Current, voltage and resistance
Current (I) is how much charge passes a point each second. I = Q ÷ t. Unit: ampere (A); 1 A = 1 coulomb per second.
Voltage (V), also called potential difference, is the push on the charges. It is the energy given to each coulomb. V = W ÷ Q. Unit: volt (V); 1 V = 1 joule per coulomb.
Resistance (R) tells how much a part opposes the current. R = V ÷ I. Unit: ohm (Ω). This is Ohm's law, V = I × R. See the full lesson on Ohm's law for the experiment and graph.
A circuit is a closed path. A source (battery) gives the push, wires carry the charge, and the parts (resistor, capacitor, coil) use the energy. If the path is broken, the current stops.
Electrical resistance
A wire's resistance depends on four things: R = ρ × L ÷ A. L is the length (longer, more R). A is the cross-section area (thicker, less R). ρ (rho) is the resistivity of the material: copper has a small ρ, so wires are made of copper; nichrome has a large ρ, so heaters use it. Temperature also matters: in metals, hotter means more resistance.
Series (one after another): R = R1 + R2 + ... Parallel (side by side): 1/R = 1/R1 + 1/R2 + ...
A resistor turns electric energy into heat: power P = V × I = I² × R. This is how a heater, an iron and a bulb filament work.
Capacitance and electrostatics
Electrostatics is about charges at rest. There are two kinds: positive and negative. Like charges push each other away; unlike charges pull towards each other. Coulomb's law: F = k × q1 × q2 ÷ r², with k = 9 × 10⁹ N·m²/C². The force is smaller when the charges are farther apart.
A capacitor is two metal plates with an insulator (a gap) between them. When a battery is connected, positive charge collects on one plate and negative on the other. The capacitor stores charge and electric energy. Capacitance C = Q ÷ V. Unit: farad (F). Common sizes are µF (10⁻⁶ F), nF and pF.
A bigger plate area and a smaller gap give a bigger capacitance. Energy stored: E = ½ C V². In a DC circuit, a capacitor lets current flow only until it is full; then it blocks DC. That is why it is used to smooth the output of a power supply and to store a charge for a camera flash.
Parallel capacitors add: C = C1 + C2. In series: 1/C = 1/C1 + 1/C2.
Inductance and magnetic phenomena
A current makes a magnetic field around the wire. Wind the wire into a coil and the field becomes strong, like a bar magnet. This is an electromagnet.
If the field through a coil changes, a voltage is made in it. This is electromagnetic induction. When the current in a coil changes, its own field changes, and the coil makes a voltage that opposes the change (Lenz's law). So current in a coil cannot jump up or drop at once; it builds slowly.
Inductance L measures this. V = L × (change in I ÷ change in time). Unit: henry (H). A coil with more turns, and an iron core, has a bigger L. Energy stored in the magnetic field: E = ½ L I².
Uses: transformers, motors, relays, loudspeakers, and filters. In DC, after a short time the current becomes steady and the coil acts like a plain wire.
Try it
At home (safe, no mains): wrap about 20 turns of wire around a nail and connect it to a 1.5 V cell for a few seconds only. The nail picks up a few pins. You made an electromagnet.
In the 3D: pick Capacitor and press "Switch on". Predict: will the current stay or die away? Pick Coil and press it again. Is the current instant or slow? Compare the two.
Key formulas and definitions
- I = Q ÷ t V = W ÷ Q R = V ÷ I
- R = ρ L ÷ A; series R = R1 + R2; parallel 1/R = 1/R1 + 1/R2
- P = V I = I² R
- Coulomb: F = k q1 q2 ÷ r² (k = 9 × 10⁹ N·m²/C²)
- C = Q ÷ V; E = ½ C V²; parallel C = C1 + C2; series 1/C = 1/C1 + 1/C2
- Coil: V = L ΔI ÷ Δt; E = ½ L I²
Worked examples
1. A charge of 120 C passes a point in 60 s. Find the current.
I = Q ÷ t = 120 ÷ 60 = 2 A.
2. A copper wire is 100 m long and has area 1 mm² (1 × 10⁻⁶ m²). Resistivity of copper is 1.7 × 10⁻⁸ Ω·m. Find its resistance.
R = ρ L ÷ A = 1.7 × 10⁻⁸ × 100 ÷ 1 × 10⁻⁶ = 1.7 Ω.
3. A 100 µF capacitor is charged to 12 V. Find the charge and the energy stored.
Q = C V = 100 × 10⁻⁶ × 12 = 1.2 × 10⁻³ C = 1.2 mC. E = ½ C V² = 0.5 × 100 × 10⁻⁶ × 144 = 7.2 × 10⁻³ J = 7.2 mJ.
4. Capacitors of 6 µF and 3 µF are joined (a) in series, (b) in parallel. Find the total capacitance.
(a) 1/C = 1/6 + 1/3 = 1/2, so C = 2 µF. (b) C = 6 + 3 = 9 µF.
5. Two charges of 1 µC each are 0.1 m apart in air. Find the force between them.
F = k q1 q2 ÷ r² = 9 × 10⁹ × 10⁻⁶ × 10⁻⁶ ÷ 0.01 = 0.9 N. Like charges, so it is a push apart.
6. The current in a 2 H coil rises from 0 to 3 A in 0.5 s. Find the average voltage across the coil and the energy stored at 3 A.
V = L ΔI ÷ Δt = 2 × 3 ÷ 0.5 = 12 V. E = ½ L I² = 0.5 × 2 × 9 = 9 J.
Common mistakes
- Thinking a capacitor passes DC for ever. In DC it passes current only while charging; then the current is zero.
- Forgetting that capacitors in series add like resistors in parallel (1/C = 1/C1 + 1/C2), and in parallel they add directly.
- Using micro (µ) and milli (m) wrongly: 1 µF = 10⁻⁶ F, not 10⁻³ F.
- Thinking a coil stops current. A coil only slows changes in current; steady DC passes through it easily.