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Electrical Circuits and Sensors · Module 1 of 10

Circuit Fundamentals and Ohm's Law

Every electrical measurement starts here. Current is the flow of charge, voltage is the push that drives it, and a resistor relates the two through one of the most reliable laws in engineering.

01

Readiness check

This module opens the course. Tick only what you can do closed-notes.

  • State that current is the rate of flow of charge.
  • Recall what a volt and an amp measure.
  • Rearrange a simple equation such as V = IR for any one variable.
  • Convert between kΩ, Ω, mA, and A.
  • Recall that power is energy per unit time.
0 or 1 weak itemsContinue with this module.
2 weak itemsReview charge and potential in Physics: Electricity and Magnetism.
3 or more weak itemsSpend time on the units and definitions below before the worked examples.
02

The core idea

Voltage pushes, current flows, and resistance opposes. Ohm's law ties the three together, and power is the rate at which the resistor turns electrical energy into heat.

V = I RP = V I = I2 R = V2 / RI = dq / dt

Current I is the rate at which charge passes a point, measured in amperes (one amp is one coulomb per second). Voltage V is the energy given to each coulomb of charge as it moves between two points, measured in volts. A resistor relates them: the current through an ideal resistor is proportional to the voltage across it, and the constant of proportionality is the resistance R in ohms. That is Ohm's law, V = IR. Because the resistor dissipates energy, the power it absorbs is P = VI, which Ohm's law lets you write three equivalent ways. These four quantities, and the one law that links them, underlie every circuit in the course.

The skill works when: you identify which two quantities are given and use the matching form of Ohm's law or the power relation.
The skill breaks down when: resistance is in kΩ but current is read as amps, or power is computed before the current is known.
The concept. The source sets up a voltage that drives current around the loop. The resistor limits the current to V/R and converts the delivered power VI into heat.
03

The skills, taught in order

Five skills build the language of circuits: the quantities, the law that links them, and the power they carry.

1.1 Charge and current

Charge q is measured in coulombs; current is its rate of flow, I = dq/dt, in amperes. A steady current of one amp moves one coulomb past a point each second. Current has a direction: by convention it flows in the direction positive charge would move, from the higher to the lower potential through a resistor.

1.2 Voltage and potential

Voltage, or potential difference, is the energy per coulomb between two points, measured in volts (one volt is one joule per coulomb). A source maintains a voltage; the circuit responds with a current. Voltage is always measured between two points, never at a single point alone.

1.3 Ohm's law and resistance

For a linear resistor the current is proportional to the voltage across it: V = IR, with R in ohms (Ω). A larger resistance means less current for the same voltage. Real resistors are linear over a wide range, which is what makes Ohm's law so dependable in practice.

QuantitySymbolUnitDefining relation
Chargeqcoulomb (C)q = ∫ I dt
CurrentIampere (A)I = dq / dt
VoltageVvolt (V)V = energy / charge
ResistanceRohm (Ω)R = V / I
PowerPwatt (W)P = V I

The five base quantities of circuit analysis. Every later module adds elements, but these definitions never change.

1.4 Electrical power

The power absorbed by a resistor is P = VI. Using Ohm's law, this is also I2R or V2/R, whichever form matches the quantities you know. Power tells you how fast a component heats up, which sets its physical size and rating.

1.5 Units and prefixes

Circuit work spans many orders of magnitude: milliamps and microamps of current, kilohms and megohms of resistance, millivolts of sensor output. Keep prefixes explicit and convert to base units before combining quantities, then report the answer in a convenient prefix.

Engineering connection: a strain gauge changes resistance by a fraction of a percent; reading that change as a voltage, in Module 9, rests entirely on Ohm's law.

04

Worked example 1: current and power in a resistor

A 12 V source is connected across a 4 Ω resistor. Find the current through the resistor and the power it dissipates.

Figure 1. The full source voltage appears across the single resistor, so the current is V/R and the power follows directly.
  1. ProblemFind the current and dissipated power for the resistor in Figure 1.
  2. Given / findV = 12 V, R = 4 Ω. Find I and P.
  3. AssumptionsIdeal source; linear resistor; all the source voltage appears across R.
  4. ModelApply Ohm's law for the current, then the power relation.
  5. EquationsI = V / RP = V I
  6. SolveI = 12 / 4 = 3 A. P = 12 × 3 = 36 W.
  7. CheckCross-check with P = I2R = 32 × 4 = 36 W, and with V2/R = 144 / 4 = 36 W. All three forms agree.
  8. ConclusionThe resistor carries 3 A and dissipates 36 W, so it must be rated above that to avoid overheating.
Result. I = 3 A and P = 36 W.
05

Worked example 2: voltage from a known current

A current of 3 A flows through a 5 Ω resistor. Find the voltage across the resistor and the power it dissipates.

Figure 2. With the current known, Ohm's law gives the voltage drop across the resistor, and the power follows from either V and I or I and R.
  1. ProblemFind the voltage across, and power in, the resistor of Figure 2.
  2. Given / findI = 3 A, R = 5 Ω. Find V and P.
  3. AssumptionsLinear resistor; steady current.
  4. ModelUse Ohm's law to find V, then the power relation in the form I2R.
  5. EquationsV = I RP = I2 R
  6. SolveV = 3 × 5 = 15 V. P = 32 × 5 = 45 W.
  7. CheckP = VI = 15 × 3 = 45 W agrees. Units: A × Ω = V, and A2 × Ω = W.
  8. ConclusionThe resistor drops 15 V and dissipates 45 W. Knowing any two of V, I, R fixes the third and the power.
Result. V = 15 V and P = 45 W.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Voltage at a pointTalking of the voltage of a single node"Between which two points?"Voltage is always a difference between two points.
Prefix slipCurrent off by a factor of 1000"Is R in ohms or kilohms?"Convert to base units before applying V = IR.
Wrong power formPower computed from quantities not yet known"Which two of V, I, R do I have?"Pick the power form that uses your known quantities.
Confusing current and voltageCurrent treated as if it pushes"What flows and what pushes?"Voltage is the push; current is the resulting flow.
07

Practice ladder

Level 1 · Direct skill

A 9 V battery drives a 3 Ω resistor. Find the current.

Show answer

I = V/R = 9/3 = 3 A.

Level 2 · Mixed concept

A resistor carries 0.5 A when 10 V is across it. Find its resistance and the power it dissipates.

Show answer

R = V/I = 10/0.5 = 20 Ω. P = VI = 10 × 0.5 = 5 W.

Level 3 · Independent problem

A 100 Ω resistor is rated for 0.25 W. What is the largest voltage you may apply across it, and the current at that point?

Show answer

From P = V2/R, V = √(PR) = √(0.25 × 100) = √25 = 5 V. The current is I = V/R = 5/100 = 0.05 A = 50 mA.

Transfer task | Real engineering

A small heater must deliver 60 W from a 12 V supply. Find the resistance it needs and the current it draws, and comment on the wire rating.

What good work looks like

R = V2/P = 144/60 = 2.4 Ω; I = P/V = 60/12 = 5 A. The 5 A current sets the wire gauge and the supply rating, the kind of check every powered design needs.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my units cancel to volts and watts in this calculation."
"Give me three V, I, R problems with one quantity missing; I will solve each."
"Just tell me the current." Choosing and applying the right form is the skill.
"Which power formula is correct?" All three are; matching one to your knowns is the point.

Portfolio task

Take a real powered component (an LED, a small motor, a heater) and, from its rated voltage and current, compute its resistance and power, then confirm the three power forms agree.

Must include: the rated values, a resistance, a power, and a units check.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. State Ohm's law.

V = IR: the voltage across a linear resistor is its resistance times the current through it.

2. What is current, physically?

The rate of flow of charge, I = dq/dt, in amperes.

3. Write power three ways.

P = VI = I2R = V2/R.

4. Why must voltage be stated between two points?

It is a potential difference: the energy per coulomb moved from one point to another.

5. What does a larger resistance do?

It allows less current for the same applied voltage.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the two worked examples from a blank page.
+3 daysSolve three new V, I, R, P problems.
+7 daysCarry Ohm's law into series and parallel networks, Module 2.
+30 daysReuse the power relation when sizing sensor and supply currents.
10

Textbook mapping

This module follows Alexander and Sadiku, Fundamentals of Electric Circuits, 4th edition. Use these references to read further.

Topic in this moduleWhere to read more
Charge, current, and voltageAlexander & Sadiku, Chapter 1
Ohm's law and resistanceAlexander & Sadiku, Chapter 2
Power and energyAlexander & Sadiku, Chapter 1

Chapter numbers refer to the 4th edition. The names of laws and quantities are standard, so any recent edition will align closely.