Read and interpret circuit schematic symbols: Visual walkthrough

This is the visual companion to the parent topic. We go slower and draw more.

Step 1 — WHAT a wire and a node are (the empty stage)

The "height" here is called electric potential — a number, measured in volts ( $V$ ), that tells you how much push charge feels at that spot (built properly in Electric Potential and Voltage). The whole reason a schematic is useful is this: shape does not matter, only which points share a node.

Figure — Read and interpret circuit schematic symbols

Look at the figure. The bent wire on the left and the straight wire on the right are, electrically, the same thing — both are one node, one plateau. The two blobs of colour mark two different nodes. Nothing has happened yet; this is our empty stage.

Step 2 — WHAT a battery does (it builds a height difference)

$E = 5 V$

$E$ — the electromotive force, i.e. the guaranteed potential difference the battery holds between its two terminals no matter what.
$5 V$ — our chosen value: the $+$ plateau sits 5 volts above the $-$ plateau.

Step 3 — WHAT a resistor does (Ohm's law as a ramp)

$V$ — the voltage drop: how much the plateau falls from the resistor's entry to its exit (volts).
$I$ — the current: how fast charge flows, measured in amperes ( $A$ ). One amp is a lot of charge per second.
$R$ — the resistance: how narrow the pipe is, in ohms ( $Ω$ ). Bigger $R$ = steeper cost per unit of current.

$R = 150 Ω$

The figure draws the resistor as a downhill ramp: charge enters high, leaves lower, and the amount it falls is exactly $V = I R$ . The steeper you want the same drop, the more current — or the bigger the $R$ .

Step 4 — WHAT an LED does (a one-way cliff of fixed height)

When it does conduct, an LED behaves very unlike a resistor: it drops an almost fixed voltage no matter the current, called its forward voltage $V_{L E D}$ .

$V_{L E D} = 2 V$

$V_{L E D}$ — the fixed cliff height the LED demands to light up. Below $2 V$ across it, it stays dark; at conduction it "eats" exactly $2 V$ .

The figure shows the LED as a fixed-height cliff in the path — not a ramp. Whatever current flows, the fall across it is $2 V$ . The arrow on the cliff shows the only allowed direction.

Step 5 — WHAT the loop is (stringing the plateaus together)

Trace the figure with your finger: leave the $+$ terminal (top plateau), fall down the resistor ramp, fall off the LED cliff, arrive back at the $-$ terminal (bottom plateau). One continuous walk, one loop, one current $I$ .

Step 6 — WHY the drops must add up (Kirchhoff's Voltage Law)

Here is the key idea that lets us solve for $I$ . Walk the whole loop and return to where you started. You are back on the same plateau — same height. So every rise and every fall around the loop must cancel exactly. This is Kirchhoff's Voltage Law (KVL).

$rise at battery E = fall in resistor V_{R} + fall at LED V_{L E D}$

$E$ — the height the battery lifts charge (a rise, $+ 5 V$ ).
$V_{R}$ — the fall across the resistor ( $= I R$ from Step 3).
$V_{L E D}$ — the fixed fall across the LED ( $= 2 V$ from Step 4).

The figure is a height profile of the walk: up by $E$ at the battery, down by $V_{R}$ across the resistor, down by $V_{L E D}$ across the LED, landing back exactly at the start line. Total up = total down.

Now substitute $V_{R} = I R$ and solve for the one unknown, $I$ :

$E = I R + V_{L E D} ⟹ I = \frac{E - V _{L E D}}{R}$

Numerator $E - V_{L E D}$ — the height left over for the resistor after the LED takes its cut.
Divide by $R$ — because $V_{R} = I R$ inverts to $I = V_{R} / R$ : the ramp converts leftover height into current.

$I = \frac{5 - 2}{150} = \frac{3}{150} = 0.02 A = 20 mA$

That is the parent topic's central result, now earned from the height picture, not quoted.

Step 7 — The degenerate cases (never leave the reader stranded)

The figure lines up all four broken cases beside the working one so you can see at a glance which walks complete the loop and which dead-end.

The one-picture summary

One height diagram compresses the whole derivation: the battery lifts charge by $E$ , the resistor ramp gives back $V_{R} = I R$ , the LED cliff gives back $V_{L E D}$ , and the walk lands exactly where it began — forcing $E = I R + V_{L E D}$ , hence $I = 20 mA$ .

Recall Feynman retelling — explain the whole walk to a 12-year-old

Imagine a water park with two flat pools: a low pool and a high pool. A pump (the battery) lifts water from the low pool up to the high pool — that lift is 5 units high, that's the EMF. Now the water wants to slide back down. On its way it must go down a gentle slide (the resistor) and then over a fixed little waterfall (the LED) that lights up when water pours over it the right way. The waterfall always takes 2 units of height. Since the water ends back in the low pool at the exact same level it started, all the drops must add back up to the 5 units the pump lifted: $5 = (slide) + 2$ , so the slide takes $3$ units. How fast the water flows depends on how gentle the slide is — a steeper narrower slide (bigger $R$ ) lets less through. With our slide, the flow works out to 20 little water-drops per moment — that's the 20 milliamps. Turn the waterfall the wrong way and no water flows; take away the slide and the water crashes down too fast and breaks the light. That's the whole circuit.

Connections

Ohm's Law — the ramp rule $V = I R$ behind Step 3
Series and Parallel Circuits — why one loop means one current (Step 5)
Kirchhoff's Voltage Law — the "land back at the same height" law (Step 6)
Electric Potential and Voltage — what "height / plateau" really means (Step 1)
Diodes and LEDs — the one-way cliff (Step 4)
Grounding and Reference Voltages — choosing the zero of height

Visual walkthrough — Read and interpret circuit schematic symbols