ENSC2003 - Lecture 3

ENSC2003 - Lecture 2
Laws & Assumptions - Slides cont.

Power

Power is the time rate of change of work: $P = \frac{d W}{d t}$
- Measured in watts

Because $V = \frac{W}{Q} \to W = Q V$ : $P = \frac{d ( Q V )}{d t}$
For constant voltage, $P = V \frac{d Q}{d t} = V \cdot I$
We can also rearrange using Ohm’s Law: $P = I^{2} R = \frac{V ^{2}}{R}$

Power associated with a circuit element can be positive or negative, depending whether element consumes or supplies energy respectively
i.e. $P =$ 0.18 Watts → element supplies 0.18 Watts to the circuit

Energy

Energy: $E = (V \cdot I) \cdot Δ t$
See full derivation here
Also rewritten using Ohm’s Law as: $E = (I^{2} R) Δ t = (\frac{V ^{2}}{R}) Δ t$

Lumped Circuits

Our simple circuit theory/laws assume that electrical interactions happen instantaneously throughout a system
We can make this assumption because if the system are physically small
A system small enough to make this assumption is a lumped circuit

For a circuit to be lumped, wavelength $λ = \frac{c}{f _{ma x}}$ of the signals travelling through the system must be bigger than the dimension $d$ of the circuit
$λ >> d$
Rule of thumb: at least 10 times → $\frac{λ}{d _{ma x}} > 10$

Example Lumped Circuit Problem:

Non-lumped circuits are known as distributed circuits
e.g. modern power systems and computer networks

A lumped circuit is one where all the terminal voltages and currents are functions of time only
A distributed circuit is one where the terminal voltages and currents are functions of position as well as time

Circuit theory is based on the assumption that all circuit elements can be lumped, this leads to two more assumptions:

Net charge inside system of lumped elements $= 0$
Electrical components interact solely through wires

Basic Resistive Circuits Analysis - Slides

Resistive Circuits

Different ways to connect electrical elements:

Series connection $-$ single node is shared by two elements exclusively

Series connection; $R_{1} + R_{2}$
Parallel connection $-$ two elements connected to same two nodes

Parallel connection; $R_{1} ∣∣ R_{2}$

A node is defined as a point in a circuit where two or more circuit elements are interconnected

Series Connection

Two elements in series:

Individual voltages evaluate as $V_{1} = I R_{1}$ and $V_{2} = I R_{2}$

For a series connection branch (multiple parts connected in series):

Total voltage: $V = I R_{e q}$
Equivalent resistance: $R_{e q} = \frac{V}{I} = (R_{1} + R_{2} + \dots + R_{N})$

Voltage Division (VD)

How do we find voltage across $R_{1}$ and $R_{2}$ ?
Total current is: $I = \frac{V}{R _{e q}} = \frac{V}{R _{1} + R _{2}}$

$V_{1} = I \times R_{1} = \frac{R _{1}}{R _{1} + R _{2}} V$
$V_{2} = I \times R_{2} = \frac{R _{2}}{R _{1} + R _{2}} V$

For the voltage drop across the $i$ th resistor in series, we get:
$V_{i} = \frac{R _{i}}{R _{1} + R _{2} + \dots + R _{N}} \cdot V$

Same current through all elements connected in series
Voltage across element is proportional to resistance

Parallel Connection

Two elements in parallel:

Individual currents evaluate as $I_{1} = V / R_{1}$ and $I_{2} = V / R_{2}$

For a parallel connection branch:

Total current: $I = V \cdot \frac{1}{R _{e q}}$
Equivalent resistance: $\frac{1}{R _{e q}} = \frac{I}{V} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots + \frac{1}{R _{N}}$

Same voltage across all elements connected in parallel
Current through element is inversely proportional to resistance

Current Division (CD)

How do we find current through $R_{1}$ and $R_{2}$ ?
Total voltage is: $V = I R_{e q} = I \frac{R _{1} R _{2}}{R _{1} + R _{2}}$

$I_{1} = \frac{V}{R _{1}} = \frac{R _{2}}{R _{1} + R _{2}} I$
$I_{2} = \frac{V}{R _{2}} = \frac{R _{1}}{R _{1} + R _{2}} I$

For the current drawn by the $n$ th resistor in parallel, we get:
$I_{n} = \frac{1/ R _{n}}{1/ R _{1} + 1/ R _{2} + \dots + 1/ R _{n}} \cdot I$

Complicated Circuit Simplification

Read through: Complicated Circuit Simplification example
- There is also a good animation in the lecture recording

The aim of the simplification is to reduce the number of resistors in the circuit down by grouping series/parallel elements together

First, identify all the nodes in the circuit diagram
Then, group resistors that are in series or parallel
Continue this method until you cannot reduce resistors anymore

Short Circuits

ENSC2003 - Lecture 4

Quartz 5

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