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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