Initial assessment of available information The user finally provided the real problem statement: “Find (i) the currents through the resistors and (ii) the potential difference between points a and b for the circuit in Fig-1”, where
• ε₁ = 3 V, ε₂ = 5 V, ε₃ = 4 V • R₁ = 100 Ω (between nodes a and the junction of ε₂–ε₃) • R₂ = 50 Ω (top branch, in series with ε₁)
All sample answers simply asked the user to supply the question; none of them solved it. There is therefore no conflict between online/off-line answers that needs reconciliation. A complete technical solution is required.
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Direct answer to the question Assuming the batteries are oriented with their positive plates toward the right (i.e. each ε raises the potential as we move from left → right):
1. Current through R₁ I₁ = (E₂) ⁄ R₁ = 5 V ⁄ 100 Ω = 0.050 A (50 mA) flowing from the ε₂–ε₃ junction toward node a.
2. Current through R₂ I₂ = (E₂ + E₃ – E₁) ⁄ R₂ = (5 V + 4 V – 3 V) ⁄ 50 Ω = 6 V ⁄ 50 Ω = 0.12 A (120 mA) flowing from node b toward node a through R₂.
3. Potential difference between a and b V_ab ≡ V(a) – V(b) = –(E₂ + E₃) = –9 V. Hence node b is 9 V higher than node a (|V_ba| = 9 V).
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Detailed problem analysis 1. Node assignment • Node a: reference (V_a = 0 V) • Node 2: junction between ε₂ and ε₃ (V₂) • Node b: right-hand end of ε₃ and R₂ (V_b) • Node t: top of ε₁ / left of R₂ (V_t)
3. Currents (passive sign convention, positive from left to right or from the first-named node to the second): I₁ = (V_a – V₂) ⁄ R₁ = (0 – 5) ⁄ 100 = –0.05 A ⇒ 50 mA from node 2 to node a. I₂ = (V_t – V_b) ⁄ R₂ = (3 – 9) ⁄ 50 = –0.12 A ⇒ 120 mA from node b to node t to node a (opposite the assumed positive direction).
5. Power check (optional sanity-test) Total power delivered by sources equals total power dissipated in resistors: P_R1 = I₁²R₁ = (0.05)²·100 = 0.25 W P_R2 = I₂²R₂ = (0.12)²·50 ≈ 0.72 W ΣP_R ≈ 0.97 W. Source powers (taking delivery positive when supplying power): P_ε₁ = ε₁·(–I₂) = 3·0.12 = 0.36 W (absorbing) P_ε₂ = ε₂·I₁ = 5·0.05 = 0.25 W (delivering) P_ε₃ = ε₃·(–I₂) = 4·0.12 = 0.48 W (absorbing) ΣP_sources = –0.97 W (equal in magnitude to ΣP_R). The energy balance confirms the calculation.
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Current information and trends Nothing in this textbook-style DC analysis is affected by very recent industry developments. Modern simulation tools (LT-Spice, KiCad/NgSpice, TINA-TI, etc.) automate exactly this nodal-analysis procedure.
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Supporting explanations and details • “Negative current” in the algebra simply tells us the actual direction is opposite to the one we initially assumed. • An ideal voltage source in parallel with a resistor (ε₂ || R₁ here) forces the resistor voltage, hence fixes its current completely independent of the rest of the network. • Because ε₁ is the smallest source, while ε₂+ε₃ in series is the largest, the top branch current ends up running against ε₁.
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Practical guidelines 1. When drawing a problem yourself, label the polarities of every source explicitly to avoid the ambiguity we had to resolve by assumption. 2. If you simulate: place a tiny (e.g. 1 mΩ) series resistor in every ideal source. That lets you inspect branch currents directly without relying on the SPICE “current through voltage source” sign convention.
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Possible disclaimers or additional notes If the actual polarities of ε₂ or ε₃ in your copy of the figure are opposite to the assumption “positive to the right”, just flip the algebraic signs of the corresponding ε and recompute. The method stays identical.
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Suggestions for further research 1. Extend the analysis to include internal resistances of the batteries; see how the currents shift. 2. Replace R₁ with a current-controlled element (e.g. a thermistor) and study the operating point with temperature feedback.
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Brief summary • I_R1 = 50 mA from the ε₂–ε₃ junction toward node a. • I_R2 = 120 mA from node b toward node a (through R₂ and ε₁). • Node b is 9 V higher than node a (V_ab = –9 V).
Those answers follow directly from Kirchhoff’s laws once the source polarities are fixed.,
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