Circuits - Physics?

Question

Hi, I'd like to learn how to do this problem, I solved it once, but I came up with the wrong answer.
Here is a diagram of the circuit; http://i11.tinypic.com/2hsao0n.jpg

We are to find the potential difference (delta V) between point A and B, and the current through the 20 ohm resistor on the right.

I'm having trouble understanding the Kirchoff Law stuff and what not too. :S
Best answer goes to whoever can help me with both ^^;

Top resistor had the resistance cut off of the image because it was too high when it got cropped...its 10 ohms.
Sorry!

Answer 1

EDITED after given the value of R:

Kirchoff's law is used to solve networks by summing currents at nodes. Your circuit has only two nodes of significance, A and B. Assume a potential difference ∆V = VA - VB and compute the currents in each leg from that. Then form equations using Kirchoffs law.

The currents flowing out of B are ∆V/(5+20) + ∆V/5 + ∆V/10 +(∆V-25)/10. Kirchoff's law states that these must sum to 0, giving the equation

∆V/(5+20) + ∆V/5 + ∆V/10 +(∆V-25)/10 = 0

Solve for ∆V

∆V*(1/25 + 1/5 + 1/10 + 1/10) - 25/10 = 0


∆V = (25/10)/(1/25 + 1/5 + 1/10 + 1/10)

The current in the leg that contains the 20-ohm resistor is ∆V/25. That current flows through both the 20-ohm and the 5-ohm resistors.

In general there are two ways to solve networks: voltage nodes and current loops. The above example shows the node method. The loop method would propose a current loop in each section of the network. You would get n equations from n loops with n current unknowns. You choose the method which gives the fewest unknowns and equations. Solving your problem with current loops would require 3 equations. The nodal method does it in one. Other circuit configurations would have a different result, and sometimes the loop method is simpler.

Answer 2

Unfortunately your drawing is incomplete.
I see three resistors between A and B and only two values.
The first step at any rate is to combine the 10 ohm and 5 ohm resistors between A and B into one resistor. You do this by summing their inverses, then taking the inverse of the result, since they are in parallel.
1/10 + 1/5 = .3
1/.3 = 3.33 ohms.
You can now redraw the picture, creating one resistor instead of two.
I will consider that the resistor without a value has a value R from here on out, you can substitute its value later.

I have uploaded a document to help:
http://www.mypchelp.com/circuit.doc

Figure 1 shows the above, combining the two resistors.
You know have a circuit with two meshes.

You can combine the 5 and 20 ohm resistors into one resistor, since they are in series you just add them together giving you one 25 ohm resistor. As seen in Figure 2.

Now if you notice, in Figure 2, you _could_ sum the two resistors in parallel. But you don't want to, because that would eliminate points A and B and you'd no longer find the appropriate resistance.

Kirchoff's voltage law KVL states that the sum of the voltages in a mesh must equal zero. So, you make two meshes and each one must be equal to zero, this gives you two equations.

So now you define your two meshes as seen in Figure 3.
Starting from the negative terminal of the power supply, and in a clockwise manner, we will form an equation for mesh 1.
We will call the current in Mesh 1, I1 and in Mesh 2, I2
You already know that multiplying current by resistance gives you voltage. We assume the positive side of the resistor is the side that accepts current, therefore thats the side that is closest to the positive side of the voltage supply.
First Mesh equation:
0=-25V+3.33ohm*(I1-I2)+Rohm*I1
Note, I used I1-I2 because we're dealing with Mesh1, and because the current through the 3.3 ohm resistor is the current I1 minus the current I2 with respect to Mesh1.
Now mesh 2:
0=3.33ohm*(I2-I1)+25ohm*I2
Note: I d I2-I1 because we are now in Mesh 2 and the current is in respect to Mesh 2.

You must now solve these two equations using simultaneous equations. You also know that the voltages on the 3.33 ohm resistor and the 25ohm resistor are equal, because they are connected in parallel. So once you know the currents, you could either multiply I2*25 ohms to get the voltage between A and B, or you can multiply (I1-I2)*3.33 to get the voltage as well. Note you need to make sure that the sign of the voltage you calculate makes sense with what the question is asking.

Assuming R is 10 ohms:
0=-25V+3.33ohm*(I1-I2)+10ohm*I1
0=3.33ohm*(I2-I1)+25ohm*I2
-25*I2-3.33*I2=-3.33*I1
(-25/3.33)*I2-I2=-I1
-7.5I2-I2=-I1
8.5*I2 = I1
Substituting in Mesh 1's equation:
25V=3.33*(8.5*I2-I2)+10*(8.5*I2)
25V=3.33*(7.5*I2)+85*I2
25V=110*I2
I2=25/110 = .227 Amps

I2*25ohms=Voltage between A and B
=5.68 Volts difference between A and B.

Answer 3

"Circuit/s" is used too many situations in UR question. ie =>what's "no longer for circuits" mean on the tip? it incredibly is complicated given something of the phrases in this question Impedance refers to AC circuit/s and is composed of aspects of: resistance capacitance inductance