VCE Physics (Year 11) · Unit 1 — Electricity · 11m read

Ohm's Law and Power in Circuits

Ever wondered why some circuits fizzle while others light up brilliantly? Picture water flowing through a pipe: the flow rate is like electric current (how much charge moves), the pressure pushing it is like voltage (the energy per charge), and obstacles blocking the flow are like resistance (anything opposing movement). This water analogy helps you understand how current, voltage, and resistance work together — but it breaks down when you try to explain things like why electricity can flow through a vacuum or how circuits store energy. Mastering these three ideas is essential for VCE physics and any real-world troubleshooting, from phone batteries to household wiring.

Learning objective: Apply Ohm's law and power equations in circuits.

By the end of this lesson you can:

• I use V = IR to solve problems.
• I compute power as P = VI = I²R = V²/R.
• I interpret I-V graphs.

Resistance is a narrow pipe. Narrow pipes slow water; resistors slow current. Power is how fast energy is delivered.

Ohm's law and power

OHM'S LAW:
• V = IR.
• V in volts, I in amperes, R in ohms.

OHMIC vs NON-OHMIC:
• OHMIC (constant R): straight line V-I graph through origin.
• NON-OHMIC (variable R): curved V-I graph (e.g., bulbs, diodes).

RESISTANCE DEPENDENCIES:
• Material (ρ, resistivity).
• Length L.
• Cross-section A.
• R = ρL/A.
• Hotter wire → higher R for metals.

POWER:
• P = VI.
• Also: P = I²R = V²/R.
• Units: watts (W) = J/s.

ENERGY:
• W = Pt = VIt = I²Rt = V²t/R.
• 1 kWh = 3.6 × 10⁶ J.

DISSIPATION:
• All R converts electrical energy to heat.
• Used intentionally in heaters, unintentionally elsewhere.

GRAPHICAL ANALYSIS:
• Gradient of V-I graph = R (if linear).
• For non-ohmic device, R = V/I at a point (not gradient).

TEMPERATURE DEPENDENCE:
• Metals: R increases with T.
• Semiconductors: R decreases with T.
• Superconductors: R = 0 below critical T.

PRACTICAL EXAMPLES:
• Light bulb: when cold, low R; when hot, R increases.
• Why it flashes on startup with high current.

COMPONENT RATINGS:
• A "60 W, 240 V" bulb: calculate operating I = P/V = 0.25 A.
• Calculate R = V/I = 960 Ω.

SERIES RESISTORS: R_total = R₁ + R₂ + …
PARALLEL RESISTORS: 1/R_total = 1/R₁ + 1/R₂ + …

CIRCUIT TROUBLESHOOTING:
• Open circuit → no current.
• Short circuit → huge current (dangerous).
• Fuse blown → protects by breaking circuit.

CAS CHECK:
• Always double-check units before plugging into formulae.

Power trio

• V = IR: Ohm's law.
• P = VI: Power basic.
• P = I²R: When I, R known.

Ohm's law and power

• Ohm's law: V = IR
• Power: P = VI = I^2 R = V^2 / R
• Energy: W = Pt

Quick reference

| Formula | Meaning |
|---|---|
| V = IR | Ohm's law |
| P = VI | Power |
| P = I²R | Power (current) |
| P = V²/R | Power (voltage) |
| E = Pt | Energy |

Worked example (Medium)

A 1500 W heater runs at 240 V. Find I, R, and energy used in 30 minutes.

1. Current. I = P/V = 1500/240 = 6.25 A.
2. Resistance. R = V/I = 240/6.25 = 38.4 Ω.
3. Energy. W = Pt = 1500 × (30 × 60) = 2,700,000 J = 2.7 MJ.

Answer: I = 6.25 A, R = 38.4 Ω, W = 2.7 MJ.

Worked example (Hard)

A 240 V kettle is rated at 2400 W. It takes 3 minutes to boil 1 L of water from 20 °C. Calculate the kettle's efficiency. ($c_{water} = 4180$ J/kg°C)

1. Useful energy. $Q_{useful} = mc\Delta T = 1 \times 4180 \times 80 = 334400$ J.
2. Total energy input. $W_{total} = Pt = 2400 \times 180 = 432000$ J.
3. Efficiency. $\eta = \frac{Q_{useful}}{W_{total}} \times 100 = \frac{334400}{432000} \times 100 \approx 77.4\%$.

Answer: Efficiency ≈ 77.4%. The remaining 22.6% is lost as heat to the kettle body, surroundings, and steam.

Common mistakes

• Forgetting to convert time to seconds before calculating energy Students use minutes directly in $W = Pt$, giving an answer 60× too small Fix: Always convert: 1 min = 60 s. Double-check units: $P$ in watts × $t$ in seconds = energy in joules.
• Using the wrong power formula for the situation Students apply $P = I^2R$ when they only know $V$ and $R$, leading to errors Fix: Choose the formula with the known quantities: $P = VI$, $P = I^2R$, or $P = V^2/R$. All three are equivalent but use the one matching your data.
• Confusing resistance and resistivity Students use $R = \rho L/A$ numbers where simple $R$ values are given Fix: $R$ (ohms) is the total resistance of a component. $\rho$ (Ω·m) is a material property. Use $R = \rho L/A$ only when calculating resistance from dimensions.

Ohm's law, I-V graphs, and non-ohmic devices

Ohm's law ($V = IR$) applies perfectly only to ohmic devices — those with constant resistance. On an I-V graph, an ohmic resistor produces a straight line through the origin, with gradient $1/R$ (or $R$ if axes are swapped).

Non-ohmic devices have curved I-V graphs because their resistance changes with conditions. A filament lamp has low resistance when cold (low current) and high resistance when hot (high current), producing a characteristic S-shaped curve. A diode conducts in one direction but blocks in the other, creating an asymmetric I-V graph.

For non-ohmic devices, resistance at a specific point is calculated as $R = V/I$ at that point — not from the gradient. This distinction trips up many students on VCAA exams.

Power dissipation is a critical concept: all resistors convert electrical energy to heat. This is intentional in heaters and unintentional in transmission lines. The formula $P = I^2R$ explains why power companies transmit electricity at high voltage (low current): for the same power delivered, halving the current reduces resistive losses by a factor of four.

The kilowatt-hour (kWh) is the practical energy unit used on electricity bills: 1 kWh = 1000 W × 3600 s = 3.6 MJ. VCAA frequently asks students to calculate energy costs, combining physics with financial literacy. The key formula is: cost = power (kW) × time (hours) × rate ($/kWh).

Practice questions

1. A 100 Ω resistor has 5 V across it. The current is:

   1. 0.02 A
   2. 0.05 A
   3. 20 A
   4. 500 A

   (1 marks)

2. A household runs a 2 kW heater for 3 hours, 60 W lamp for 5 hours and 800 W washing machine for 45 minutes. Calculate total energy used in kWh, and cost at $0.32/kWh.

   (4 marks)

Summary

• Use V = IR to solve problems.
• Compute power as P = VI = I²R = V²/R.
• Interpret I-V graphs.