GRADE 11 FINAL PHYSICS PAPER

ELECTRIC CIRCUITS

1. Basic Concepts

Term

Definition

Unit

Current (I)

Rate of flow of charge

Ampere (A)

Potential difference (V)

Work done per unit charge

Volt (V)

Resistance (R)

Opposition to current flow

Ohm (Ω)

Charge (Q)

Quantity of electricity

Coulomb (C)

EMF (ε)

Energy supplied per unit charge by source

Volt (V)

Key formulas:

  • Q = I·t

  • V = W/Q

  • R = V/I (Ohm's Law)

  • EMF: ε = I(R + r) where r = internal resistance

2. Ohm's Law

V = IR (at constant temperature)

  • Ohmic conductors: V-I graph is a straight line through origin

  • Non-ohmic conductors (e.g., filament bulb): graph curves

3. Series Circuits

  • Current: I₁ = I₂ = I₃ (same everywhere)

  • Voltage: V_total = V₁ + V₂ + V₃

  • Resistance: R_total = R₁ + R₂ + R₃

4. Parallel Circuits

  • Voltage: V₁ = V₂ = V₃ (same across each branch)

  • Current: I_total = I₁ + I₂ + I₃

  • Resistance: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃

For two resistors: R_total = (R₁ × R₂)/(R₁ + R₂)

5. Internal Resistance & EMF

Real batteries have internal resistance (r).

  • ε = I(R + r) or ε = V_terminal + I·r

  • V_terminal = ε − I·r (terminal voltage drops as current increases)

  • Lost volts = I·r

6. Electrical Power & Energy

Formula

Use

P = VI

General

P = I²R

When current known

P = V²/R

When voltage known

E = Pt

Energy consumed

Cost of electricity: kWh = (Power in kW) × (time in hours); Cost = kWh × tariff

7. Worked Example

Problem: A battery with ε = 12 V and r = 0.5 Ω is connected to two resistors in series: R₁ = 4 Ω and R₂ = 6 Ω. Find:
(a) total current (b) terminal voltage (c) power dissipated in R₁

Solution:

  • R_total = 4 + 6 = 10 Ω

  • (a) I = ε/(R + r) = 12/(10 + 0.5) = 1.14 A

  • (b) V_terminal = ε − Ir = 12 − (1.14)(0.5) = 11.43 V

  • (c) P = I²R₁ = (1.14)² × 4 = 5.20 W