Unit 1: How is energy useful to society?

VICPhysicsSyllabus dot point

How are electric circuits analysed using Ohm's law and conservation of energy?

Electric current, voltage and resistance, Ohm's law $V = IR$, series and parallel circuits, electric power $P = VI$, energy in circuits, and household electricity

A focused answer to the VCE Physics Unit 1 key knowledge point on electric circuits. Charge, current, voltage, resistance and Ohm's law $V = IR$; series and parallel resistance combinations; electric power $P = VI = I^2 R = V^2/R$; energy use and household electricity ($E = Pt$, billing in kWh).

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What this dot point is asking

VCAA wants you to apply Ohm's law and energy conservation to series and parallel circuits, calculate electric power, and apply the same to household electricity contexts.

Charge, current, voltage, resistance

Charge qq. Measured in coulombs (C). The elementary charge e=1.6×1019e = 1.6 \times 10^{-19} C.

Current II. Rate of flow of charge: I=q/tI = q / t. Measured in amperes (A). 1 A = 1 C/s.

Voltage / potential difference VV. Energy per unit charge between two points: V=E/qV = E / q. Measured in volts (V). 1 V = 1 J/C.

Resistance RR. Opposition to current flow. Measured in ohms (Ω\Omega).

Ohm's law

For an ohmic conductor, current is proportional to voltage:

V=IRV = I R

Equivalently: I=V/RI = V/R or R=V/IR = V/I.

Non-ohmic devices (diodes, filament bulbs at high current) deviate from Ohm's law but it remains a useful first approximation for many circuit elements.

Series circuits

Resistors in series: same current through each, voltages add.

Rtotal=R1+R2+R3+R_{\text{total}} = R_1 + R_2 + R_3 + \ldots

Vtotal=V1+V2+V3+V_{\text{total}} = V_1 + V_2 + V_3 + \ldots

The same current II flows through each resistor.

Parallel circuits

Resistors in parallel: same voltage across each, currents add.

1Rtotal=1R1+1R2+1R3+\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots

For two resistors: Rp=R1R2/(R1+R2)R_p = R_1 R_2 / (R_1 + R_2).

The same voltage VV is across each resistor.

Mixed circuits

For circuits with both series and parallel sections, simplify step by step:

  1. Identify the parallel combinations and replace with their equivalent.
  2. Identify the series sums.
  3. Continue until one equivalent resistance remains.
  4. Apply Ohm's law to find total current.
  5. Work back to find individual currents and voltages.

Electric power

Power dissipated by a resistor:

P=VI=I2R=V2RP = V I = I^2 R = \frac{V^2}{R}

Three equivalent forms; choose the one with the quantities you know.

For a power source delivering current II at voltage VV, the power delivered is P=VIP = V I. By energy conservation, this equals the total power dissipated in the circuit.

Energy and household electricity

Total energy delivered or dissipated over time tt:

E=Pt=VItE = P t = V I t

Measured in joules (J) or kilowatt-hours (kWh). 1 kWh = 3.6 MJ.

Household electricity is billed in kWh. A 100 W bulb left on for 10 hours uses 1 kWh.

AC vs DC

Household electricity in Australia is alternating current (AC) at 230 V, 50 Hz. The 230 V is RMS (root mean square); the peak is about 325325 V.

DC sources (batteries) have constant voltage. AC sources have voltage varying sinusoidally with time.

For Unit 1, treat household electricity as RMS-equivalent DC for calculations involving energy use.

Safety

  • Wires have low resistance; a short circuit gives very high current and can cause fire.
  • Fuses and circuit breakers limit current to safe levels.
  • Earth wires provide a low-resistance path to ground for fault currents.
  • Residual current devices (RCDs) detect imbalances between live and neutral and trip rapidly.

Worked example: household power

A 1500 W heater is used for 4 hours per day. (a) Energy use per day in kWh. (b) Cost at 25 cents per kWh.

(a) E=Pt=1.5 kW×4 h=6E = P t = 1.5 \text{ kW} \times 4 \text{ h} = 6 kWh per day.

(b) Cost per day: 6×0.25=1.506 \times 0.25 = 1.50 dollars per day.

Worked example: brightness of bulbs in series vs parallel

Two identical 60 W (at 240 V) bulbs are connected: (a) in series across 240 V, (b) in parallel across 240 V. In which case does each bulb dissipate more power?

Each bulb has R=V2/P=2402/60=960ΩR = V^2 / P = 240^2 / 60 = 960 \Omega.

(a) Series. Total R=1920ΩR = 1920 \Omega. Current I=240/1920=0.125I = 240/1920 = 0.125 A. Power per bulb =I2R=0.1252×960=15= I^2 R = 0.125^2 \times 960 = 15 W.

(b) Parallel. Each bulb has 240 V across it. Power per bulb =V2/R=60= V^2/R = 60 W.

In parallel, each bulb is at full design brightness. In series, each gets only quarter the design power.

Common errors

Adding parallel resistances directly. Use 1/R=1/R1+1/R21/R = 1/R_1 + 1/R_2, not R=R1+R2R = R_1 + R_2.

Wrong power formula. Use the form that matches your knowns: P=VIP = VI if you know VV and II; P=I2RP = I^2 R if you know II and RR; P=V2/RP = V^2/R if you know VV and RR.

Confusing voltage and current. Voltage is across components; current is through them. In series, current is the same; in parallel, voltage is the same.

Energy units in calorimetry context. kWh and J both measure energy but differ by a factor of 3.6 million.

In one sentence

Electric circuits obey Ohm's law V=IRV = IR in ohmic conductors, with series resistances adding (R=R1+R2+R = R_1 + R_2 + \ldots) and parallel combinations following 1/R=1/R1+1/R2+1/R = 1/R_1 + 1/R_2 + \ldots; power dissipated by a resistor is P=VI=I2R=V2/RP = VI = I^2 R = V^2/R, and energy is E=PtE = Pt; household electricity in Australia is 230 V AC at 50 Hz and energy is billed in kilowatt-hours.

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

Year 11 SAC4 marksA circuit has a $12$ V battery connected to a $4 \Omega$ resistor in series with a parallel combination of $6 \Omega$ and $12 \Omega$. (a) Find the total resistance. (b) Find the current from the battery. (c) Find the power dissipated by the $4 \Omega$ resistor.
Show worked answer →

(a) Total resistance. Parallel combination: 1/Rp=1/6+1/12=2/12+1/12=3/121/R_p = 1/6 + 1/12 = 2/12 + 1/12 = 3/12, so Rp=4ΩR_p = 4 \Omega.

Series total: Rtotal=4+4=8ΩR_{\text{total}} = 4 + 4 = 8 \Omega.

(b) Current. I=V/R=12/8=1.5I = V / R = 12 / 8 = 1.5 A.

(c) Power. P=I2R=(1.5)2×4=9P = I^2 R = (1.5)^2 \times 4 = 9 W.

Markers reward the parallel combination, the series sum, Ohm's law for current, and the power calculation.

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