Solve coupled junctions and interpret a signed bridge current
A bridge joins the midpoint of two source-to-return arms.
Learning outcome: Predict bridge-current sign for right lower resistances 2, 6 and 8 ohm before solving. Retain both node residuals, a loop voltage account and total power. At balance, vary Rb and explain what remains unchanged and why.
Starting recap
Integrate current or resistance, preserve node topology and derive transient state with a signed energy account.
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Try
At right lower R = 6 ohm, the ratios 2/4 and 3/6 match. Both midpoint potentials are 8 V and bridge current is zero for every supported positive Rb. Left arm current is 2 A, right arm current 4/3 A, and source power is 40 W.
Open the experimentPredict
A bridge current solves to a negative value relative to a chosen arrow. What should you do?
Experiment
Predict bridge-current sign for right lower resistances 2, 6 and 8 ohm before solving. Retain both node residuals, a loop voltage account and total power. At balance, vary Rb and explain what remains unchanged and why.
Solve a bridge without false series reduction
Worked teaching model · synthetic values · Top node 12 V, bottom 0 V. Left upper/lower arms 2/4 ohm; right upper 3 ohm and selected lower arm. Selected bridge joins the two midpoint nodes. Conventional bridge current is positive left to right. One ideal source, no conflicting parallel sources.
Reset keeps your prediction, captures and practice history.
Read the node connections before calculating
The source fixes T relative to B; its branch is represented by the labeled boundary nodes. Bridge current is positive L to R. An arm is not a simple series pair when bridge current leaves its midpoint. Component orientation and drawing lengths are schematic; source polarities are specified in the caption and node labels.
| First node | Second node | Element |
|---|---|---|
| T: 12 V | L | 2 Ω |
| L | B: 0 V | 4 Ω |
| T: 12 V | R | 3 Ω |
| R | B: 0 V | 6 Ω |
| L | R | 2 Ω; I positive L → R |
Common schematic symbols and conventions
The symbol key is not a wiring recommendation. An open switch removes a conducting path; an ideal closed switch supplies a zero-drop connection. A current arrow is a chosen conventional-charge reference, not electron drift. A junction dot signifies connection; a crossing without a dot need not connect.
Potential around top-left-bridge-right-top loop
Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.
Potential around top-left-bridge-right-top loop: numerical coordinates
| ordered node index, not distance | potential (V) |
|---|---|
| 0 | 12 |
| 1 | 8 |
| 2 | 8 |
| 3 | 12 |
| Quantity | Value | Unit |
|---|---|---|
| Left midpoint potential | 8 | V |
| Right midpoint potential | 8 | V |
| Bridge current positive left-to-right | 0 | A |
| Left upper current | 2 | A |
| Left lower current | 2 | A |
| Right upper current | 1.33333 | A |
| Right lower current | 1.33333 | A |
| Left junction residual | 0 | A |
| Right junction residual | 0 | A |
| Source power | 40 | W |
| Total resistive power | 40 | W |
A signed negative bridge result means the true current opposes the chosen reference arrow, not that charge conservation failed.
At left, (12−VL)/2 = VL/4 + (VL−VR)/Rb. At right, (12−VR)/3 + (VL−VR)/Rb = VR/Rlower. Solve the coupled equations rather than grouping arms that have a midpoint branch as simple series. The balanced ratio gives zero bridge current even when both arm currents remain nonzero.
Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.
A signed negative bridge result means the true current opposes the chosen reference arrow, not that charge conservation failed.
At left, (12−VL)/2 = VL/4 + (VL−VR)/Rb. At right, (12−VR)/3 + (VL−VR)/Rb = VR/Rlower. Solve the coupled equations rather than grouping arms that have a midpoint branch as simple series. The balanced ratio gives zero bridge current even when both arm currents remain nonzero.
Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.
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Public practice checks a specific claim. It does not establish independent mastery.
Which source, topology, initial condition or instrument assumption would invalidate this inference?
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