Current Electricity: Circuits, Ohm’s Law & Resistance
CSEC Physics: Electrical Energy & Circuits
Essential Understanding: Current electricity involves the flow of electric charge through conductors. Understanding circuits, resistance, and the relationship between voltage, current, and resistance is fundamental to how all modern technology works – from smartphones to the power grid.
Core Concepts: Understanding Electric Current
Electric Current ($I$)
Definition: The rate of flow of electric charge through a conductor.
Formula: $$I = \frac{Q}{t}$$
- $I$: Current (Amperes, A)
- $Q$: Charge (Coulombs, C)
- $t$: Time (seconds, s)
Unit: Ampere (A) or Amperes
Note: 1 Ampere = 1 Coulomb/second
Potential Difference / Voltage ($V$)
Definition: The work done to move a unit charge between two points in a circuit. It is the “push” that drives current through the circuit.
Formula: $$V = \frac{W}{Q}$$
- $V$: Potential Difference (Volts, V)
- $W$: Work done (Joules, J)
- $Q$: Charge (Coulombs, C)
Unit: Volt (V)
Note: 1 Volt = 1 Joule/Coulomb
Electrical Resistance ($R$)
Definition: The opposition that a conductor offers to the flow of electric current.
Formula: $$R = \frac{V}{I}$$
- $R$: Resistance (Ohms, $\Omega$)
- $V$: Potential Difference (Volts, V)
- $I$: Current (Amperes, A)
Unit: Ohm ($\Omega$ – Greek letter omega)
Electrical Power ($P$)
Definition: The rate at which electrical energy is transferred or consumed in a circuit.
Formulas:
- $$P = VI$$
- $$P = I^2R$$
- $$P = \frac{V^2}{R}$$
Unit: Watt (W)
Ohm’s Law
The most fundamental relationship in DC circuit analysis:
The Ohm’s Law Triangle: Cover the quantity you want to solve for to see the formula.
Interactive Circuit Lab
Circuit Builder & Ohm’s Law Explorer
Objective: Explore how voltage, current, and resistance are related. Adjust the voltage and resistance to see how current changes in the circuit.
Voltage (V)
12 V
Current (I)
1.20 A
Resistance (R)
10 Ω
Calculation: Using $I = \frac{V}{R}$
I = 12 / 10 = 1.20 A
Factors Affecting Resistance
The resistance of a conductor depends on four main factors:
Resistance Formula
Where:
- $\rho$ (rho) = Resistivity of the material ($\Omega \cdot m$)
- $L$ = Length of the conductor (m)
- $A$ = Cross-sectional area ($m^2$)
Length ($L$)
Relationship: Resistance is directly proportional to length.
$$R \propto L$$
Explanation: Longer conductors have more atoms for electrons to collide with, increasing resistance.
Cross-Sectional Area ($A$)
Relationship: Resistance is inversely proportional to area.
$$R \propto \frac{1}{A}$$
Explanation: Thicker wires provide more pathways for current, reducing resistance.
Material (Resistivity $\rho$)
Explanation: Different materials have different resistivities.
- Good conductors: Copper, Silver, Aluminum ($\rho$ very low)
- Resistors: Carbon, Constantan ($\rho$ moderate)
- Insulators: Rubber, Plastic, Glass ($\rho$ very high)
Temperature
Relationship: For most conductors, resistance increases with temperature.
Explanation: Higher temperature causes more lattice vibrations, increasing electron collisions.
Exception: For some materials (thermistors), resistance decreases with temperature.
Key Observations:
- Resistance is directly proportional to length ($R \propto L$)
- Resistance is inversely proportional to cross-sectional area ($R \propto \frac{1}{A}$)
- Thicker wires (larger diameter) have lower resistance
Series Circuits
In a series circuit, components are connected one after another in a single loop. There is only one path for the current to flow.
Series Circuit Rules
1. Current ($I$): The current is the same throughout all components.
$$I_{\text{total}} = I_1 = I_2 = I_3 = \dots$$
2. Voltage ($V$): The total voltage is the sum of individual voltage drops.
$$V_{\text{total}} = V_1 + V_2 + V_3 + \dots$$
3. Resistance ($R$): The total resistance is the sum of individual resistances.
$$R_{\text{total}} = R_1 + R_2 + R_3 + \dots$$
Series Circuit Diagram
Worked Example: Series Circuit
Problem: Two resistors of 4Ω and 6Ω are connected in series to a 20V battery. Calculate: (a) the total resistance, (b) the current in the circuit, and (c) the voltage across each resistor.
$$R_{\text{total}} = R_1 + R_2 = 4\Omega + 6\Omega = 10\Omega$$
$$I = \frac{V}{R} = \frac{20V}{10\Omega} = 2A$$
$$V_1 = I \times R_1 = 2A \times 4\Omega = 8V$$
$$V_2 = I \times R_2 = 2A \times 6\Omega = 12V$$
Check: $V_1 + V_2 = 8V + 12V = 20V$ ✓
Series Circuit Characteristics
- If one component fails (open circuit), the entire circuit stops working
- Total resistance is always greater than the largest individual resistance
- Voltage drops add up to the source voltage
Parallel Circuits
In a parallel circuit, components are connected across the same voltage source, providing multiple paths for current to flow.
Parallel Circuit Rules
1. Voltage ($V$): The voltage is the same across all parallel branches.
$$V_{\text{total}} = V_1 = V_2 = V_3 = \dots$$
2. Current ($I$): The total current is the sum of currents in individual branches.
$$I_{\text{total}} = I_1 + I_2 + I_3 + \dots$$
3. Resistance ($R$): The reciprocal of total resistance is the sum of reciprocals.
$$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$$
For two resistors in parallel:
$$R_{\text{total}} = \frac{R_1 \times R_2}{R_1 + R_2}$$
Parallel Circuit Diagram
Worked Example: Parallel Circuit
Problem: Three resistors of 2Ω, 3Ω, and 6Ω are connected in parallel across a 12V battery. Calculate: (a) the total resistance, (b) the current through each resistor, and (c) the total current.
$$\frac{1}{R_{\text{total}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$$
$$R_{\text{total}} = 1\Omega$$
$$I_1 = \frac{V}{R_1} = \frac{12}{2} = 6A$$
$$I_2 = \frac{V}{R_2} = \frac{12}{3} = 4A$$
$$I_3 = \frac{V}{R_3} = \frac{12}{6} = 2A$$
$$I_{\text{total}} = I_1 + I_2 + I_3 = 6A + 4A + 2A = 12A$$
Check: $I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12}{1} = 12A$ ✓
Parallel Circuit Characteristics
- If one component fails, other branches continue working
- Total resistance is always less than the smallest individual resistance
- Each branch gets the full source voltage
Combined Series-Parallel Circuits
Many real circuits contain both series and parallel components. The key is to simplify step by step.
Worked Example: Combined Circuit
Problem: A 4Ω resistor is connected in series with a parallel combination of two 6Ω resistors. The circuit is connected to a 24V battery. Calculate the total current.
Two 6Ω resistors in parallel:
$$R_{\text{parallel}} = \frac{6 \times 6}{6 + 6} = \frac{36}{12} = 3\Omega$$
$$R_{\text{total}} = R_{\text{series}} + R_{\text{parallel}} = 4\Omega + 3\Omega = 7\Omega$$
$$I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{24V}{7\Omega} \approx 3.43A$$
CSEC Examination Mastery Tip
Step-by-Step Approach for Complex Circuits:
- Identify which components are in series and which are in parallel
- Start simplifying from the furthest point from the battery working back
- Calculate equivalent resistances step by step
- Find total current using Ohm’s Law ($I = \frac{V}{R_{\text{total}}}$)
- Work backwards to find currents and voltages in individual components
- ALWAYS draw a circuit diagram with your values labeled
Comparing Series and Parallel Circuits
| Characteristic | Series Circuit | Parallel Circuit |
|---|---|---|
| Current ($I$) | Same through all components | Splits between branches; total = sum of branch currents |
| Voltage ($V$) | Split between components; total = sum of voltage drops | Same across all components |
| Resistance ($R$) | Total = sum of individual resistances ($R_{\text{total}} > R_{\text{any}}$) | Reciprocal formula; total < smallest resistance ($R_{\text{total}} < R_{\text{any}}$) |
| Failure Mode | One component fails → entire circuit stops | One component fails → other branches still work |
| Common Use | String lights, switches, fuses | Household wiring, car tail lights, appliances |
CSEC Practice Arena
Test Your Understanding
Or using formula: $$R_{\text{total}} = \frac{12 \times 12}{12 + 12} = \frac{144}{24} = 6\Omega$$
CSEC Past Paper Questions
CSEC Past Paper Practice
The following questions are based on actual CSEC Physics examination patterns. Practice these to prepare for your exams!
(a) Total Resistance:
$$R_{\text{total}} = R_1 + R_2 = 3\Omega + 6\Omega = 9\Omega$$
(b) Current:
$$I = \frac{V}{R} = \frac{9V}{9\Omega} = 1A$$
(c) Potential Difference across 3Ω resistor:
$$V = IR = 1A \times 3\Omega = 3V$$
(a) Ohm’s Law:
The current flowing through a conductor is directly proportional to the potential difference across it, provided the temperature remains constant.
Or: $$V = IR$$
(b) New Resistance:
Since $$R \propto L$$ (resistance is directly proportional to length), and length doubles from 2m to 4m:
New resistance = $$4\Omega \times 2 = 8\Omega$$
(a) Two differences:
- In series, current is the same through all components; in parallel, current splits.
- In series, total resistance equals sum; in parallel, total resistance is less than smallest.
- In series, if one component fails, all stop; in parallel, others continue.
$$I_{\text{total}} = 3 \times 0.5A = 1.5A$$
(b)(ii) Resistance of each lamp:
$$R = \frac{V}{I} = \frac{12V}{0.5A} = 24\Omega$$
(a) Resistivity:
The resistivity of a material is the resistance of a unit length of the material having unit cross-sectional area. It is a property of the material itself.
(b)(i) Twice the length:
Since $$R \propto L$$, new resistance = $$2\Omega \times 2 = 4\Omega$$
(b)(ii) Half the cross-sectional area:
Since $$R \propto \frac{1}{A}$$, if area is halved, resistance doubles.
New resistance = $$2\Omega \times 2 = 4\Omega$$
Key Examination Insights
Common Mistakes to Avoid
- Confusing series and parallel formulas
- Forgetting that voltage is the same in parallel (not current)
- Not showing working for calculations
- Using the wrong units (always check units before calculating)
- Forgetting that adding resistors in parallel decreases total resistance
Success Strategies
- Always draw circuit diagrams with labeled values
- Write formulas before substituting values
- Check your answers using Kirchhoff’s Laws
- Remember: $V = IR$ is your best friend
- For parallel, start with $\frac{1}{R_{\text{total}}}$ when you have different values
Electrical Energy and Power
Power Formulas
Energy: $E = P \times t$ (Joules) or $E = VIt$ (Joules)
Worked Example: Electrical Power
Problem: A kettle has a power rating of 2kW when connected to a 240V supply. Calculate (a) the current drawn by the kettle (b) the resistance of the heating element.
Power $P = 2kW = 2000W$
Voltage $V = 240V$
Using $P = VI$, rearrange to get $I = \frac{P}{V}$
$$I = \frac{2000W}{240V} = 8.33A$$
Using $P = \frac{V^2}{R}$, rearrange to get $R = \frac{V^2}{P}$
$$R = \frac{240^2}{2000} = \frac{57600}{2000} = 28.8\Omega$$