Signs Around the Loop — Voltages Only
A lab board carries four sealed grey components, k, x, y and z, each with its own LED voltmeter and its + and − ends marked. They sit on a square and its diagonal: x along the top, k on the left, z on the right and y on the diagonal, while the bottom side is a plain wire. Only two voltages are known — the meters on x and y are covered with masking tape. Two sliders set V_k and V_z, and a switch picks the direction to walk the loops. Press ▶ and a test charge walks the right-hand triangle, then the left-hand one, then the outer square, under a fence that rises and falls with the potential; each time a loop closes, the tape comes off the meter that loop has just pinned down. The walk plays once and stops, and the timeline takes you back to any moment.
Bu simülasyon nasıl kullanılır
- Voltage of component k, V_k — 0 to 10 V, starting at 2 V.
- Voltage of component z, V_z — 0 to 10 V, starting at 4 V. Every change re-computes the walks and keeps the moment on screen; nothing moves until you press ▶.
- Loop direction — clockwise or anticlockwise: every sign in the walk flips, the answers do not.
- 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.
Nelere dikkat etmeli
- At the starting values the first loop gives V_y = 4 V and the second V_x = 2 V; as the tape comes off, the meters read 4.000 and 2.000.
- At the end of every walk the fence comes back to the height it started from: the potential changes round a loop always add to zero.
- Switch to anticlockwise: every chip and every term changes sign, and V_x and V_y stay exactly the same.
- Raise V_k above V_z: V_x turns negative and the meter on x shows a minus sign — its + end is now the lower one.
- V_y always equals V_z, whatever V_k is: y and z both join b to the pair c–d, which a plain wire holds at one potential.
Arkasındaki fizik
Kirchhoff's loop rule says that the potential changes round any closed loop add up to zero, ΣV = 0: a charge that goes round and comes back to where it started has neither gained nor lost energy. When only voltages are given, no current is needed. Treat every component as a battery that pushes from its − end to its + end, and choose a direction round the loop: walking through a component with its push (from − to +) counts +V, walking against it (from + to −) counts −V, and a plain wire counts nothing. Start with the loop that has only one unknown. Clockwise round the right-hand triangle b → c → d → b, z is crossed from + to − and y from − to +: −V_z + V_y = 0, so V_y = V_z = 4 V. Round the left-hand triangle b → d → a → b: −V_y + V_k + V_x = 0, so −4 + 2 + V_x = 0 and V_x = 2 V. The outer square b → c → d → a → b gives −V_z + V_k + V_x = −4 + 2 + 2 = 0: it brings nothing new and checks both answers, because it is the two triangles added together, with the shared diagonal cancelling. Walk anticlockwise instead and every sign flips, −V_y + V_z = 0, but the equations and the answers stay the same. In general V_y = V_z and V_x = V_z − V_k; a negative answer just means that the component's + end is really the lower one.