Current Runs Downhill — the Junction Rule Meets Ohm's Law
Three gold power resistors on a lab board meet at a junction C. Posts A, B and D lead off to the rest of a circuit under the bench, and an earth terminal on the board is the 0 V reference. Sliders set the potentials of B, C and D, measured from earth, and all three resistances; from them the junction rule and Ohm's law fix the one unknown, the potential of A. Press ▶ and a multimeter's red probe hops from earth to D, C and B and last of all to A, its display showing each potential the moment the tip lands, while a fence above each resistor shows the potential sloping downhill the way the current runs. The probing plays once and stops, and the timeline takes you back to any moment.
इस सिमुलेशन का उपयोग कैसे करें
- Potential of B, V_B — 0 to 12 V, starting at 4 V.
- Potential of the junction C, V_C — 0 to 20 V, starting at 10 V.
- Potential of D, V_D — 0 to 12 V, starting at 6 V.
- Resistor R₁ between A and C — 0.5 to 10 Ω, starting at 2 Ω.
- Resistor R₂ between C and B — 0.5 to 10 Ω, starting at 3 Ω.
- Resistor R₃ between C and D — 0.5 to 10 Ω, starting at 4 Ω. Every change re-solves the network and keeps the moment on screen; nothing moves until you press ▶.
- Play/pause, back to the start, a timeline you can drag to any moment, and ½×, ¼×, ⅛× slow motion.
- Drag anywhere on the scene to look around.
क्या देखें
- At the starting values the probe reads 6 V on D, 10 V on C and 4 V on B, then 16 V on A — exactly what the junction rule and Ohm's law predict.
- Every arrow and every stream of beads runs downhill: from A at 16 V into C at 10 V, and out of C to B at 4 V and D at 6 V. The fences show the same slopes.
- Raise R₁ to 10 Ω: the same 3 A now needs a 30 V drop, and A climbs to 40 V.
- Lower V_C to 2 V, below both B and D: current now flows into C from B and D, the arrows flip, and A falls to −1.33 V, below C, with 1.67 A leaving C through R₁.
- Set V_B, V_C and V_D all to 10 V: no current flows anywhere, and A sits at 10 V too.
इसके पीछे की भौतिकी
Through a resistor, conventional current flows from the higher potential to the lower one, and Ohm's law gives its size: I = (V_high − V_low)/R, or V_ab = V_a − V_b = IR. With B at 4 V, C at 10 V and D at 6 V, the current in the 3 Ω resistor is (10 − 4)/3 = 2 A, flowing out of C towards B, and the current in the 4 Ω resistor is (10 − 6)/4 = 1 A, out of C towards D. Charge does not pile up at a junction (ΣI_in = ΣI_out), so 2 + 1 = 3 A must arrive at C through the 2 Ω resistor from A — and to drive 3 A through 2 Ω, A must sit 3 × 2 = 6 V above C. In one line: (V_A − V_C)/R₁ = (V_C − V_B)/R₂ + (V_C − V_D)/R₃, so (V_A − 10)/2 = 2 + 1 and V_A = 16 V. The signs take care of every other case: if C is lower than B, the current in that branch flows into C and its term turns negative; if more current reaches C from B and D than leaves it, the balance must leave through R₁ and A ends up below C. The sim holds A, B and D at their potentials with ideal sources from earth and solves the whole network exactly, and C comes back at precisely the potential you set.