Before the exercises, look at this. It is the "staircase" idea drawn as a real circuit: three nodes stacked by a battery, and we slide the "floor = 0 V" label to different steps.
Notice: the gaps between steps never change (that red arrow is always the same length). Only the numbers written on each step change when we move the floor. That is the whole subject.
WHAT we do: apply the definition of a reference node.
By definition the reference node is fixed at 0 V, so ==VC=0 V==.
The shorthand VA always means the difference between A and the reference:
VA=VA−VC=VA−0.WHY: a lone node voltage is meaningless without a "from where" — the reference supplies that "from where."
Recall Solution 1.2
One node. Every ground symbol in a single circuit is the ==same 0 V electrical node==. They are drawn separately only to avoid crossing wires everywhere — a drawing convenience, not three different points.
Recall Solution 1.3
False. A resistor responds only to the potential difference across it:
I=RVtop−Vbottom.
The absolute value at one terminal alone tells you nothing about the current.
WHAT: apply I=(VX−Vref)/R.
WHY the −0 trick: the grounded end is 0 V, so the numerator collapses to just VX:
I=RVX−0=40008=0.002 A===2 mA==.
Recall Solution 2.2
WHAT: this is a voltage divider with equal resistors, so the middle sits exactly halfway.
VB=0 (it is ground).
VA=12 V (full battery above ground).
Equal resistors split the drop in half: VM=212===6 V==.
Recall Solution 2.3
WHAT: we shift every node by the same amount so that M becomes 0. Previously VM=6, so subtract 6 from all:
VA=12−6===+6 V==,VM=6−6=0,VB=0−6===−6 V==.WHAT IT LOOKS LIKE: look again at the figure above — we slid the floor to the middle step, so the top is +6 and the bottom drops below the floor to −6. A dual-rail ±6 V supply is just a 12 V battery with ground at its midpoint.
Check:VA−VB=6−(−6)=12 V — the same physical difference as in 2.2. ✓
WHAT: re-referencing shifts every node by a constant c=−6. Take any resistor carrying I=(VA−VB)/R. Under the shift:
I′=R(VA+c)−(VB+c)=RVA−VB=I.
The ccancels. So the current is unchanged: I=12/(R1+R2) in both references.
WHY it matters: this cancellation is precisely why we are allowed to move ground freely — see Kirchhoff's Voltage Law, whose loop sums are also built from differences.
Recall Solution 3.2
WHAT: use the difference directly — ground never enters, because neither end is ground.
IP→Q=RVP−VQ=10003−(−5)=10008===8 mA==.WHY carefully with the sign: the drop across the resistor is 3−(−5)=8 V, not 3−5=−2. Subtracting a negative adds.
Recall Solution 3.3
WHAT: to make Q the floor, subtract its old value −5 from every node — i.e. add +5 to all.
VQ=−5+5=0,VP=3+5===+8 V==.
Current: I=(VP−VQ)/R=(8−0)/1000===8 mA==, identical to Exercise 3.2. ✓
The number written on P changed (3→8), but the physical current did not.
WHAT: put the two equal resistors in series across the 9 V battery, then declare the midpoint the ground. This is a voltage-divider used as a rail splitter.
Top node: Vtop===+4.5 V==
Midpoint (ground): 0 V
Bottom node: Vbottom===−4.5 V==WHY it works: equal resistors put the midpoint exactly halfway (4.5 V above the bottom). Calling that midpoint "0" makes the top +4.5 and the bottom −4.5. The battery still only supplies a single 9 V difference; ground placement did all the "dual-rail" magic.
Recall Solution 4.2
WHAT: shifting all nodes by c gives readings 10+c,4+c,c. We want their sum =0:
(10+c)+(4+c)+(0+c)=14+3c=0⇒c=−314≈−4.667.
New readings:
V1=10−314=316≈==5.333 V==,V2=4−314=−32≈−0.667 V,V3=−314≈−4.667 V.This is the "centroid" reference — no single existing node becomes exactly 0; instead the average of all nodes becomes 0. That is a legitimate choice too: reference is any potential we declare zero, not necessarily a labelled node. (It matches the physics of Electric Potential — only differences are fixed, the offset is ours.)
(a) Not earthed: the only path to earth is through the person. Treating the fault as 230 V across the person's 2000Ω:
Iperson=2000230=0.115 A===115 mA==.
This is far above the ~30 mA that can stop a heart — lethal.
(b) Earthed through 0.5Ω: now there is a low-resistance path in parallel with the person. Almost all fault current takes the easy route:
Iearth=0.5230===460 A==.WHY it protects:460 A is enormous — it instantly trips the breaker/fuse, cutting the supply within milliseconds. And because the earth wire is 0.5Ω vs the person's 2000Ω, the current divides overwhelmingly through the wire (it is a current divider — the person sees a tiny fraction). See Earthing and Electrical Safety.
Recall Solution 5.2
(a) With 5 nodes there are naively 5 unknown potentials. But the equations only ever contain differencesVi−Vj. If (V1,…,V5) is a solution, so is (V1+c,…,V5+c) for any c — an entire family of solutions. The system is underdetermined by exactly one free constantc.
(b) Fixing one node to 0 V removes that free constant. Remaining unknowns:
n−1=5−1===4==.
This is precisely why Nodal Analysisalways begins by picking a reference node — it makes the linear system solvable.
Recall Solution 5.3
WHAT: power in a resistor is P=(VA−VB)2/R (from P=VI and I=(VA−VB)/R).
Under a reference shift, every node gains the same c:
P′=R((VA+c)−(VB+c))2=R(VA−VB)2=P.
The c cancels inside the difference before it is squared, so P′=P. Power is invariant under choice of ground. Physics (currents, powers, differences) never depends on the label; only the individual node numbers do.
Why is a 5-node network underdetermined before choosing ground?
The equations contain only differences, so any global shift +c is also a solution — one free constant remains until a node is fixed to 0 V.
How do you build a ±4.5 V supply from a 9 V battery?
Two equal resistors in series across the battery; declare the midpoint ground → top +4.5 V, bottom −4.5 V.
Is power dissipation reference-dependent?
No — P=(VA−VB)2/R; the shift +c cancels inside the difference, so P is invariant.
Why does earthing a case protect a person?
It adds a low-resistance parallel path, so fault current is huge and trips the breaker; the current divides overwhelmingly through the wire, not the person.
