Step 1 · Core path

Week 1 circuit accounting.

See what changes, predict before motion, and use meters to account for every volt and amp. The path moves from one complete loop to series, parallel, and safe component ratings.

How this path works

Predict, run, compare, explain

Each mission isolates one relationship. Commit what you expect, run the circuit, compare the readings, then reuse the idea in a changed context.

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Complete loop

A return path decides whether current exists

Start with the physical condition every later equation assumes: charge needs an unbroken path out of the source and back again.

Mission experiment

Interactive

Predict, then close the switch

A 9 V source drives a lamp represented here by a 330 Ω teaching load. Predict both switch states before the circuit moves.

Switch open Lamp dark; ammeter reads 0.0 mA Lamp lit; ammeter reads about 27.3 mA Lamp dark, but current still moves through the connected wire

After you close it Lamp lit; ammeter reads about 27.3 mA Lamp stays dark; ammeter remains at 0.0 mA Lamp lights from voltage, but the current remains zero

Why? Sustained current needs an unbroken path out of the source, through the load, and back. The battery sends current down one wire; the return wire is optional. Voltage flows into the lamp and gets used up before it reaches the gap.

JavaScript unlocks the switch. The state table and worked answer remain available below.

State table and worked answer

One loop with a 9 V source and 330 Ω teaching load
Switch	Lamp	Current	Voltage across gap
Open	Dark	0.0 mA	9.0 V
Closed	Lit	27.3 mA	About 0 V

Open does not mean “no voltage.” It means no complete path, so current is zero. Closing the gap completes the path and gives I = 9 V / 330 Ω = 27.3 mA.

Control current

Change one variable at a time

First calculate one fixed point. Then sweep resistance and voltage separately so the inverse and proportional relationships cannot blur together.

Quick check

Prediction

Ohm's law: source and load set current together

A 12 V battery drives a 6 Ω resistor in a complete loop. Predict the current.

Predict the current, then reveal

I = V / R = 12 V / 6 Ω = 2 A

The source voltage and resistor value together set the current. The resistor does not consume current; it converts electrical energy into heat.

Key idea: the supply does not choose a fixed current; current changes when either voltage or resistance changes.

Practice lab · Resistance sweep

Interactive

Hold voltage steady. Sweep resistance.

Keep the source at 9 V. Commit both endpoint currents, then move from 330 Ω to 1000 Ω and watch current respond.

Predict both endpoints before moving the resistor

At 9 V and 330 Ω

At 9 V and 1000 Ω

Enter milliamps. Both values and one reason are required before the sweep unlocks.

What relationship should the sweep show? With voltage fixed, I = V/R, so more resistance means less current. The 9 V battery carries one fixed current, so changing the resistor barely matters. A larger resistor pulls more current because both numbers increase together.

JavaScript unlocks the slider. Static values and the worked answer remain available below.

Static values and worked answer

Ohm's law values with voltage fixed at 9.0 V
Voltage	Resistance	Current
9.0 V	330 Ω	27.3 mA
9.0 V	470 Ω	19.1 mA
9.0 V	680 Ω	13.2 mA
9.0 V	1000 Ω	9.0 mA

330 Ω: 9/330 = 27.3 mA. 1000 Ω: 9/1000 = 9.0 mA.

With voltage fixed, current is inversely proportional to resistance. Roughly triple resistance and current falls to roughly one third.

Practice lab · Voltage sweep

Interactive

Hold resistance steady. Double voltage.

Keep resistance at 330 Ω. Commit the current at 4.5 V and 9 V, then sweep the source and test whether equal voltage steps make equal current steps.

JavaScript unlocks the source slider. Static values and the worked answer remain available below.

Static values and worked answer

Ohm's law values with resistance fixed at 330 Ω
Voltage	Resistance	Current
4.5 V	330 Ω	13.6 mA
6.0 V	330 Ω	18.2 mA
7.5 V	330 Ω	22.7 mA
9.0 V	330 Ω	27.3 mA

4.5/330 = 13.6 mA. 9/330 = 27.3 mA.

With resistance fixed, current is proportional to voltage. Doubling voltage doubles current, so the V–I graph is a straight line through the origin.

Share in series

One current, several voltage drops

A series loop has only one path. The current is the same everywhere, while each resistor takes a share of the source voltage.

Quick check

Prediction

KVL: the loop ledger must balance

A 12 V source feeds two series resistors: 4 Ω followed by 8 Ω. Predict the voltage across the 8 Ω resistor.

Predict the 8 Ω drop, then reveal

I = 12 / (4 + 8) = 1 A, so V8 = I x 8 = 8 V.

The 4 Ω resistor drops 4 V and the 8 Ω resistor drops 8 V. Around the loop, +12 V - 4 V - 8 V = 0.

Key idea: in series, the same current flows through both resistors; the larger resistor drops more voltage.

Series equivalent

Prediction

Series resistors add to one equivalent

A 4 Ω and an 8 Ω resistor sit in series. Predict the single equivalent resistance.

Predict the equivalent resistance, then reveal

R = 4 Ω + 8 Ω = 12 Ω

The same current must pass through both elements, so their opposition and voltage drops add.

Transfer: replace the 8 Ω resistor with 12 Ω. Predict the new current from the same 12 V source.

Split in parallel

One voltage, several branch currents

Parallel branches share both end nodes, so each receives the full voltage. Their currents split by resistance and add again at the source.

Quick check

Prediction

KCL: branch currents split and rejoin

A 12 V source is applied across three parallel resistors: 12 Ω, 6 Ω, and 4 Ω. Predict the total source current.

Predict the total source current, then reveal

Each branch sees 12 V: 1 A + 2 A + 3 A = 6 A.

KCL says current entering the node equals current leaving it.

Key idea: parallel branches share voltage; current divides by branch resistance and adds back at the node.

Parallel equivalent

Prediction

Parallel resistance drops below either branch

Two 6 Ω resistors sit in parallel. Predict the single equivalent resistance.

Predict the equivalent resistance, then reveal

1/R = 1/6 + 1/6 → R = 3 Ω

A second path makes it easier for total current to flow, so equivalent resistance falls below the smallest branch.

Transfer: open one of the two branches. What happens to equivalent resistance and source current?

Design safely

Current and voltage together become heat

Reuse Ohm's law to calculate power, then choose a component rating above the load rather than exactly equal to it.

Safety check

Prediction

Power: ratings decide whether parts survive

A 12 V source is connected across a 100 Ω resistor. Common parts are rated 0.25 W, 0.5 W, 1 W, and 2 W. Predict the smallest safe resistor rating.

Predict the resistor rating, then reveal

I = 12 / 100 = 0.12 A. P = V x I = 1.44 W.

A 1 W resistor is below the heat load, and an exact 1.44 W rating has no margin. Choose 2 W from the listed values.

Transfer: halve the source to 6 V. Compute the new power and decide whether a 0.5 W resistor survives.

Optional extension: sources in series, parallel, and under load

These checks extend the resistive model without blocking the six-mission beginner spine.

Source extension

Optional

Batteries in series: voltages stack

Two 12 V batteries are connected in series, positive to negative. Predict the total voltage they supply.

Predict the total voltage, then reveal

V = 12 V + 12 V = 24 V

Sources in series stack their voltage rises.

Source extension

Optional

Batteries in parallel: voltage holds

Two identical 12 V batteries are connected in parallel, positive to positive. Predict the voltage across the load.

Predict the load voltage, then reveal

V = 12 V

Identical sources in parallel keep the same voltage and share load current. Only matched sources should be paralleled.

Source extension

Optional

Real battery: terminal voltage sags

A 12 V battery has 1 Ω of internal resistance and drives a 5 Ω load. Predict the terminal voltage across the load.

Predict the terminal voltage, then reveal

I = 12 / (1 + 5) = 2 A; Vterminal = 12 - 2 x 1 = 10 V

A real source loses some voltage across its internal resistance under load.

Later path · stored energy

Capacitors need a model before a combination rule

Capacitor series and parallel drills are deliberately deferred until charge storage, voltage change over time, and RC behavior have been made visible. They no longer interrupt this resistive-DC beginner sequence.

Prove the model

Account for one safe load

Bring the relationships together without a new rule: find current, power, and a safe rating from one source and one load.

Exit proof

Calculate before revealing

A 12 V source feeds a 100 Ω resistor. Predict current, power, and the minimum listed rating before opening the answer.

Reveal the exit proof

I = 12 / 100 = 0.12 A. P = 12 x 0.12 = 1.44 W. Minimum listed safe rating = 2 W.

The part survives because its rating is above the heat load, not because the voltage is familiar.

Continue the core path

When these DC accounting habits are stable, move to sine waves, RMS, impedance, and phasors.

Continue to phasors

Optional branch: control theory once a project needs feedback.

Companion material

Read deeper

The chapter keeps longer visual examples, mixed-circuit work, and relay/contact control notes.

Open chapter