Preview

DC Circuits — Ohm's Law

Q: In a series circuit with V=12V, R₁=10Ω, R₂=20Ω — what is the total current?

You can work it out this way: find R_eq first (series), then use I = V / R_eq.

Q: Switch to parallel with the same values. How does total current change and why?

Parallel R_eq is less than either resistor — more current flows from the same voltage.

Q: In parallel mode, if R₁=10Ω and R₂=30Ω, what fraction of total current flows through R₁?

Each branch current: I = V/R. Ratio of currents = R₂/R₁ (inverse of resistance).

Series and parallel resistors with live current visualization

Here is a bet you can price in one click. A 6-volt battery drives two 100-ohm resistors wired in series, and the current reads exactly 0.0300 amperes. Now rewire the SAME two resistors side by side — parallel — same battery, same components, nothing added and nothing removed. Most people reason: two resistors are two resistors; however you arrange them, they should impede about the same. Some reason worse: parallel looks more tangled, so maybe it resists MORE. The panel will read 0.1200 amperes — four times the current. Nothing about the resistors changed; what changed is the OFFER you made the charges. Series is one narrow road through both tollbooths: resistances add, 200 ohms total, and every electron pays both tolls. Parallel is two roads open at once: each resistor spans the full 6 volts and draws its own current as if the other weren't there, so the battery sees less resistance than EITHER branch alone — 50 ohms, below the smallest resistor in the circuit, always. The textbook states it flat: more current flows from the source than would flow for any of them individually, so the total resistance is lower. This lab lets you watch the negotiation in three currencies at once — the current row, the pace of the yellow electron stream, and the glow of each resistor pricing the power it burns. By the end you'll read wiring diagrams the way the battery does: not 'how many components' but 'how many paths'.

What you'll be able to do

Price a rewiring with the panel's own numbers: the same 6 V battery and the same two 100 Ω resistors drive 0.0300 A in series (R_total = 200.00 Ω) but 0.1200 A in parallel (R_total = 50.00 Ω) — four times the current, because the second path ADDS a lane, never a roadblock
Read the series loop as one indivisible current: 'Since there is only one path for the charges to flow through, the current is the same through each resistor' — the electron stream keeps one shared pace before, between, and after both resistors, and swapping R₁ with R₂ changes nothing the panel can detect
Predict which resistor runs hottest from the wiring alone: series hands P = I²R to the LARGER resistance (300 Ω outglows 200 Ω on the Simple Series preset), parallel hands P = V²/R to the SMALLER one — the glow flip is the fastest visual test of whether you understand both power identities

Formulas

V = IR

Ohm's Law

R_{series} = R_1 + R_2

Series equivalent resistance — resistors add directly

\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}

Parallel equivalent resistance — reciprocals add

P = IV = I^2R = \frac{V^2}{R}

Power dissipated

Make a prediction

Series baseline: 6 V across 100 Ω + 100 Ω — the current row reads 0.0300 A. Now flip the SAME two resistors to parallel (one button, nothing else changes). What does the total current from the battery do?

No grading here — commit to a guess, then scroll down and test it yourself.

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Run the gate live: from the Series Baseline (0.0300 A, R_total 200.00 Ω), press Parallel and watch three things jump at once — R_total collapses to 50.00 Ω, the current row quadruples to 0.1200 A, and the electron stream visibly speeds up. Flip back and forth until the ×4 stops surprising you.
Audit the series loop for 'weakening' current: in series mode, follow the yellow stream all the way around — one shared pace before R₁, between the resistors, and back to the battery; nothing thins downstream. Then swap the values of R₁ and R₂ and confirm every panel row holds: the single path makes order invisible.
Catch the glow flip: on Simple Series (9 V, 200+300 Ω) the LARGER resistor glows brighter — same I through both, so P = I²R favors big R. Now take Parallel Resistors (6 V, 300‖300 Ω) and drag R₁ down to 100 Ω: the SMALLER resistor takes over the glow, because each branch spans the full 6 V and P = V²/R favors small R. Same boxes, opposite champions.
Break the loop honestly: toggle the Switch OFF and watch the current row hit 0.0000 A and the stream freeze — no closed path, no current, in either wiring. Then tour Heavy Load (9 V, 10+10 Ω): 0.4500 A, the brightest drift in the tour — and note the panel stays finite; a true short circuit is outside this rig's slider range.

Scroll for the debrief ↓

Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: It quadruples to 0.1200 A — parallel opens a second path, and R_total drops to 50 Ω, below either resistor alone

Parallel wiring is a second ROAD, not a second roadblock. Each 100 Ω resistor now spans the full 6 V and draws its own 0.0600 A independently — 'more current flows from the source than would flow for any of them individually, so the total resistance is lower' (OpenStax §10.2, verbatim). The reciprocal sum gives R_total = 50.00 Ω — below the smallest branch, which is the theorem this lab's machine gate certifies — and I = 6/50 = 0.1200 A, exactly four times the series baseline (equal resistors make the ratio a clean (2R)²/R² = 4). Options B and C are the documented classic (McDermott & Shaffer 1992): counting components instead of counting paths. The series number was never about 'two resistors' — it was about ONE path through 200 Ω; break the single-path assumption and the arithmetic flips from addition to reciprocals.

Quiz (0/3)

In a series circuit with V=12V, R₁=10Ω, R₂=20Ω — what is the total current?

Switch to parallel with the same values. How does total current change and why?

In parallel mode, if R₁=10Ω and R₂=30Ω, what fraction of total current flows through R₁?

You can now

Price a rewiring with the panel's own numbers: the same 6 V battery and the same two 100 Ω resistors drive 0.0300 A in series (R_total = 200.00 Ω) but 0.1200 A in parallel (R_total = 50.00 Ω) — four times the current, because the second path ADDS a lane, never a roadblock
Read the series loop as one indivisible current: 'Since there is only one path for the charges to flow through, the current is the same through each resistor' — the electron stream keeps one shared pace before, between, and after both resistors, and swapping R₁ with R₂ changes nothing the panel can detect
Predict which resistor runs hottest from the wiring alone: series hands P = I²R to the LARGER resistance (300 Ω outglows 200 Ω on the Simple Series preset), parallel hands P = V²/R to the SMALLER one — the glow flip is the fastest visual test of whether you understand both power identities