It has been suggested that you redraw the circuit, but you seem to be having trouble with that. I've attached a schematic showing how you should redraw the circuit to get rid of that "ring" in the center. The ring is just a wire and everything connected to it is just connected together.

After you redraw the schematic, you should probably just solve the whole thing. Then you'll have all the currents and you can calculate the voltages across various resistors with just Ohm's law.

I'm going to do more of the work for you than would be usual, because you seem to be having a lot of trouble with this problem.

To do this is complicated, and you will have to solve a system of 5 equations in 5 unknowns. Have you studied methods for solving systems of simultaneous linear equations? To solve 5 such equations by hand can be a daunting task, and I would recommend not trying to do it by hand, but use a calculator or software program that can do this.

You said you have studied KVL, so using the loop, or mesh, method would be the appropriate thing for you to do. See:

http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

In post #6 Georacer showed how to designate 5 loops (or meshes). To proceed with the solution you need to designate the loop currents with some symbolic value; I labeled the currents I1, I2, I3, I4 and I5. You can see those labels on the attached schematic, though I haven't actually drawn loops like Georacer did. The currents in all my loops are assumed to travel clockwise.

Next you need to derive the loop equations. The way you do this is to add up all the voltages around a particular loop. Take the first loop, for example, where the current is I1. The voltage across R2 is R2*I1; the voltage across R3 is R3*I1. R5 is slightly different because it has two currents, I1 and I2, in it. Therefore the voltage across R5 is (I1-I2)*R5. There's a minus sign in front if I2 because the current I2 is in the opposite direction of I1. Remember, all the currents are in the clockwise direction. Finally, the voltage across R12 is (I1-I5)*R12

When you add up all the voltages around the loop and equate their sum to zero, you get the first loop equation:

(I1)*R2 + (I1)*R3 + (I1-I2)*R5 + (I1-I5)*R12 = 0

By rearranging and collecting terms, this can be re-written so coefficients of the currents are gathered together. To help with the overall solution, put zero placeholders in front of currents that aren't flowing in loop 1:

Eq1: (R2+R3+R5+R12)*I1 - (R5)*I2 + (0)*I3 + (0)*I4 - (R12)*I5 = 0

The reason for putting the zero placeholders in there is to make each equation include all 5 currents; you'll need to do this in order to use standard linear algebra techniques to solve the system of 5 equations in 5 unknowns that you'll end up with.

The equations for loops 4 and 5 will include the 12 volt source like this:

R11*(I4-I3) + R10*I4 + R14*I4 + R13*I4 + 12 = 0

R1*I5 + R12*(I5-I1) - 12 = 0

After rearranging, these two equations become (move the 12 volt quantities to the right-hand side; this is necessary for the later linear algebra solution):

Eq4: (0)*I1 + (0)*I2 - (R11)*I3 + (R10+R11+R13+R14)*I4 + (0)*I5 = -12

Eq5: -(R12)*I1 + (0)*I2 + (0)*I3 + (0)*I4 + (R1+R12)*I5 = 12

It's very important to get the minus signs right!

I've given you equations 1, 4 and 5; I'll leave it to you to derive equations 2 and 3.

After you derive all 5 equations, substitute numerical values for the symbols for the various resistances and put the 5 equations in matrix form like this:

\( \left[ \begin{array}{5}2200 & -1000 & 0 & 0 & -1000\\-1000 & 2200 & -1000 & 0 & 0\\0 & -1000 & 2200 & -1000 & 0\\0 & 0 & -1000 & 1300 & 0\\-1000 & 0 & 0 & 0 & 1100\end{array}\right]\left[ \begin{array}{1}I_1\\I_2\\I_3\\I_4\\I_5\end{array}\right]=\left[ \begin{array}{1}0\\0\\0\\-12\\12\end{array}\right]\)

Then you can solve the system with a suitable calculator or software program.

The result is:

I1 = .010582
I2 = .0027519
I3 = -.0045282
I4 = -.012714
I5 = .020530

These numbers compare almost exactly with values on your original schematic. You can now use Ohm's law to calculate the voltage across any resistor since you now have the currents through them. Don't overlook the fact that the current in the 1000 ohm resistors is the difference of two of the currents you got from the KVL solution. For example, the voltage across R5 is given by (I1-I2)*R5.

 

Since I suggested the redraw, and provided you with my start ... I'm going to give you the whole thing.

 

@JoeJester

In my opinion, even though the mesh analysis is correct, it is a major detour from the straighter solution. Since we are already given most of the currents and voltages, all that remains to do is caclulate the current through R4 and R6 by Ohm's Law and the currents towards the center with 5 simple applications of the Kirchoffs Current Law. Analytical methods are an overkill in this case.

 

You asked the OP what the question was, and he replied:

"...use the reults on the ring circuit diagram to confirm the above laws ohms law and kirchoffs law."

He isn't being asked to just "solve" the circuit. He is apparently being asked to "confirm the above laws ohms law and kirchoffs law."

The results on the original circuit he posted appear to have been derived from a circuit simulator. To "confirm" those results would seem to me to require an independent solution of the network using, for example, KVL, and then comparison of the results obtained with those given in the original schematic.

 

He isn't being asked to just "solve" the circuit. He is apparently being asked to "confirm the above laws ohms law and kirchoffs law."

I didn't think of it that way. Yes, if this is the case, JoeJester is on the right track.
I thought the way I did judging by the low level of knowledge of the OP which led me to believe that the exercise was actually very trivial.
It's up to the OP, and the OP only, to confirm one opinion and reject the other. However, he seems to have lost interest to the exercise long ago...

 

thanks for the detail reply it has helped a lot! i have got all the voltages over the resistors but i am not 100% sure i have done R5 ( 0.03v), R7 ( 0.01246v ), R11 ( 0.057571v ), R12 ( 0.217v ) right. i have added all the voltages together in my circuit and this only come to 9.76v and this should be 12v ( what goes in comes out ) with this formula (I1-I2)*R5 how do i put it in my calculator i tryed it as a takeaway but the answer that come back was to high of a high voltage.

 

thanks for the detail reply it has helped a lot! i have got all the voltages over the resistors but i am not 100% sure i have done R5 ( 0.03v), R7 ( 0.01246v ), R11 ( 0.057571v ), R12 ( 0.217v ) right.

These voltages are all too low. For example, the voltage across R5 is given by (I1-I2)*R5 = (.010582-.0027519)*1000 = (.0078301)*1000 = 7.8301 volts. The voltage across R7 = (I2-I3)*R7 = (.0027519-(-.0045282))*1000 = (.0072801)*1000 = 7.2801 volts, and so forth for R11 and R12.

i have added all the voltages together in my circuit and this only come to 9.76v and this should be 12v ( what goes in comes out ) with this formula (I1-I2)*R5 how do i put it in my calculator i tryed it as a takeaway but the answer that come back was to high of a high voltage.

Exactly which voltages are you adding up with the expectation that the sum will be 12.000 volts?

 

thanks for the reply and all the help i sorted were i was going rong last night