Microcontrollers, Robotics, and Circuit Analysis

Posted on Dec 19, 2025 in Industrial Electronics and Automation Engineering

Microcontrollers and Robotics Fundamentals

Microcontrollers vs. Personal Computers

147: A microcontroller is a programmable integrated circuit on a single chip that incorporates all the basic functions of a computer, including the processing unit, memory, and input/output peripherals. It essentially functions as a computer with limited capabilities on a single chip. It differs from a Personal Computer (PC) in several key ways:

  • Microcontrollers are cheaper and have lower power consumption (allowing for battery use).
  • They are designed to perform one specific task instead of running hundreds of programs.
  • They possess significantly less memory.
  • They connect directly to sensors and motors without needing adaptation modules.

These devices are ubiquitous in modern life, controlling everyday items such as washing machines, refrigerators, alarm clocks, television remote controls, and ABS systems in cars.

Automated Mechanisms and Robots

146: An automated mechanism is a device or machine designed to perform a fixed, specific task whose operation cannot be changed (examples include car windscreen wipers). A robot is a more advanced, programmable automated machine that captures information about its surroundings, processes this information, and then acts on it.

Robots are widely used today in industry (assembly lines, moving goods) and for tasks like extinguishing fires and detecting explosives. To control the processes a robot performs, it is connected to a computer (or has one built-in) whose microprocessors process the input signals and calculate commands for the robot’s components (motors, lights, etc.). However, microprocessors cannot communicate directly with sensors or give direct commands to motors, which necessitates an intermediary: a controller board. This electronic circuit acts as an interface between the computer and the devices being controlled by adapting the input and output signals, with Arduino controller boards being the most popular type currently.

Electronics Exercises

Exercise 29: Series Circuit

Problem Statement

Given the circuit shown, calculate:

  • a) The total equivalent resistance.
  • b) The total current in the circuit.
  • c) The current through each resistor.
  • d) The voltage across each resistor.
  • e) The power supplied by the battery and the total power dissipated.

Circuit Data

  • Voltage source: 9 V
  • Resistors: 20 kΩ, 30 kΩ, 40 kΩ

Solution

a) Total Equivalent Resistance: In a series circuit, resistances add up.
Req = 20 kΩ + 30 kΩ + 40 kΩ = 90 kΩ
Result: 90,000 Ω

b) Total Current:
I = V / Req = 9 V / 90,000 Ω = 0.0001 A
Result: 0.0001 A (0.1 mA)

c) Current Through Each Resistor: In a series circuit, current is the same everywhere.
Result: 0.0001 A through each resistor

d) Voltage Across Each Resistor: (V = I × R)

  • 20 kΩ: 2 V
  • 30 kΩ: 3 V
  • 40 kΩ: 4 V

(2 + 3 + 4 = 9 V, consistent with the source)

e) Power Supplied and Dissipated:
P = V × I = 9 V × 0.0001 A = 0.0009 W
Result: 0.0009 W (0.9 mW)

Exercise 30: Series–Parallel Circuit

Problem Statement

Calculate all the currents in the circuit.

Circuit Data

  • Voltage source: 10 V
  • Series resistor: 100 Ω
  • Parallel resistors: 60 Ω, 20 Ω, 10 Ω, 150 Ω

Solution

Parallel Block Equivalent Resistance:
1 / Rp = 1/60 + 1/20 + 1/10 + 1/150
1 / Rp ≈ 0.1734
Rp ≈ 5.767 Ω

Total Equivalent Resistance:
Req = Rs + Rp = 100 Ω + 5.767 Ω ≈ 105.77 Ω

Total Current:
IT = 10 V / 105.77 Ω ≈ 0.0945 A

Voltage Across the Parallel Block:
Vp = IT × Rp ≈ 0.0945 A × 5.767 Ω ≈ 0.545 V

Currents in Each Branch:

  • 100 Ω (series): 0.0945 A
  • 60 Ω: 0.545 V / 60 Ω ≈ 0.0091 A
  • 20 Ω: 0.545 V / 20 Ω ≈ 0.0273 A
  • 10 Ω: 0.545 V / 10 Ω ≈ 0.0545 A
  • 150 Ω: 0.545 V / 150 Ω ≈ 0.0036 A

Exercise 31: Circuit Analysis A and B

Circuit A Data

  • Voltage source: 200 V
  • Series resistor: 100 Ω
  • Parallel resistors: 25 Ω, 30 Ω, 30 Ω

Solution A

a) Equivalent Resistance:
1 / Rp = 1/25 + 1/30 + 1/30 ≈ 0.1066
Rp ≈ 9.38 Ω
Req = 100 Ω + 9.38 Ω = 109.38 Ω

b) Total Current:
IT = 200 V / 109.38 Ω ≈ 1.828 A

c) Currents Through Each Resistor:
Voltage across the parallel block: Vp = IT × Rp ≈ 1.828 A × 9.38 Ω = 17.15 V

  • 100 Ω: 1.828 A
  • 25 Ω: 17.15 V / 25 Ω ≈ 0.686 A
  • 30 Ω (first): 0.572 A
  • 30 Ω (second): 0.572 A

d) Voltages:

  • 100 Ω: 1.828 A × 100 Ω ≈ 182.8 V
  • Parallel block resistors: 17.15 V

e) Power:
P = V × IT = 200 V × 1.828 A ≈ 365.6 W

Circuit B Data

  • Voltage source: 9 V
  • Series resistors: 1 kΩ, 2 Ω, 5 Ω
  • Parallel resistors: 3 Ω, 4 Ω

Solution B

a) Equivalent Resistance:
Rp = (3 × 4) / (3 + 4) = 12 / 7 ≈ 1.714 Ω
Req = 1000 Ω + 1.714 Ω + 2 Ω + 5 Ω ≈ 1008.71 Ω

b) Total Current:
IT = 9 V / 1008.71 Ω ≈ 0.00892 A

c) Currents Through Each Resistor:
Series resistors (1 kΩ, 2 Ω, 5 Ω): 0.00892 A
Voltage across the parallel block: Vp = IT × Rp ≈ 0.0153 V

  • 3 Ω: 0.0153 V / 3 Ω ≈ 0.0051 A
  • 4 Ω: 0.0153 V / 4 Ω ≈ 0.0038 A

d) Voltages:

  • 1 kΩ: 0.00892 A × 1000 Ω ≈ 8.92 V
  • Parallel resistors: 0.0153 V
  • 2 Ω: 0.0178 V
  • 5 Ω: 0.0446 V

e) Power:
P = 9 V × 0.00892 A ≈ 0.080 W

Exercise 32: Circuit with Two Sources

Problem Statement

Given a circuit with two power sources, calculate:

  • a) The total equivalent resistance.
  • b) The total equivalent voltage.
  • c) The total current.
  • d) The current through each resistor.
  • e) The voltage across each resistor.

Circuit Data

  • Sources: 9 V and 11 V (opposite polarity)
  • Series resistors: 100 Ω, 20 Ω
  • Parallel resistors: 60 Ω, 10 Ω, 150 Ω

Solution

Simplification of Voltage Sources: The sources oppose each other.
Veq = 11 V – 9 V = 2 V

Series Resistance:
Rs = 100 Ω + 20 Ω = 120 Ω

Parallel Block Equivalent Resistance:
1 / Rp = 1/60 + 1/10 + 1/150 ≈ 0.12334
Rp ≈ 8.107 Ω

a) Total Equivalent Resistance:
Req = Rs + Rp = 128.11 Ω

b) Total Equivalent Voltage:
Veq = 2 V

c) Total Current:
IT = 2 V / 128.11 Ω ≈ 0.0156 A

d) Currents Through Each Resistor:
Series resistors: 0.0156 A
Voltage across the parallel block: Vp = 0.0156 A × 8.107 Ω ≈ 0.1265 V

  • 60 Ω: 0.0021 A
  • 10 Ω: 0.0127 A
  • 150 Ω: 0.0008 A

e) Voltages:

  • 100 Ω: 1.56 V
  • 20 Ω: 0.31 V
  • Parallel resistors: 0.1265 V

Exercise 33: Power Source Associations

Problem Statement

From the given circuit, identify the possible associations of power sources.

Solution

The circuit shows a main 20 V source feeding two parallel branches, each one containing a 680 Ω resistor and several power sources (9 V and 10 V in the first branch; 9 V and 9 V in the second branch).

  1. Series-Aiding Association: This occurs when two sources are connected in series with the positive terminal of one to the negative terminal of the other, increasing the total voltage.
    • Example: In the first branch, the 9 V and 10 V sources in series-aiding give 19 V.
    • Example: In the second branch, the two 9 V sources in series-aiding give 18 V.
  2. Series-Opposing Association: This occurs when two sources are connected in series with positive-to-positive or negative-to-negative terminals, causing the voltages to subtract.
    • Example: In the first branch, the 10 V and 9 V sources in series-opposing give a net voltage of 1 V.
  3. Parallel Association: No power sources are directly in parallel with each other, because each branch includes a resistor separating the voltage sources.

Exercise 34: Parallel Resistance Equations

Problem Statement

Show how we can use the following equations in a parallel resistor configuration:

  • a) For two resistors in parallel
  • b) For three resistors in parallel

Solution (Conceptual)

In a parallel circuit, the voltage across each branch is the same, and the total current is the sum of the individual branch currents.

a) Two Resistors in Parallel

Formula: RT = (R1 × R2) / (R1 + R2)

Explanation:

  1. Start with the general reciprocal formula: 1 / RT = 1 / R1 + 1 / R2
  2. Combine the fractions using a common denominator (R1 × R2): 1 / RT = (R1 + R2) / (R1 × R2)
  3. Invert both sides to obtain: RT = (R1 × R2) / (R1 + R2)

This is known as the “product over sum” formula.

b) Three Resistors in Parallel

Formula: 1 / RT = 1 / R1 + 1 / R2 + 1 / R3

Explanation:

  1. Apply Kirchhoff’s Current Law: Itotal = I1 + I2 + I3
  2. Substitute Ohm’s law (I = V / R): V / RT = V / R1 + V / R2 + V / R3
  3. Since the voltage is the same in a parallel circuit, cancel V: 1 / RT = 1 / R1 + 1 / R2 + 1 / R3

Exercise 35: Component Identification

Problem Statement

Identify the names of the following electronic components and draw their electrical symbols.

Solution

(Only the names are provided; you should draw the symbols in your notes.)

  • a) Relay: An electromechanical device that works as a switch controlled by a coil.
  • b) LDR (Light Dependent Resistor): A resistor whose resistance varies with light intensity.
  • c) BJT Transistor (Bipolar Junction Transistor): An active device used to amplify or switch signals.
  • d) Thermistor: A resistor whose resistance changes with temperature.
  • e) LED (Light Emitting Diode): A diode that emits light when current flows through it.
  • f) Potentiometer: A three-terminal variable resistor used to adjust voltage or resistance.

Exercise 36: Component Classification

Problem Statement

Classify the following components as passive or active.

Solution

Active components can control or amplify electrical current or energy. Passive components only consume, dissipate, or store energy.

  • Diode – ACTIVE (controls current direction)
  • Transistor – ACTIVE (switches or amplifies current)
  • Capacitor – PASSIVE (stores energy in an electric field)
  • LDR – PASSIVE (a type of resistor)
  • Thermistor – PASSIVE (a type of resistor)
  • Potentiometer – PASSIVE (variable resistor)

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