Circuit Query

I’m betting that you tried to attach a couple of images, but because they weren't of one of the three file types: GIF, JPEG, or PNG, it failed to appear (or you forgot to attach Fig:1 and the "2nd Fig"

...and figure 3 [didn't finish reading ]

To Find R in the given circuit(Fig:1), the circuit was redrawn as in the 2nd Fig. But as my understanding,the current flow will be as in the 3rd fig.Plz tell me which one is correct.

2nd Fig:

3rd Fig:

Yes Steve, You are right. I used to upload the wrong format & also there is no option to upload more than one image file.Thanks for indicating.

you are correct. current direction will be as shown in fig 3
value of R would be 336/49 ohms, i think.

No right.

1st of all the key to solving these sorts of problems, and they get mind blowing, is to simplify, simplify, simplify. You have three basic screwdrivers, Norton, Thevinen, Superposition (not used here).

Look at the circuit, you have a voltage divider with the 10 & 1 Ohm resistors. Redraw it. The Thevinen source is now 10.91V, Thevinen impedance is now 0.91 Ohms. Look at it again, the 2 Ohms is in series with that impedance and the load we call 6//R. So now the source impedance is 2.91 Ohms. Now you have one node voltage at 6//R and one unknown branch current call it Ix.

Ix = ((10.91 - ((Ix + 1)*(2.91))/6 => Ix = 0.90A, now you can determine the node voltage Vx and R which is 5.38V/1A = 5.38 Ohms.

Another method is to use Norton. With the source impedance of 2.91 Ohms, that becomes parallel with the 6 Ohms ( 1.96 Ohms), the source current will be 10.91V/2.91 Ohms = 3.75A, Therefore Ix = 3.75A - 1A, Va = 2.75A*1.95Ohms = 5.39V, R = 5.39V/1A

Mr. Nidhi,

The diference concerning to the current direction assignmento that one do when solving a a circuit problem like that will not disturb too much the correct result. Only the signal of the current may change but its module will be correct. For exemple: When you solve the mesh equation (tension) or nod equation (current), if the current was assigned in the correct diretion it will resul with the plus signal (+) and if it result with a minus signal (-) this means that the correct direction is the contrary of the initial assignment.

Best regards

To solve this problem you do not need mesh equations. It seems like at first, but you do not. You do use loop analysis but no need for mesh. This looks like a homework problem to me. They are 4 possible loops but you do not need them.
Loop 1 12 = (I1+I2) + 10(I1)
Loop 2 12 = (I1+I2) + 6(I2-1) + 2(I2)
Loop 3 12 = (I1+I2) + 1(R) + 2(I2)

And one that alot of people miss
loop 4 6*(I2-1) = 1*R
You have 3 unknowns and 4 equations so it can be solved but because of Loop 4 you really only have 2 unknowns so you can just use loop 1 and 2 to solve.

Using simultaneous equations - solve for the unknown.

Good Luck

and yes fig 3 is correct

I agree that the , "Student," should learn Kirchoff's rules, this is how a SPICE program solves it. On the other hand that doesn't allow a quick insight into the problem as a whole.

If a student never leads the art of simplification, Norton, Thevinen, then likely they will not be employed in the technical arena. Wouldn't get past the 1st interview.

There are 3 current loops, true, but one is known, the one defined as 1-Amp, but then again the resistance defined as R is an unknown so 3 unknowns means 3 independent equations are required. Correctly pictured in Figure 3.

Thanks for all of your replies.Still i'm not familiar with thevinen,norton etc etc.. I'm just studying from the beginning. It is actually a solved problem. R value is 5.38 Ohms. To solve R,they have redrawn the circuit which is shown in the 2nd fig. I felt the current directions were wrong.Since I had studied in the previous lessons, following the polarities (- to + as in 12V) the adjacent resistors will have the polarity mark as in the 3rd fig. Apart from this i never know this acts as the voltage divider or not. It would be better if I read the replies once again after understanding the theorems you mentioned.Thanks..