This page is a drill . The parent note built the idea that voltage is a difference and ground is a chosen 0 V every case that idea can throw at you — positive and negative node voltages, ground at the top / middle / bottom, a node that IS ground, a floating (undefined) node, a limiting case, a real-world word problem, and an exam twist.
Think of a ruler you can slide up and down a wall. The marks on the ruler (the differences between nodes) never move relative to each other. Choosing ground = choosing where the ruler's "0" mark sits on the wall. Every example below just slides that ruler — and we confirm nothing physical ever changes.
Before any example, here is the full list of case-classes this topic contains. Every cell gets covered by at least one worked example.
Cell
Case class
Covered by
A Ground at the bottom — all node voltages ≥ 0
Ex 1
B Ground in the middle — some voltages go negative (sign case)
Ex 2
C Ground at the top — everything below is negative
Ex 3
D A floating node — no reference path, voltage undefined (degenerate)
Ex 4
E Zero source / limiting value (V src → 0 Ex 5
F Unequal resistors — divider ratio with re-referencing
Ex 6
G Real-world word problem — dual-rail supply for an op-ampEx 7
H Exam twist — two ground symbols, "trap" questionEx 8
I Current invariance check across a reference change
Ex 9
The workhorse circuit is a simple two-resistor divider. Study it once:
Definition Our reference dividerA source of V src R 1 R 2 A M B I I = R 1 + R 2 V src , V M − V B = I R 2 physical — they do not know or care where ground is.
B V src = 9 V R 1 = R 2 = 1 k Ω B V A , V M , V B
Forecast: Guess the three numbers before reading on. (Hint: equal resistors split evenly.)
Set the reference. V B = 0 Why this step? Every node voltage is shorthand for "difference from ground", so we must first pin ground to 0 Current in the loop. I = 1000 + 1000 9 = 4.5 mA Why this step? Current is the physical quantity; it depends on the total 9 V Climb from ground upward. V M = V B + I R 2 = 0 + ( 4.5 mA ) ( 1 k Ω ) = 4.5 V V A = V M + I R 1 = 4.5 + 4.5 = 9 V Why this step? Ground is our start line; we add each resistor's voltage drop as we walk up.
Verify: V A − V B = 9 − 0 = 9 V ≥ 0
M M V A , V M , V B
Forecast: One node must now be negative. Which one, and by how much?
Slide the ruler. We are subtracting 4.5 V M = 4.5 0 Why this step? Re-referencing = adding one constant c c = − 4.5 Apply the shift. V A = 9 − 4.5 = + 4.5 V V M = 4.5 − 4.5 = 0 V B = 0 − 4.5 = − 4.5 V Why this step? The additive constant cancels in every difference, so this is guaranteed valid.
Verify: V A − V B = 4.5 − ( − 4.5 ) = 9 V ± 4.5 V
The picture below shows this re-referenced circuit — note how the same divider now reads + 4.5 − 4.5 M Voltage Dividers .
A A V M , V B
Forecast: With the top pinned to 0 lower . Sign?
Shift by c = − 9 (old V A = 9 0 Why this step? Same slide-the-ruler move; the constant is whatever forces the chosen node to 0 Compute. V A = 0 V M = 4.5 − 9 = − 4.5 V V B = 0 − 9 = − 9 V Why this step? Adding − 9
Verify: V A − V B = 0 − ( − 9 ) = 9 V Example 3 of the parent note exactly (the forecast-then-verify one). All lower nodes are negative — normal, not an error.
Worked example What is the voltage of a disconnected node?Start from the divider with ground at B cut the wire between R 1 A A nothing — it dangles in the air. Ask: what is V A
Forecast: Can you write down a single number for V A
Trace the path from A After the cut there is no resistor, wire, or source connecting A B Why this step? A node voltage is defined only as the difference V A − V ground Conclude: V A It is not 0 9 no determined value . Meanwhile the current through the cut branch is I = 0 A Why this step? This is the true degenerate case: a floating node is under-determined exactly the way the whole circuit would be if we pinned no ground at all. No path ⇒ no equation ⇒ no answer.
Verify: Current through an open branch = 0 A V A why floating nodes are avoided in design and why we always give every node a reference path.
Worked example Let the source shrink to zeroSame divider, but sweep V src → 0 B V M
Forecast: As the battery weakens, where does the midpoint voltage head?
General midpoint. With ground at B V M = V src ⋅ R 1 + R 2 R 2 Why this step? This formula holds for any V src Take the limit. As V src → 0 V M → 0 ⋅ 2 1 = 0 V I → 0 Why this step? With no driving voltage, all nodes collapse to the same potential — which we've called 0
Verify: Plug V src = 0 V M = 0 V A − V B = 0 I = 0 no source ⇒ every node equals ground .
Worked example Divider ratio survives re-referencingV src = 12 V R 1 = 3 k Ω R 2 = 1 k Ω B M
Forecast: With ground at M V A V B
Ground at B I = 3000 + 1000 12 = 3 mA V M = I R 2 = ( 3 mA ) ( 1 k Ω ) = 3 V V A = 12 V Why this step? The small bottom resistor drops little, so the tap sits low — physical, reference-free.Re-reference to M Shift by c = − 3 V A = 12 − 3 = + 9 V V M = 0 V B = 0 − 3 = − 3 V Why this step? Same constant subtraction; the divider ratio is baked into the differences , which do not move.
Verify: V A − V B = 9 − ( − 3 ) = 12 V asymmetric (+ 9 − 3
Worked example Powering an op-amp from one batteryAn op-amp chip needs both a + V − V 0 9 V ±
Forecast: What voltage does each rail sit at, and where does the chip's "0 V
Build a divider. Put two equal resistors across the 9 V M Why this step? Equal resistors halve the battery, creating a natural midpoint at 4.5 V Declare the tap as ground. Wire the op-amp's ground pin to M Why this step? Re-referencing to the midpoint makes the top + 4.5 V − 4.5 V ±
Verify: Rails are + 4.5 V − 4.5 V 9 V weak reference — real designs buffer M
Worked example Trap: "how many nodes?"A schematic shows node A + 5 V C + 2 V two separate ground symbols drawn in different corners. An exam asks: "What is the voltage between the two ground symbols, and how many distinct nodes have a voltage of 0 V
Forecast: Is the answer "two nodes at 0 0
Identify the grounds. Every ground symbol inside one circuit is the same electrical node: the multiple symbols are just a drawing convenience so we don't have to draw long wires back to a single point. Electrically, all ground symbols are wired together at 0 V Why this step? If you treat them as separate nodes you will invent a fake voltage difference that does not exist — that is the trap.Compute the requested voltage. Both symbols are the same 0 V V gnd1 − V gnd2 = 0 − 0 = 0 V one node at 0 V Why this step? Same node ⇒ zero difference by definition.
Verify: 0 − 0 = 0 V 0 V Kirchhoff's Voltage Law : a loop between the two symbols sums to 0
Worked example Prove the current never movedTake Example 6's divider (12 V 3 k Ω 1 k Ω B M node voltages each time.
Forecast: The two computations use different node numbers — will the current come out the same?
Ground at B I = R 1 V A − V M = 3000 12 − 3 = 3 mA Why this step? Current uses the difference across R 1 Ground at M I = R 1 V A − V M = 3000 9 − 0 = 3 mA Why this step? The nodes changed to + 9 0 difference is still 9 V c = − 3
Verify: Both give 3 mA Nodal Analysis .
Recall Every cell, one line eachGround at bottom → all voltages? ::: All ≥ 0 ± → 0 → 0 V + 9 − 3 ± 4.5 9 V 0 V
Slide the ruler, marks don't move. Every re-reference is one subtraction c all nodes; every difference — and therefore every current — survives untouched.
Voltage Dividers — Examples 2, 6, 7 are re-referenced dividers.Nodal Analysis — Example 9 is the invariance that makes reference choice free.Kirchhoff's Voltage Law — the loop-sum logic behind the two-ground trap (Ex 8).Electric Potential — why only differences drive current.Earthing and Electrical Safety — physical ground vs the analysis reference. 
