Year 8 Science | Victorian Curriculum 2.0
Electrical circuits - voltage & current
Topic 09 | Physical sciences | Answer key

Year 8 answers

Fluency

Definitions and units

    1. (a) Ampere (A), (b) volt (V), (c) ohm (Ω\Omega).
    2. (a) Ammeter, (b) voltmeter.
    3. Ammeter: in series (in line with the component, so current flows through it). Voltmeter: in parallel (across the component).
    4. A cell is one unit; a battery is two or more cells joined together.
    5. Any two of: bulb (light + heat), resistor (heat), motor (movement), buzzer/speaker (sound), heater element (heat).
Fluency

Ohm's law

    1. I=12/3=4I = 12/3 = 4 A.
    2. V=0.5×20=10V = 0.5 \times 20 = 10 V.
    3. R=9/0.3=30R = 9/0.3 = 30 Ω\Omega.
    4. R=240/10=24R = 240/10 = 24 Ω\Omega.
    5. R=1.5/0.1=15R = 1.5/0.1 = 15 Ω\Omega.
Fluency

Series and parallel

    1. 9/3=39/3 = 3 V across each.
    2. 1212 V across each branch.
    3. The circuit is broken; all other bulbs go out.
    4. The others keep working (each branch is still complete).
    5. Appliances need the same full voltage, and if one device turns off or breaks, others must keep working.
Reasoning

Explain

    1. High resistance means almost no current flows through the voltmeter itself, so it measures the voltage across the component without changing the circuit.
    2. Ammeters have near-zero resistance. Connecting directly across a battery gives a huge current (short circuit) that can overheat wires and damage the ammeter.
    3. In series, each bulb only gets a share of the voltage, so less current flows through both bulbs and each is dimmer. In parallel each bulb still sees the full voltage, so it glows at full brightness.
    4. A simple loop: cell (positive terminal) →\to switch →\to ammeter →\to bulb →\to back to negative terminal. All components in a single series loop.
Problem solving

Applied contexts

    1. R=V/I=5/2=2.5R = V/I = 5/2 = 2.5 Ω\Omega.
    2. R=240/0.25=960R = 240/0.25 = 960 Ω\Omega. Power P=VI=240×0.25=60P = VI = 240 \times 0.25 = 60 W.
    3. (a) Each bulb: I=12/4=3I = 12/4 = 3 A. (b) Total: 3+3=63 + 3 = 6 A.
    4. The Christmas lights are on their own circuit; the smoke alarm is on a different parallel branch (usually battery-powered). A fault in one parallel branch does not affect another.
Reasoning

Challenge

    1. (a) Rtotal=2+3+4=9R_{\text{total}} = 2 + 3 + 4 = 9 Ω\Omega. (b) I=9/9=1I = 9/9 = 1 A. (c) V4=IR=1×4=4V_4 = IR = 1 \times 4 = 4 V.
    2. R=12/5=2.4R = 12/5 = 2.4 Ω\Omega per headlight. In parallel, each still draws 55 A, so total current =10= 10 A.
    3. A short circuit has near-zero resistance. By I=V/RI = V/R, a tiny resistance gives a huge current. The wire then heats rapidly (P=VIP = VI is large), can melt insulation, and start a fire.
    4. Put bulbs 1 and 2 in parallel with each other, controlled together by a first switch. Put bulb 3 on its own branch (also parallel to the rest) controlled by a second switch. Because every branch is in parallel with the battery, each branch operates independently at full voltage.
Year 8 Science study companion | Answer key