Key Concepts: In series: current is the same throughout; voltage is shared. In parallel: voltage is the same across each branch; current splits. Series resistance: $R_{total} = R_1 + R_2$. Parallel resistance: $1/R_{total} = 1/R_1 + 1/R_2$.
Section A — Circuit Concepts
1. State the symbols used in circuit diagrams for: a cell, a battery, a lamp, a resistor, an ammeter, and a voltmeter. [3]
2. Describe how current and voltage behave in a series circuit. [3]
3. Describe how current and voltage behave in a parallel circuit. [3]
4. State one advantage of using a parallel circuit for domestic lighting compared to series. [2]
Section B — Series Circuit Calculations
5. Two resistors of $3\,\Omega$ and $5\,\Omega$ are connected in series to a 12 V supply. [5]
a) Calculate the total resistance.
b) Calculate the current through the circuit.
c) Calculate the voltage across each resistor.
6. Three resistors of $4\,\Omega$, $6\,\Omega$ and $2\,\Omega$ are in series with a 24 V supply. Calculate the current and the voltage across the $6\,\Omega$ resistor. [3]
Section C — Parallel Circuit Calculations
7. Two resistors of $6\,\Omega$ and $12\,\Omega$ are connected in parallel to a 12 V supply. [5]
a) Calculate the total resistance.
b) Calculate the total current from the supply.
c) Calculate the current through each branch.
8. Explain why adding more resistors in parallel decreases the total resistance of the circuit. [2]
Total marks: 26
Mark Scheme
1. Correct symbols for: cell (long and short line), battery (multiple cells), lamp (circle with cross), resistor (rectangle), ammeter (circle with A), voltmeter (circle with V) [3 — 0.5 mark each, award in pairs]
2. Current is the same at every point in the series circuit [1]; voltage is shared/divided between components [1]; voltages add up to equal the supply voltage [1] [3]
3. Voltage across each branch is the same as the supply voltage [1]; current splits at junctions [1]; currents in branches add up to equal the total current [1] [3]
4. Components operate independently [1]; if one lamp fails the others remain on [1] [2]
5. a) $R_{total} = 3 + 5 = 8\,\Omega$ [1]; b) $I = V/R = 12/8 = 1.5\,\text{A}$ [2]; c) $V_{3\Omega} = 1.5 \times 3 = 4.5\,\text{V}$; $V_{5\Omega} = 1.5 \times 5 = 7.5\,\text{V}$ [2] [5]
6. $R_{total} = 4 + 6 + 2 = 12\,\Omega$ [1]; $I = 24/12 = 2\,\text{A}$ [1]; $V_{6\Omega} = 2 \times 6 = 12\,\text{V}$ [1] [3]
7. a) $1/R_{total} = 1/6 + 1/12 = 2/12 + 1/12 = 3/12$; $R_{total} = 4\,\Omega$ [2]; b) $I_{total} = 12/4 = 3\,\text{A}$ [1]; c) $I_{6\Omega} = 12/6 = 2\,\text{A}$; $I_{12\Omega} = 12/12 = 1\,\text{A}$ [2] [5]
8. Each additional parallel branch provides an extra path for current [1]; more paths means total resistance decreases (more current flows for the same voltage) [1] [2]