Lesson 1

Voltage and current

Build circuits 1 and 2 from the printed diagrams before reading past the figures. Get every meter reading matching, then come back here.

Diagram 1: five small circuits — a single AA battery with a voltmeter, two AA batteries wired in parallel, two AA batteries wired in series, a battery with one resistor, and a battery with two resistors in series, each with a voltmeter attached. — Diagram 1: Voltage — five circuits, one AA battery each unless noted

Diagram 2: the same resistor circuits from diagram 1, now read with ammeters instead of voltmeters, plus a fourth circuit with two resistors and two ammeters. — Diagram 2: Ammeter — the same circuits, measured a different way

Voltage is measured across, current is measured through

That’s the whole difference between these two diagrams. Same batteries, same resistors, same layout — diagram 1 puts a meter across something, diagram 2 puts a meter in the path of something.

Look at what that difference does to the readings. In diagram 1, two AA batteries in parallel still read 1.50 V — adding a second cell in parallel gives you more current capacity, not more voltage. Wired in series instead, the same two cells read 3.00 V. Voltage sources add when you stack them end to end; they don’t add when you just wire them side by side.

The last two circuits in diagram 1 put one resistor, then two resistors in series, across a battery. The two-resistor circuit reads 1.12 V — less than the battery’s 1.50 V. That’s not a meter error. Getting to the bottom of that number is the point of this lesson.

Same circuit, other meter

Diagram 2 rebuilds the one-resistor and two-resistor circuits from diagram 1, but swaps the voltmeter for an ammeter. The one-resistor circuit (300 Ω) reads 4.99 mA. The two-resistor circuit (100 Ω + 200 Ω) also reads 4.99 mA on both ammeters — one on either side of the two resistors.

Sit with that for a second: the current is the same everywhere in that loop, on both sides of both resistors. Current doesn’t get “used up” as it flows past a resistor — voltage does.

OHM’S LAW

Three quantities describe every circuit here: voltage (V, in volts), current (I, in amps), and resistance (R, in ohms). They’re related by

V = I × R

Check it against the numbers already on the page. The one-resistor circuit in diagram 2 has a 300 Ω resistor and reads 4.99 mA (0.00499 A) across a 1.50 V battery:

1.50 V ÷ 300 Ω = 0.005 A = 5 mA

That’s the meter’s 4.99 mA, off only by simulator rounding. Ohm’s law didn’t just describe the circuit after the fact — it predicted the photograph.

Now use it on the two-resistor circuit. Both resistors carry the same 4.99 mA (you measured that above). What voltage should each one drop?

V = I × R = 0.00499 A × 100 Ω ≈ 0.499 V

V = I × R = 0.00499 A × 200 Ω ≈ 0.998 V

Diagram 2’s voltmeters read 499 mV and 998 mV — matching. And 0.499 V + 0.998 V ≈ 1.50 V, the full battery voltage, split between the two resistors in proportion to their resistance. That’s the 1.12 V from diagram 1 explained: a two-resistor divider never gives you back the full supply voltage on either resistor alone.

Questions

Two batteries in parallel and two batteries in series both use two AA cells. Why doesn’t the parallel pair read 3.00 V too?
The two-resistor circuit had the same current through both resistors but different voltages across them. Could a circuit ever have the same voltage across two different resistors instead? What would that take?