Current Flow in Redrawn Circuit for Finding R—Which Diagram Is Correct?

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#1 21663101 22 Jul 2012 10:21

Steve Lawson Steve Lawson

Anonymous

Post #1
21663101 22 Jul 2012 10:21

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"
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#2 21663102 22 Jul 2012 10:23

Steve Lawson Steve Lawson

Anonymous

Post #2
21663102 22 Jul 2012 10:23

...and figure 3 [didn't finish reading ]
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#3 21663100 22 Jul 2012 10:23

Shri Nidhi Shri Nidhi

Anonymous
Post #3
21663100 22 Jul 2012 10:23

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.

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#4 21663103 22 Jul 2012 10:25

Shri Nidhi Shri Nidhi

Anonymous

Post #4
21663103 22 Jul 2012 10:25

2nd Fig:
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#5 21663104 22 Jul 2012 10:29

Shri Nidhi Shri Nidhi

Anonymous

Post #5
21663104 22 Jul 2012 10:29

3rd Fig:
#6 21663105 22 Jul 2012 10:33

Shri Nidhi Shri Nidhi

Anonymous

Post #6
21663105 22 Jul 2012 10:33

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.
#7 21663106 22 Jul 2012 11:22

Kvks kumar Kvks kumar

Anonymous

Post #7
21663106 22 Jul 2012 11:22

you are correct. current direction will be as shown in fig 3
value of R would be 336/49 ohms, i think.
#8 21663107 22 Jul 2012 15:41

Earl Albin Earl Albin

Anonymous

Post #8
21663107 22 Jul 2012 15:41

No right.
#9 21663108 22 Jul 2012 15:54

Earl Albin Earl Albin

Anonymous

Post #9
21663108 22 Jul 2012 15:54

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
#10 21663109 23 Jul 2012 09:54

Geraldo Lopes Serodio Geraldo Lopes Serodio

Anonymous

Post #10
21663109 23 Jul 2012 09:54

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
#11 21663110 24 Jul 2012 06:08

David Adams David Adams

Anonymous

Post #11
21663110 24 Jul 2012 06:08

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
#12 21663111 24 Jul 2012 10:15

Earl Albin Earl Albin

Anonymous

Post #12
21663111 24 Jul 2012 10:15

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.
#13 21663112 24 Jul 2012 10:22

Earl Albin Earl Albin

Anonymous

Post #13
21663112 24 Jul 2012 10:22

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.
#14 21663113 24 Jul 2012 10:53

Shri Nidhi Shri Nidhi

Anonymous

Post #14
21663113 24 Jul 2012 10:53

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..
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Topic summary

✨ The discussion centers on determining the correct current flow direction in a redrawn circuit used to find an unknown resistance R. The original circuit (Fig:1) was redrawn (2nd Fig) for simplification, but the current direction proposed by the user aligns with that shown in the 3rd Fig. Multiple contributors confirm that the current direction in Fig 3 is correct, noting that the sign of current in circuit analysis indicates direction but does not affect magnitude. The problem involves applying circuit simplification techniques such as Thevenin and Norton equivalents, voltage dividers, and loop analysis. Calculations show R to be approximately 5.38 Ohms. The discussion emphasizes the importance of simplification methods over mesh equations for quick insight and solving, and clarifies that the current direction assignment is a convention that can be corrected by sign interpretation in equations. The user acknowledges unfamiliarity with Thevenin and Norton theorems but recognizes their utility after reviewing the explanations.
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Current Flow in Redrawn Circuit for Finding R—Which Diagram Is Correct?

FAQ

TL;DR: The correct redrawn diagram is Figure 3, and the solved value is R ≈ 5.38 Ω; “simplify, simplify, simplify.” [Elektroda, Earl Albin, post #21663108]

Why it matters: This FAQ helps students and hobbyists quickly decide current direction and compute R using Thevenin/Norton without overcomplicating homework-style circuits.

Quick Facts

Correct current direction: Figure 3. This aligns loop relationships used to solve R. [Elektroda, David Adams, post #21663110]
Thevenin result for left network: Vth ≈ 10.91 V, Zth ≈ 0.91 Ω. [Elektroda, Earl Albin, post #21663108]
Norton form: In ≈ 3.75 A with ≈ 1.95–1.96 Ω parallel source resistance. [Elektroda, Earl Albin, post #21663108]
Solved resistor: R ≈ 5.38 Ω at 1 A branch, V ≈ 5.38 V. [Elektroda, Earl Albin, post #21663108]
Study focus: “learn Kirchoff’s rules… Norton, Thevinen” for insight and employability. [Elektroda, Earl Albin, post #21663111]

Which redrawn circuit has the correct current direction?

Figure 3 is correct. It matches loop relationships that lead to a solvable set without sign conflicts, confirming the chosen current polarities. [Elektroda, David Adams, post #21663110]

What is the final value of R in this problem?

R ≈ 5.38 Ω. Using simplification and source transformation yields a node voltage near 5.38 V across a 1 A branch, so R = 5.38 Ω. [Elektroda, Earl Albin, post #21663108]

How can I solve for R fast using Thevenin?

Form a Thevenin for the 12 V, 10 Ω, 1 Ω side: Vth ≈ 10.91 V, Zth ≈ 0.91 Ω.
Add the 2 Ω in series: Zs ≈ 2.91 Ω feeding 6 Ω || R.
Solve the node current/voltage, then compute R = V/1 A ≈ 5.38 Ω. [Elektroda, Earl Albin, post #21663108]

Can I use Norton instead of Thevenin here?

Yes. Convert to a Norton source: In ≈ 3.75 A with ≈ 1.95–1.96 Ω in parallel. Subtract the known 1 A branch to get Ix, compute node voltage, then R = V/1 A ≈ 5.39 Ω. [Elektroda, Earl Albin, post #21663108]

Do I need mesh equations to solve this?

No. Loop insights suffice. One post lists four loops but shows you can solve using relationships and simplification instead of full mesh equations. [Elektroda, David Adams, post #21663110]

Does choosing the “wrong” current direction ruin the answer?

No. If your assumed direction is opposite, the solution returns a negative current. The magnitude remains correct; flip the arrow to fix direction. [Elektroda, Geraldo Lopes Serodio, post #21663109]

Why do experts stress simplification first?

It gives quick insight and fewer equations. As one expert put it, the key is to “simplify, simplify, simplify.” This often outperforms brute-force meshes on paper. [Elektroda, Earl Albin, post #21663108]

Where is the voltage divider in this circuit?

Across the 12 V source, the 10 Ω and 1 Ω act as a divider. Their reduction produces Vth ≈ 10.91 V with Zth ≈ 0.91 Ω for the left side. [Elektroda, Earl Albin, post #21663108]

What are Thevenin and Norton equivalents in plain terms?

They are simplification tools. Replace a network by a single source plus an equivalent resistance to make hand analysis faster and clearer. [Elektroda, Earl Albin, post #21663111]

How do the 10 Ω and 1 Ω resistors affect the source seen by R?

They set the Thevenin pair for the left network: Vth ≈ 10.91 V and Zth ≈ 0.91 Ω. That then combines with the 2 Ω series path before the 6 Ω || R load. [Elektroda, Earl Albin, post #21663108]

Can Kirchhoff’s rules and SPICE confirm this result?

Yes. Kirchhoff-based equations are how SPICE solves circuits. They confirm the simplified approach and the R ≈ 5.38 Ω outcome. [Elektroda, Earl Albin, post #21663111]

Someone said “No right.” What’s the final consensus?

Later posts resolve the ambiguity: Figure 3’s current directions align with the solvable loop set and yield the correct R value. [Elektroda, David Adams, post #21663110]

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