# How Do You Calculate Current and Voltage in a Complex Circuit Diagram?

• kazekko
In summary, the current in the 37 Ohm resistor is 12.5 volts and the potential difference between points A and B is 3.33 volts.
kazekko

## Homework Statement

Given the diagram:

http://img517.imageshack.us/img517/5995/p2809altbe2.gif

where R = 37 Ohms

a) Find the current in the 37 Ohm resistor

b) Find the potential difference between points A and B

## Homework Equations

DV = IR, Kirchhoff's rules

## The Attempt at a Solution

a) I = 25 V / 2 = 12.5 V (at the first (right) junction) , there for I thought I at the 37 Ohm resistor would be 12.5 V / 37 Ohms = 0.34, but it doesn't.

b) I combined the middle resistor and the lower 5 Ohm resistor to get ((1/10)+(1/5))^-1 = 3.33 Ohms and made the V at "b" 12.5 V ... but I don't know where to go from there.

Thank you so much for your time! -- I'm really stuck.

Last edited by a moderator:
Are you familiar with Thevenin's theorem? There are many ways to compute the current in the 37 Ohm resistor but this circuit lends itself to Thevenin's theorem. So:

step 1) disconnect the R from the circuit.
step 2) calculate the equivalent resistance looking into the circuit. There are a couple of ways to do this Vopencircuit/Ishortcircuit will give it to you. But I like the method of open circuiting all current sources and short circuiting all voltage sources (ideal) and then calculating the total resistance Rth. Now you have everything you need to replace everything to the left of the R with the Thevenin equivalent circuit which will be a voltage source with value Vopencircuit and a series resistance of Rth.
step 3) reattach your load resistor R and calculate the current in the now trivial circuit.

hope this helps you a bit...

That's still beyond me at this point, but I do like the sound of it and should read about it. Thanks for your help!

Meanwhile, I found an answer to this question in another book. What they suggested was turning the picture on its side to make 3 normal-looking horizontal loops, mostly in parallel but with R and 5 Ohm in series. Then the problem is simple an you don't even need Kirchhoff's rules (which I thought you must, and were confusing me).

I "got the answers" by doing it this way but really need someone to help me visualize how this is possible, because I'm not really seeing it!

Hope this helps anyone with a similar problem.

## 1. What is an overlapping loop circuit?

An overlapping loop circuit is a type of electrical circuit in which the output of one loop is connected to the input of another loop, creating a continuous loop of electricity. This allows for multiple devices to be connected and powered by the same circuit.

## 2. How does an overlapping loop circuit differ from a traditional circuit?

An overlapping loop circuit differs from a traditional circuit in that it has multiple loops that are interconnected, whereas a traditional circuit typically has a single loop with one power source and multiple devices connected in parallel.

## 3. What are the advantages of using an overlapping loop circuit?

The main advantage of an overlapping loop circuit is that it allows for a more efficient use of power. Since the loops are connected, the electricity does not have to travel as far, reducing the amount of resistance and energy loss. Additionally, if one device in the circuit fails, the rest of the devices can still receive power from the other loops.

## 4. How is an overlapping loop circuit useful in scientific experiments?

An overlapping loop circuit is commonly used in scientific experiments to power multiple devices simultaneously. This is particularly useful in experiments that require precise timing or synchronization between devices.

## 5. Are there any potential drawbacks to using an overlapping loop circuit?

One potential drawback of an overlapping loop circuit is that it can be more complex and difficult to troubleshoot if there are issues with the circuit. Additionally, if one loop in the circuit is overloaded, it can affect the entire circuit and cause other devices to malfunction.

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