Read and interpret circuit schematic symbols: Worked examples

We build every needed idea from the ground up. Before touching a single example, let us make sure every symbol we will use is earned.

Symbols we will lean on (earned before use)

The scenario matrix

Every worked example below is tagged with the cell it covers. Together they fill the whole grid.

Cell	Scenario class	The "gotcha" it tests
A	Normal series loop (battery–R–LED)	Baseline: apply KVL + Ohm
B	Two resistors in series	Same current shared; add resistances
C	Two resistors in parallel	Same voltage shared; reciprocal add
D	Degenerate: open switch	Zero current, undefined-looking but defined
E	Degenerate: $R = 0$ (short)	Division-by-zero danger, infinite current
F	Sign/orientation: LED backwards	Diode blocks — current is zero, not negative
G	Crossing wires (dot vs no dot)	Topology flips the whole answer
H	Limiting case: $R \to \infty$	Current collapses to zero smoothly
I	Real-world word problem	Choose a resistor for a target current
J	Exam twist: mixed series+parallel	Reduce step-by-step, then KVL

Worked examples

Cell A — the baseline series loop

Worked example Battery → resistor → LED → back

A battery of $E = 9 V$ drives a resistor $R = 350 Ω$ in series with an LED that drops $V_{L E D} = 2 V$ . Find the current $I$ .

Forecast: guess — will $I$ be near $0.02 A$ , or wildly bigger? Jot a number before reading on.

Read the figure: follow the blue "I" arrows — current leaves the battery's long ( $+$ ) line at top-left, crosses the orange zigzag resistor, drops through the green LED at the bottom, and returns to the short ( $-$ ) line. That single unbroken path is the loop we will balance.

Mark the source direction. Conventional current leaves the long ( $+$ ) line (top-left of the figure). Why this step? It fixes which way we walk the loop, so every drop has the right sign.
Write KVL around the loop. Starting at $+$ : rise $E$ , drop across $R$ , drop across LED, back to $-$ . $E - I R - V_{L E D} = 0$ Why this step? The loop must balance to zero — the battery's push is fully spent on the two drops.
Solve for $I$ . The resistor absorbs whatever the LED does not. $I = \frac{E - V _{L E D}}{R} = \frac{9 - 2}{350} = 0.02 A = 20 mA$ Why this step? Ohm's law $V_{R} = I R$ turns the leftover voltage into a current.

Verify: $V_{R} = I R = 0.02 \times 350 = 7 V$ . Then $7 + 2 = 9 V = E$ . ✓ Loop closes. Units: $V /Ω = A$ . ✓

Cell B — two resistors in series

Worked example Same current, resistances add

$E = 12 V$ pushes through $R_{1} = 200 Ω$ then $R_{2} = 400 Ω$ (end-to-end on one wire). Find $I$ and the drop across each.

Forecast: which resistor drops more voltage — the big one or the small one?

Read the figure: the two orange zigzags sit end-to-end on the top rail, joined only at the gray "middle node" dot (the junction dot defined above). There is exactly one path, so the blue "I" arrow labels the one current that threads through both.

Recognise series from the drawing. They sit end-to-end on a single wire, sharing only one middle node (the gray dot). Why this step? Series ⇒ same current everywhere (charge can't pile up). This picks the add rule. See Series and Parallel Circuits.
Combine into one equivalent resistor. $R_{e q} = R_{1} + R_{2} = 200 + 400 = 600 Ω$ Why this step? One resistor is easier; KVL on the reduced loop gives the shared current directly.
Find the current. $I = \frac{E}{R _{e q}} = \frac{12}{600} = 0.02 A$
Find each drop with Ohm's law. $V_{1} = I R_{1} = 0.02 \times 200 = 4 V, V_{2} = I R_{2} = 0.02 \times 400 = 8 V$ Why this step? Bigger resistance ⇒ bigger share of the voltage (it's harder to push through).

Verify: $V_{1} + V_{2} = 4 + 8 = 12 V = E$ . ✓ The larger $R_{2}$ took the larger drop, as forecast.

Cell C — two resistors in parallel

Worked example Same voltage, reciprocals add

$E = 12 V$ across $R_{1} = 200 Ω$ and $R_{2} = 300 Ω$ drawn side-by-side between the same two nodes. Find $R_{e q}$ and the total current.

Forecast: will $R_{e q}$ be bigger or smaller than the smaller resistor, $200 Ω$ ?

Read the figure: both orange resistors hang between the same top rail and the same bottom rail (see the gray junction dots on each rail). They share both endpoints, so they feel the identical voltage; the blue "I_total" arrow is the current the battery supplies to both branches together.

Recognise parallel from the drawing. Both resistors touch the same top node and the same bottom node. Why this step? Parallel ⇒ same voltage across each (both hang between the identical two potentials). This picks the reciprocal rule.
Combine with the reciprocal formula. $\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} = \frac{1}{200} + \frac{1}{300} = \frac{3 + 2}{600} = \frac{5}{600}$ $R_{e q} = \frac{600}{5} = 120 Ω$ Why this step? Two paths let current flow more easily ⇒ lower total resistance (smaller than either branch).
Total current. $I = \frac{E}{R _{e q}} = \frac{12}{120} = 0.1 A = 100 mA$

Verify: Branch currents: $I_{1} = 12/200 = 0.06 A$ , $I_{2} = 12/300 = 0.04 A$ . Sum $= 0.06 + 0.04 = 0.1 A = I$ . ✓ And $120 Ω < 200 Ω$ , as forecast.

Cell D — degenerate: the open switch

Cell E — degenerate: $R = 0$ (a short circuit)

Worked example Remove the resistor entirely

The battery $E = 9 V$ connects to the LED through a wire with no resistor ( $R = 0$ ). What does the current formula say, and what really happens?

Forecast: what happens to $I = (E - V_{L E D}) / R$ as $R \to 0$ ?

Push $R$ toward zero in the formula. $I = \frac{E - V _{L E D}}{R} R \to 0 \frac{7}{0} \to \infty$ Why this step? Dividing a fixed leftover voltage by an ever-smaller resistance blows the current up without bound.
Interpret physically. An ideal LED can't limit current itself (its drop stays clamped near $V_{L E D}$ , it isn't a resistor), so the current runs away and the LED burns out. Why this step? This is exactly why the parent note said "the resistor sets $I$ " — it's the safety valve.

Verify: Sanity check the trend with a shrinking $R$ : at $R = 350$ , $I = 0.02$ ; at $R = 35$ , $I = 0.2$ ; at $R = 3.5$ , $I = 2 A$ . Each 10× smaller $R$ makes $I$ 10× bigger. ✓ The limit is genuinely unbounded.

Cell F — sign/orientation: the LED backwards

Worked example Diode pointing the wrong way

Take Cell A's loop but flip the LED so its triangle points against the current ( $- \to +$ direction).

Forecast: does the current reverse (go negative) or stop?

Read the figure: the green LED triangle now points against the direction the battery wants to push current (see the red caption). A diode is a valve that only opens one way, so this backward triangle seals the loop — the red " $I = 0$ " mark states the result.

Recall the diode rule. A diode conducts only when current tries to flow with its triangle arrow (the forward direction we earned above). See Diodes and LEDs. Why this step? Orientation, not just presence, decides whether the loop can carry current.
Current direction vs arrow. The battery wants current in the " $+ \to -$ " sense, but the flipped LED blocks that direction. Why this step? A blocked valve is like an open switch — no complete conducting path.
Conclude. $I \approx 0 A (LED reverse-biased \Rightarrow dark)$ Why this step? The answer is zero, not a negative current. A diode doesn't run backwards; it simply stops.

Verify: With $I = 0$ , resistor drop $= 0$ , so nearly all $9 V$ sits across the reverse LED. KVL: $9 - 0 - 9 = 0$ . ✓ Consistent with a blocked (open-like) branch.

Cell G — crossing wires: dot vs no dot

Worked example One drawing, two circuits

Two wires form a $+$ shape at the centre. Case (i): no dot — they cross over. Case (ii): a junction dot — they join. The vertical wire carries a battery loop; the horizontal wire carries a separate resistor. How many nodes in each case?

Forecast: does adding a single dot change whether current can flow between the two wires?

Read the figure: on the left the blue vertical wire and orange horizontal wire cross with no dot — they are two separate nets (gray caption). On the right a solid black junction dot sits at the crossing, merging them into a single node (green caption). Compare the two centres: that one dot is the entire difference.

Apply the connection rule. Wires are joined only at a junction dot; a bare crossing is electrically separate. Why this step? This is the single most misread detail in all of schematic reading.
Count nodes, case (i) no dot. The vertical loop is one circuit; the horizontal wire is a different, unconnected net. The resistor sees no source ⇒ its current is $0$ . Why this step? Without a shared node, no voltage difference is applied to the horizontal resistor, so nothing drives current through it.
Count nodes, case (ii) with dot. Now the horizontal wire taps into the vertical net at the shared node, so it sits at that net's voltage. The resistor is connected and can carry current if its other end reaches a different potential. Why this step? One dot merged two separate nets into one, opening a real current path where before there was none.

Verify: In (i), the isolated resistor has no closed loop with a source ⇒ $I = 0$ . In (ii), if that resistor bridges two points at $9 V$ and $0 V$ through $R = 450 Ω$ , then $I = 9/450 = 0.02 A$ . ✓ The dot flipped $0 \to 20 mA$ .

Cell H — limiting case: $R \to \infty$

Cell I — real-world word problem

Worked example Pick the resistor for an LED

You have a $5 V$ supply and a red LED that drops $V_{L E D} = 2 V$ and should run at $I = 15 mA = 0.015 A$ . What series resistor $R$ do you need?

Forecast: roughly how many hundred ohms — under $100$ , a few hundred, or thousands?

Write KVL and solve for $R$ . The resistor must drop the leftover voltage $E - V_{L E D}$ at the target current. $R = \frac{E - V _{L E D}}{I} = \frac{5 - 2}{0.015} = \frac{3}{0.015} = 200 Ω$ Why this step? We invert Ohm's law: knowing the desired current and the voltage it must absorb, we get the resistance directly.
Round to a real part. $200 Ω$ is a standard stock value, so no rounding is needed here. Why this step? Real resistors come only in fixed catalogue values; always check the nearest available one so the built current stays near target.

Verify: Plug back: $I = (5 - 2) /200 = 3/200 = 0.015 A = 15 mA$ . ✓ Exactly the target current. Units: $V / A = Ω$ . ✓

Cell J — exam twist: mixed series + parallel

Worked example Reduce, then KVL

$E = 12 V$ . A resistor $R_{1} = 90 Ω$ is in series with a parallel pair $R_{2} = 200 Ω ∥ R_{3} = 300 Ω$ . Find the total current the battery supplies.

Forecast: will the parallel pair count as more or less than $200 Ω$ of resistance?

Read the figure: trace the blue "I" arrow — it leaves the battery, passes the single series resistor $R_{1}$ (top orange zigzag), then reaches the gray junction dot where the current splits into the two vertical resistors $R_{2}$ and $R_{3}$ that share the same two rails. The green caption states the reduction we are about to do: collapse the pair to $120 Ω$ , then add $R_{1}$ .

Collapse the parallel pair first. $R_{2}$ and $R_{3}$ share the same two nodes ⇒ reciprocal rule. $\frac{1}{R _{23}} = \frac{1}{200} + \frac{1}{300} = \frac{3 + 2}{600} = \frac{5}{600} \Rightarrow R_{23} = 120 Ω$ Why this step? Reducing the inner combination first turns the puzzle into a plain series loop we already know how to solve.
Add the series resistor. $R_{e q} = R_{1} + R_{23} = 90 + 120 = 210 Ω$ Why this step? Series ⇒ resistances add; now it is a single-resistor loop.
Total current by KVL. $I = \frac{E}{R _{e q}} = \frac{12}{210} \approx 0.05714 A \approx 57.1 mA$ Why this step? One equivalent resistor + one source ⇒ Ohm's law gives the battery's current in a single division.

Verify: Drop on $R_{1}$ : $0.05714 \times 90 = 5.143 V$ . Remaining across the pair: $12 - 5.143 = 6.857 V$ . Branch currents: $6.857/200 = 0.03429$ and $6.857/300 = 0.02286$ ; sum $= 0.05714 A = I$ . ✓ And the pair ( $120 Ω$ ) is indeed less than $200 Ω$ , as forecast.

Active recall

Recall Test yourself on the tricky cells

Open switch: is the current zero or undefined? ::: Exactly zero — the loop is broken.
A backwards LED gives what current? ::: Zero (it blocks), not a negative current.
As $R \to \infty$ , the current does what? ::: Smoothly approaches zero (like an open switch).
$R = 0$ across an LED causes what? ::: A short circuit — runaway current, LED burns out.
Adding a junction dot at a crossing does what? ::: Merges two nets into one, opening a current path.
To combine parallel resistors you add the reciprocals.

Connections

Ohm's Law — $V = I R$ , used in every example
Kirchhoff's Voltage Law — the loop-balance behind each current
Series and Parallel Circuits — cells B, C, J
Electric Potential and Voltage — why nodes share a voltage
Diodes and LEDs — cell F, one-way conduction
Grounding and Reference Voltages — the 0 V reference in these loops
Parent topic