Ohm's Law: Current, Voltage, and Resistance

1 · Predict

A 9 V battery drives current through a single resistor. If you use a bigger resistor, does more or less current flow?

2 · Set Up

  1. Open the single-resistor preset and press Reset. The battery supplies a fixed 9 V.
  2. Enable the current readout on the resistor.
  3. Set the resistor's value for each trial and record the current.

3 · Collect Data

Resistance R (Ω)Power dissipated P (W)Current I (A)
50
100
200

Plot current I (y-axis) against 1/R (x-axis) for your three trials. Is the line straight through the origin?

4 · Analyze

  1. For one trial, compute I = E/R using E = 9 V. Compare to the table.
  2. Explain why doubling the resistance halves the current, using Ohm's law I = E/R at fixed voltage.

5 · Extend

  1. Your power column (P = E²/R) shrinks as resistance grows, even though voltage stays fixed. Explain why a bigger resistor draws less power from the same battery.
  2. What would happen to the current (and the battery) if R were set to nearly 0 Ω? Why do real circuits avoid this 'short circuit' condition?

The Physics Behind This Experiment

Ohm's Law

The current through a resistor is proportional to the voltage across it and inversely proportional to its resistance: I = V/R. Here V equals the full battery EMF, 9 V.

Electrical Power (from V and R)

Power dissipated by a resistor can be computed directly from voltage and resistance without needing current: P = V²/R.

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