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

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 $q$

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

Current $I$

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

Voltage / potential difference $V$

Energy per unit charge between two points: $V = E / q$. Measured in volts (V). 1 V = 1 J/C.

Resistance $R$

Opposition to current flow. Measured in ohms ($\Omega$).

Ohm's law

For an ohmic conductor, current is proportional to voltage:

Equivalently: $I = V/R$ or $R = V/I$.

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.

The same current $I$ flows through each resistor.

Parallel circuits

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

For two resistors: $R_p = R_1 R_2 / (R_1 + R_2)$.

The same voltage $V$ 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:

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

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

Energy and household electricity

Total energy delivered or dissipated over time $t$:

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 $325$ 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) $E = P t = 1.5 \text{ kW} \times 4 \text{ h} = 6$ kWh per day.

(b) Cost per day: $6 \times 0.25 = 1.50$ 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 $R = V^2 / P = 240^2 / 60 = 960 \Omega$.

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

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

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

Examples in context

Example 1. Residential rooftop solar inverter circuit. A typical Melbourne household has a $6.6$ kW rooftop PV array feeding a string inverter that outputs $230$ V AC. On the DC side, $20$ panels in series at $40$ V give a string voltage of $800$ V; at peak production the string current is about $8$ A. Ohm's law applied to the $30$ m DC cable run with $0.3$ $\Omega$ resistance gives a voltage drop $V = IR = 8 \times 0.3 = 2.4$ V, or $0.3\%$ of the string voltage. Energy delivered over a $5$ hour solar day is $W = Pt = 6600 \times 5 = 33$ kWh, displacing approximately $25$ kg of CO$_2$ from grid power.

Example 2. Christmas-light fault on a Bourke Street tree. A series chain of $50$ Christmas LEDs at $3$ V each totals $150$ V across a $230$ V supply with a current-limiting capacitor. If one LED short-circuits, the remaining $49$ now share the same voltage, raising individual stress from $3.00$ V to $3.06$ V, a $2\%$ overvoltage. If two shorts occur, the dropper capacitor reactance is unchanged so current stays the same but $3.13$ V per LED accelerates failure. This series-chain failure mode is why modern displays use parallel banks of three LEDs with bypass diodes, so a single failure does not cascade through the strand.

Try this

Q1. State Ohm's law and define the SI unit of resistance. [2 marks]

• Cue. $V = IR$; $1$ ohm = $1$ volt per ampere.

Q2. A toaster element of resistance $30$ $\Omega$ is connected to a $230$ V supply. Calculate the current drawn and the energy consumed in $4$ minutes. [4 marks]

• Cue. $I = 230/30 = 7.67$ A; $P = VI = 1763$ W; $W = Pt = 1763 \times 240 = 4.23 \times 10^5$ J.

Q3. Refer to the rooftop solar example. (a) Calculate the resistance of a string drawing $8$ A from an $800$ V source under ideal load. (b) Determine the power dissipated in a $0.3$ $\Omega$ cable carrying that current. (c) Explain why DC cabling losses are minimised by raising the string voltage. [2+2+2 marks]

• Cue. (a) $R = 100$ $\Omega$. (b) $P = I^2 R = 19.2$ W. (c) For fixed power, raising $V$ lowers $I$ so $I^2 R$ losses fall as the square of current.