Learn Index · Waves, Electricity and Magnetism Grade 9–10

Electric circuits and Ohm’s law

Ohm’s law says voltage equals current times resistance, V = I × R.

Read

Charge and current

Electric current is the rate of flow of charge:

I = Q ÷ t

  • I = current (A)
  • Q = charge (coulombs, C)
  • t = time (s)

One electron carries 1.6 × 10⁻¹⁹ C, so 1 A means about 6.24 × 10¹⁸ electrons pass a point every second.

Voltage is the energy per unit charge: V = E ÷ Q. A 9 V battery gives each coulomb 9 J of energy.

Ohm’s law

Ohm’s law, named for Georg Ohm (1827):

V = IR

A conductor that obeys Ohm’s law has a straight-line V–I graph, and its slope is R. These are ohmic conductors. Bulbs and diodes are non-ohmic: a bulb’s filament resistance rises as it heats.

What affects resistance

R = ρL ÷ A

  • ρ (rho) = resistivity of the material (Ω·m)
  • L = length
  • A = cross-sectional area
Material Resistivity at 20 °C (Ω·m)
Silver 1.6 × 10⁻⁸
Copper 1.7 × 10⁻⁸
Aluminum 2.7 × 10⁻⁸
Nichrome 1.1 × 10⁻⁶
Glass about 10¹⁰ – 10¹⁴

Copper is used for wiring because it combines low resistivity with reasonable cost.

Series circuits

Quantity Rule
Current Same through all components
Voltage V_T = V₁ + V₂ + V₃
Equivalent resistance R_T = R₁ + R₂ + R₃

Parallel circuits

Quantity Rule
Voltage Same across all branches
Current I_T = I₁ + I₂ + I₃
Equivalent resistance 1/R_T = 1/R₁ + 1/R₂ + 1/R₃

For two resistors: R_T = (R₁R₂) ÷ (R₁ + R₂)

Kirchhoff’s laws

  • Kirchhoff’s current law (junction rule): the current entering a junction equals the current leaving it. This follows from conservation of charge.
  • Kirchhoff’s voltage law (loop rule): around any closed loop, the voltage rises (from sources) equal the voltage drops (across resistors). This follows from conservation of energy.

Solving a combination circuit

A 12 V battery is connected to R₁ = 4 Ω in series with a parallel pair R₂ = 6 Ω and R₃ = 3 Ω.

  1. Parallel part: R₂₃ = (6 × 3) ÷ (6 + 3) = 2 Ω
  2. Total: R_T = 4 + 2 = 6 Ω
  3. Total current: I_T = 12 ÷ 6 = 2 A
  4. Voltage across R₁: V₁ = 2 × 4 = 8 V
  5. Voltage across the parallel pair: 12 − 8 = 4 V
  6. I₂ = 4 ÷ 6 = 0.67 A; I₃ = 4 ÷ 3 = 1.33 A
  7. Check: 0.67 + 1.33 = 2 A ✓

Electrical power and energy

Electrical power:

P = VI = I²R = V² ÷ R

E = Pt

Example: A 60 W bulb on 120 V draws I = 60 ÷ 120 = 0.5 A and has R = 120² ÷ 60 = 240 Ω.

Energy cost: a 1.5 kW heater running 4 hours uses 6 kWh. At 12¢/kWh, that’s 72¢.

Power lines carry electricity at very high voltages (hundreds of kilovolts) to reduce current. Since power lost as heat is I²R, lower current means much less waste.

Internal resistance and EMF

Real batteries have internal resistance (r). The full voltage of the source is its EMF (ε). Under load:

V_terminal = ε − Ir

The more current drawn, the more the terminal voltage drops. That’s why a car’s headlights dim when the starter motor runs.

Alternating and direct current

DC (direct current) AC (alternating current)
Direction One way Reverses many times a second
Source Batteries, solar cells Generators, wall outlets
North American mains 120 V at 60 Hz

The "War of the Currents" in the 1880s–1890s pitted Thomas Edison (DC) against George Westinghouse and Nikola Tesla (AC). AC won for power grids because transformers can easily change its voltage.

Common mistakes

  • Adding parallel resistances directly.
  • Assuming V = IR applies to non-ohmic devices at all currents.
  • Forgetting that current is the same everywhere in a series circuit.

Try it yourself

  1. A 220 Ω resistor carries 0.05 A. Find the voltage and power.
  2. Find the equivalent resistance of 10 Ω, 20 Ω and 20 Ω in parallel.
  3. A 12 V battery with 0.5 Ω internal resistance powers a 5.5 Ω load. Find the current and terminal voltage.
  4. How many kWh does a 900 W microwave use in 20 minutes?

(Answers: 11 V and 0.55 W; 5 Ω; 2 A and 11 V; 0.3 kWh)

Quick recap

  • I = Q/t; V = E/Q; V = IR.
  • Resistance depends on material, length, area and temperature.
  • Series resistances add; parallel combinations reduce resistance.
  • Kirchhoff’s laws come from conservation of charge and energy.
  • P = VI = I²R = V²/R; high-voltage transmission reduces losses.