between the positive and negative terminals of a power source, bypassing the components in the circuit. This results in a very low resistance and a large amount of current flowing through the circuit, potentially damaging the components and causing a fire hazard.

5. How can I troubleshoot a circuit problem?

To troubleshoot a circuit problem, you can use a multimeter to measure the voltage, current, and resistance at different points in the circuit. This can help you identify any faulty components or connections. You can also use a breadboard to test different configurations and isolate the problem. Additionally, double-checking your calculations and making sure all components are properly connected can also help troubleshoot a circuit problem.

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