IP Combined Science Notes (Lower Sec, Year 1-2): 10) Electricity & Magnetism Essentials
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Electrical and magnetic phenomena underpin everything from school lab circuits to train systems. Treat circuit diagrams and field lines with the same rigour as algebraic manipulation.
Learning targets
- State charge conventions and apply Ohm's law \( V = IR \).
- Compare series and parallel circuits in terms of current, voltage, and resistance.
- Calculate electrical power and energy consumption using \( P = VI \) and \( E = Pt \).
- Describe magnetic field patterns around magnets and current-carrying conductors; explain electromagnet applications.
1. Current, voltage, and resistance
| Quantity | Definition | Unit |
| Current \( (I) \) | Rate of flow of charge, \( I = \frac{Q}{t} \). | \( \pu{A} \) |
| Voltage \( (V) \) | Energy transferred per unit charge. | \( \pu{V} \) |
| Resistance \( (R) \) | Opposition to current flow, \( R = \frac{V}{I} \). | \( \pu{\Omega} \) |
Worked example — Ohm's law
A resistor draws \( \pu{0.35 A} \) when \( \pu{5.6 V} \) is applied.
\[ R = \frac{5.6}{0.35} = 16 , \pu{\Omega}. \]
If the voltage doubles and resistor obeys Ohm's law, current doubles to \( \pu{0.70 A} \).
2. Series vs parallel
| Property | Series | Parallel |
| Current | Same through each component. | Splits; sum of branch currents equals total. |
| Voltage | Sum of component voltages equals supply. | Voltage across each branch equals supply. |
| Resistance | Add component resistances in a single loop. | Sum reciprocals of branch resistances, then invert. |
Worked example — Parallel combination
Two resistors \( \pu{120 \Omega} \) and \( \pu{180 \Omega} \) in parallel.
Series rule: add each resistor value (e.g. R_total = R1 + R2 + R3 + R4).
Parallel rule: add reciprocals and invert (e.g. 1/R_total = 1/R1 + 1/R2 + 1/R3 + 1/R4).
1/R_total = 1/120 + 1/180 = 0.00833 + 0.00556 = 0.0139.
R_total = 71.9 \( \pu{\Omega} \).
If supply voltage \( \pu{9.0 V} \):
- Branch currents: \( I_1 = \frac{9.0}{120} = 0.075 , \pu{A} \); \( I_2 = \frac{9.0}{180} = 0.050 , \pu{A} \).
- Total current \( = 0.125 , \pu{A} \), consistent with \( I = \frac{9.0}{71.9} \).
3. Electrical power & safety
- \( P = VI = I^2 R = \frac{V^2}{R} \).
- Household energy usage: \( E = Pt \) (convert to \( \pu{kWh} \)).
- Fuse rating slightly above normal operating current; earth wire provides low-resistance path for fault current.
Worked example — Energy cost
A \( \pu{1.2 kW} \) heater runs for \( \pu{3.5 h} \).
\[ E = 1.2 \times 3.5 = 4.2 , \pu{kWh}. \]
If cost is \( $0.26 \) per \( \pu{kWh} \), total cost \( = $1.09 \) (rounded to 2 d.p.).
4. Magnetism & electromagnetism
- Magnetic field lines emerge from north pole, enter south pole.
- Around straight current-carrying conductor: concentric circles; direction given by right-hand grip rule (thumb points conventional current, fingers curl in field direction).
- Solenoids produce uniform fields similar to bar magnets; adding iron core strengthens field.
Applications
- Electromagnets in relays: small current energises coil, closing switch in external circuit.
- Electric bells: current magnetises core, attracting armature, breaking circuit to produce oscillations.
- DC motors: force on current-carrying coil in magnetic field causes rotation (Fleming's left-hand rule).
Try it yourself
- Sketch current and voltage distributions for three identical bulbs connected (a) in series, (b) in parallel to the same battery. Predict relative brightness.
- A \( \pu{240 V} \) appliance draws \( \pu{5.0 A} \). Determine required power rating and suggest a suitable fuse.
- Describe how reversing current direction affects the magnetic field around a solenoid. Propose one lab method to demonstrate the reversal.
Return to the overview at https://eclatinstitute.sg/blog/ip-combined-sciences-lower-sec-notes/IP-Combined-Science-Lower-Sec-00-Overview when you revise the full sequence.
