Day 4

Created at: 2025-01-15

Today I'm finally starting chapter 1!

I have a degree in Electrical Engineering, but I've finished that a while ago and am probably rusty. No bad in getting a refresher.

Interesting Quotes

A mid-century computer (the IBM 650) cost $300,000, weighed 2.7 tons, and
contained 126 lamps on its control panel; in an amusing reversal, a
contemporary energy-efficient lamp contains a computer of greater capability
within its (physical) base, and costs about $10.
It is often the case that the panel controls and cabinet hardware of an
instrument cost more than the electronics inside.

Voltage

Intuitive Explanation: What you apply to a circuit to make currents flow.

Enhanced Definition: The voltage between two points is the cost in energy (work) to move positive charge from a more negative point (lower potential) to a more positive point (higher potential). Equivalently, the energy released when a unit charge moves "downhill" from the higher potential to the lower.

Symbols: Often denoted by V, sometimes by E.

Nicknames: Potential Difference or Electromotive Force.

Unit of Measure: Volt (V)

Current

Symbol: I

Definition: The rate of flow of electric charge past a point.

Unit of Measure: Ampere (or amp).

Fun fact: A current of 1 amp equals a flow of 1 coulomb (6*10^18 electrons) of charge per second.

Conventions: In a circuit, current flows from a more positive point to a more negative one, even though the actual electron flow is in the opposite direction!

Important Rules

  1. Kirchhoff's current law (KCL): The sum of currents into a point in a circuit equals the sum of the currents out. Such a point is often referred to as a node.

2. Kirchhoff's voltage law (KVL): Parallel connections have the same voltage across them. This also means that the sum of the voltage drops around any closed circuit is zero.

3. The power (energy per unit time) consumed by a circuit device is: P = VI. - P is in watts. - A 55W bulb connected at 220V plug has 0.25 amps through it. - When dealing with periodic signals, P=VI deal with _average power_.

4. Ohm's law: R = V/I.

Resistors

R(series) = R1 + R2

R(parallel) = (R1*R2) / (R1 + R2)

Alternatively...

R(parallel) = 1 / (1/R1 + 1/R2)

Tip: By parallelising resistors you always get a smaller resistance.

Trivia:

Exercises

1.1 You have a 5k resistor and a 10k resistor. What is their combined
resistance (a) in series and (b) in parallel?

(a): 15k (b): 50k/15k = 3.333k

1.2 If you place a 1 ohm resistor across a 12 volt car battery, how much power
will it dissipate?

P = VI V = 12V I = V/R = 12A P = 144W

1.3 Prove the formulas for series and parallel resistors.

Given that:

  1. Kirchhoff's voltage law (KVL): Parallel connections have the same voltage across them. This also means that the sum of the voltage drops around any closed circuit is zero.
  2. Kirchhoff's current law (KCL): The sum of currents into a point in a circuit equals the sum of the currents out. Such a point is often referred to as a node.
  3. Ohm's law: R = V/I.

Imagine a circuit that has one voltage source and two resistors, all connected in series. Where:

- V1         = voltage across R1.
- V2         = voltage across R2.
- I1         = current across R1.
- I2         = current across R2.
- V          = power supply.
- R(series)  = resulting resistance of the circuit in series.
- I          = corrent through power supply.

From 1., V = V1 + V2 From 2., I1 = I2 = V From 3., R(series) = V/I

From ohm's law:

V1 = R1 * I1 = R1 * I
V2 = R2 * I2 = R2 * I

V = R1 * I + R2 * I
R(series) = V/I = (R1 * I + R2 * I)/I

Therefore
R(series) = R1 + R2

Now for the parallel case let's imagine that V, R1, and R2 are connected in parallel.

From 1., V = V1 = V2 From 2., I = I1 + I2 From 3., R(parallel) = V/I

I1 = V1/R1 = V/R1
I2 = V2/R2 = V/R2

R(parallel) = V/I = V/(V/R1 + V/R2)

Therefore
R(parallel) = 1/(R1 + R2)