Apply superposition theorem: Visual walkthrough — Hardware deep dive

Everything below rests on one habit from Linearity and homogeneity: in these circuits, double the push → double the response, and two pushes → the sum of each response. Keep that sentence in your pocket.

Step 1 — What a resistor even promises (Ohm's law as a straight line)

WHY this matters. Look at the figure: plot $v$ against $i$ and you get a straight line through the origin. Straight-through-origin is the whole game. It means proportional (twice the current → twice the voltage) and, as we'll see, additive. That straightness is the reason superposition exists.

PICTURE. The blue line is $v = R i$ ; its steepness is $R$ . There are no bends — no bend is exactly what "linear" means.

Step 2 — The circuit we will solve, and the one number we want

WHY name a node and a target first. Superposition adds contributions to one chosen quantity. If we never fix what we're adding up and which direction is positive, the signs turn to mush (the parent's fourth "mistake"). So: positive $V_{A}$ = node $A$ above ground. Locked in.

PICTURE. The yellow dot is node $A$ . Trace the two paths out of it: left through $R_{1}$ toward the battery, down through $R_{2}$ toward ground, and the red arrow of the 3 A pump feeding in.

Step 3 — Turn off the pump: the battery acts alone

WHY a gap and not a wire. A current source's whole personality is the current it forces. Set that to zero and it forces nothing to flow — which is precisely what an empty gap does. (A wire would let current flow freely; that's the opposite.)

What remains is beautifully simple: the 12 V battery, then $R_{1}$ , then $R_{2}$ to ground, in a single loop. Current has exactly one path, so the same current threads both resistors — they are in series. That is the setup for a voltage divider: $V_{A, 1} = 12 \cdot fraction of the drop across R_{2} \frac{R _{2}}{R _{1} + R _{2}} = 12 \cdot \frac{6}{4 + 6} = 7.2 V$

Reading it term by term: the total 12 V is shared between the two resistors in proportion to their sizes. $R_{2}$ takes the fraction $\frac{6}{10}$ , so node $A$ (which sits at the top of $R_{2}$ ) sits at $7.2$ V.

PICTURE. The gap where the pump was is drawn in red. The blue arrow marks the single loop current; the yellow bracket shows $R_{2}$ 's share of the voltage.

Recall Why is this a divider and not something harder?

Because with the pump open, only ONE loop exists, so KCL is trivial (same $i$ everywhere) and only KVL is left. ::: One loop ⇒ series ⇒ voltage divides by resistance ratio.

Step 4 — Turn off the battery: the pump acts alone

WHY a wire and not a gap. A voltage source insists on a fixed voltage; zero that and it insists on 0 V between its terminals — two points forced to the same potential is exactly a wire.

Watch what the wire does to the topology. $R_{1}$ used to run from node $A$ toward the battery; now its far end is wired straight to ground. So $R_{1}$ and $R_{2}$ both run from node $A$ to ground — they are now in parallel. All 3 A must leave node $A$ through this parallel pair:

V_{A,2} = \underbrace{3}_{\text{amps injected}}\times \underbrace{2.4}_{\text{ohms seen}} = 7.2\text{ V}$$ Term by term: the pump forces 3 A out of node $A$; that current sees the combined resistance $2.4\,\Omega$ on its way to ground; Ohm's law then says the node rises to $V=IR = 7.2$ V above ground. **PICTURE.** The battery is replaced by a green wire, folding $R_1$'s left end down to ground so the two resistors sit side by side — clearly parallel. The red arrow shows the 3 A splitting into them. ![[deepdives/dd-hardware-1.2.11-d2-s04.png]] > [!mistake] "Shorting the battery removes $R_1$." > **Why it feels right:** the battery is gone, so its branch seems gone. **Why it's WRONG:** the wire *stays*; it just relocates $R_1$'s end to ground. $R_1$ still carries current and still matters. **Fix:** replace the *source* with a wire, keep every resistor. --- ## Step 5 — Add the two pictures (this is superposition) > [!intuition] WHAT > Each source, acting alone, lifted node $A$ by a definite amount: > $$V_A = \underbrace{V_{A,1}}_{\text{battery's lift}} + \underbrace{V_{A,2}}_{\text{pump's lift}} = 7.2 + 7.2 = 14.4\text{ V}$$ **WHY we may just add.** From Step 1 every element is a straight line, and [[Kirchhoff's Voltage and Current Laws|KCL and KVL]] are themselves *sums equal to zero* — linear equations. Solving a linear system for a **sum of inputs** gives the **sum of the individual solutions** (the parent's $A\mathbf{x}=\mathbf b$ argument). Both contributions are measured with the *same* positive-up convention we fixed in Step 2, and both came out **positive**, so the algebraic sum is a plain addition. Had one source pushed node $A$ *below* ground, its term would carry a minus sign — that is the whole meaning of "add **algebraically**". **PICTURE.** Two stacked bars, blue ($7.2$) and green ($7.2$), stacking to the yellow total $14.4$ V. Height literally adds. ![[deepdives/dd-hardware-1.2.11-d2-s05.png]] --- ## Step 6 — Sanity check: the direct nodal solution lands on the same number > [!intuition] WHAT > Never trust a theorem you can independently check. Do the circuit *all at once* with [[Nodal and mesh analysis|nodal analysis]]. KCL at node $A$ — total current *leaving* the node is zero: > $$\underbrace{\frac{V_A-12}{4}}_{\text{leaving toward battery via }R_1} + \underbrace{\frac{V_A}{6}}_{\text{leaving to ground via }R_2} - \underbrace{3}_{\text{injected by pump}} = 0$$ **WHY this must agree.** Nodal analysis makes no assumptions beyond the same linear laws superposition used, so if algebra is consistent, the numbers coincide. Multiply through by 12: $$3(V_A-12) + 2V_A - 36 = 0 \;\Rightarrow\; 5V_A = 72 \;\Rightarrow\; V_A = 14.4\text{ V}\;\checkmark$$ **PICTURE.** Three current arrows meeting at node $A$: two draining out (through $R_1$, $R_2$) and one feeding in (the 3 A). Their signed sum is zero — that single equation *is* the whole circuit. ![[deepdives/dd-hardware-1.2.11-d2-s06.png]] --- ## Step 7 — The degenerate cases (never leave a scenario unshown) > [!intuition] WHAT — four sanity limits > Push each knob to an extreme and confirm the pictures still make sense. > [!example] Case A — pump off ($3\text{ A}\to 0$) > Only Step 3 survives: $V_A = 7.2 + 0 = 7.2$ V. A dead current source *is* the open circuit, so the divider result stands alone. > [!example] Case B — battery off ($12\text{ V}\to 0$) > Only Step 4 survives: $V_A = 0 + 7.2 = 7.2$ V. A dead voltage source *is* the wire; the parallel-resistance result stands alone. > [!example] Case C — both off > $V_A = 0$. No sources, no lift — the node rests at ground. The theorem returns $0+0$, exactly right. > [!example] Case D — $R_2 \to 0$ (short node $A$ to ground) > Divider: $V_{A,1}=12\cdot\frac{0}{4+0}=0$. Parallel: $R_1\|0 = 0$, so $V_{A,2}=3\times 0 = 0$. Node $A$ is welded to ground: $V_A = 0$. Both mechanisms agree that a grounded node cannot rise. **WHY show these.** A recipe you only trust for "nice" numbers is not trusted at all. Zero-source and zero-resistance are exactly where sign or divide-by-zero errors hide — and here every limit stays finite and sensible. **PICTURE.** A small panel of all four cases with their $V_A$ values, so no reader meets an untested scenario. ![[deepdives/dd-hardware-1.2.11-d2-s07.png]] --- ## Step 8 — The one forbidden move: power does not stack > [!intuition] WHAT > Tempting to also add the *power* delivered to $R_2$ source-by-source. Let's test it. $R_2$ current from each source alone (parent Example 3): $I_1 = 7.2/6 = 1.2$ A, $I_2 = 1.2$ A. > - **Current superposes:** $I = 1.2 + 1.2 = 2.4$ A ✓ (matches $14.4/6$). > - **Power does NOT:** true power is $I^2 R_2 = 2.4^2\cdot 6 = 34.56$ W. But adding per-source powers gives $1.2^2\cdot6 + 1.2^2\cdot6 = 8.64+8.64 = 17.28$ W — **half the truth.** **WHY the mismatch.** Power depends on the *square* of current, and squaring bends the straight line: $(I_1+I_2)^2 = I_1^2 + 2I_1 I_2 + I_2^2$. That extra **cross term** $2I_1I_2 = 2(1.2)(1.2)\cdot 6 = 17.28$ W is exactly the gap. See [[Why power does not superpose]]. **PICTURE.** A parabola $P=I^2R$ (curved, red) beside the straight line of Step 1. Marking $I_1$, $I_2$ and $I_1{+}I_2$ shows the powers falling *below* the summed guess — the curvature made visible. ![[deepdives/dd-hardware-1.2.11-d2-s08.png]] > [!mnemonic] Superpose the flow, not the fury > **Currents and voltages add. Power squares.** Get $I$ or $V$ first by superposition, *then* compute $P$ **once**. --- ## The one-picture summary > [!intuition] The whole derivation in a single frame > Left: battery alone → divider → $7.2$ V. Middle: pump alone → parallel → $7.2$ V. Right: stack them → $14.4$ V, confirmed by nodal analysis. Below, the parabola warns: power refuses to stack. ![[deepdives/dd-hardware-1.2.11-d2-s09.png]] > [!recall]- Feynman retelling — the pool with two hoses > Node $A$ is the water level in a pool. The battery is one hose; on its own it raises the level to a certain mark — that's the divider giving $7.2$ V. The current pump is a second hose; on its own it raises the level to another mark — the parallel-resistance calculation, also $7.2$ V. Because water levels *just add*, the two hoses together raise it to $14.4$ V, and if you'd instead measured it all at once (nodal analysis) you'd read the same $14.4$. Turn a hose off and you're back to one mark; turn both off and the pool sits empty at ground; plug the drain wide open ($R_2\to0$) and no hose can raise it at all. The one thing that *won't* add up is "splashiness" — power — because splashing grows with the *square* of the flow, and squares carry a hidden cross-term two hoses share. So: add the water, never the splash. > [!recall] Active recall — cover the answers > - Why is a resistor's $v$–$i$ graph being a straight line the key to superposition? > - When the battery is killed, why do $R_1$ and $R_2$ become parallel? > - Both contributions to $V_A$ were $+7.2$ V — what would a *negative* contribution have meant physically? > - Currents added to $2.4$ A correctly, but powers didn't add. Name the missing term. --- ## Connections - [[Apply superposition theorem]] - [[Linearity and homogeneity]] - [[Kirchhoff's Voltage and Current Laws]] - [[Voltage divider and current divider]] - [[Nodal and mesh analysis]] - [[Thevenin and Norton equivalents]] - [[Why power does not superpose]]