Luigi Galvani (1737–1798)
Alessandro Volta (1745–1827)
Galvani’s "life force" claim questioned.
André-Marie Ampère (1775-1836)
Georg Simon Ohm (1789-1854)
Example: If
For resistance (
Therefore,
The resistivity of a section of circuit is given as:
Where
Material | Resistivity ( |
---|---|
Wood (dry) | |
Rubber | |
Glass | |
Human Skin (dry) | |
Salt Water | |
Iron | |
Aluminum | |
Copper |
Table 1 Resistivity of common materials.
The voltage drop can be calculated using Ohm's Law:
Where
- All resistors are aligned in a single circuit - The same current exists throughout - Voltage divides across components - Higher total resistance |
![]() |
- Multiple paths for current - Current divides between paths - The same voltage across branches - Lower total resistance |
![]() |
Series Circuits
Parallel Circuits
Resistance in Series Circuits
Total resistance:
Resistance in Parallel Circuits
Total resistance:
Power and Work
Therefore:
A 100W device uses 100 joules of energy each second
Total energy = power × time (joules = watts × seconds)
Three equivalent formulae for electrical power:
Device | Voltage (V) | Power (W) | Notes |
---|---|---|---|
iPhone Charger | 5 | 20 | USB-C Power Delivery |
MacBook Pro | 20 | 140 | Via USB-C/MagSafe |
Refrigerator | 120 | 150 | Energy-efficient model |
Gaming PC | 120 | 750 | Under full load |
Microwave | 120 | 1,000 | Standard household unit |
Electric Kettle | 120 | 1,500 | Boils water in ~4 minutes |
Air Conditioner | 240 | 3,500 | Central home unit |
Tesla Model 3 | 400 | 250,000 | Peak power during acceleration |
Boeing 787 | 115 | 1,000,000 | Main electrical system |
A circuit contains three resistors in series:
a) What is the total resistance?
b) What is the current through each resistor?
c) What is the voltage drop across each resistor?
a) Approach: For resistors in series, add individual resistances.
b) Approach: With total resistance known, apply Ohm's law to find current.
c) Approach: In series circuits, use Ohm's law to find voltage drop across each resistor.
A circuit has a 9 V battery connected to a 3
a) What is the current in the circuit?
b) What is the power dissipated by the resistor?
c) How much energy is converted to heat in 5 minutes?
Solution:
a) Approach: Apply Ohm's law directly to find current.
b) Approach: Use power formula with known current and resistance.
c) Approach: Convert power to energy by multiplying by time in seconds.
Two resistors are connected in parallel:
a) What is the equivalent resistance?
b) What is the total current from the battery?
c) What is the current through each resistor?
Solution:
- a) Approach: For parallel circuits, add the reciprocals of individual resistances, then take the reciprocal of that sum.
b) Approach: Once you have the equivalent resistance, apply Ohm's law with the total voltage to find total current.
c) Approach: In parallel circuits, each component receives the full voltage. Apply Ohm's law to each resistor separately.
Category | Concept | Formula | Units/Notes |
---|---|---|---|
Fundamentals | Current | A (Amperes) | |
Electric Potential | V (Volts) | ||
Ohm's Law | Basic Form | V = A × Ω | |
Resistance Form | Ω (Ohms) | ||
Power | Voltage-Current | W (Watts) | |
Current-Resistance | W = A² × Ω | ||
Voltage-Resistance | W = V² ÷ Ω | ||
Circuit Analysis | Series Resistance | Total increases | |
Parallel Resistance | Total decreases |
 The ampere was defined relative to the force between parallel current-carrying conductors: - Two conductors 1m apart - Each carrying 1 ampere - Experience a force of 2×10⁻⁷ N per meter ---
# Kirchhoff's Laws <div class='flexbox'> <div> 1. **Kirchhoff's Current Law (KCL)**: The total current entering a junction equals the total current leaving it. - $$\sum I_{in} = \sum I_{out}$$ - * Conservation of charge: charge cannot be created or destroyed. 2. **Kirchhoff's Voltage Law (KVL)**: The sum of the potential differences (voltage) around any closed loop in a circuit is zero. - $$\sum V = 0$$ - * Conservation of energy: energy supplied equals energy consumed. - *Voltage drop across resistors equals voltage rise across sources (batteries).** 3.

<div class='important-info'>
<div class='important-info'>
<div class='important-info'>

|Unit|Unit Relationships| |-|-| | Resistance (unit) | $1~\Omega = \frac{1~V}{1~A}$ | 1 Ohm = 1 Volt / 1 Ampere | | Power (unit) | $1~W = 1~V \cdot 1~A$ | 1 Watt = 1 Volt × 1 Ampere | | Power (unit) | $1~W = 1~\frac{J}{s}$ | 1 Watt = 1 Joule per second |