Learn DC and AC circuit analysis, power factor and three-phase basics, safe measurement on battery circuits, simulation, transformers, motors and systematic troubleshooting.

Electrical Circuits

## Why it matters

Every electrical job, from wiring a house to designing a power supply, rests on four quantities: voltage, current, resistance and power. If you can move between them confidently, you can size a resistor, check whether a wire or fuse is big enough, estimate a battery's run time and spot a reading that does not make sense. This lesson gives you the vocabulary and the two formulas you will use in every lesson after it.

## The four quantities

**Voltage (V, volts)** is the electrical "push": the energy given to each unit of charge as it moves between two points. A voltage is always measured *between* two points, so "the voltage at a wire" really means "the voltage between that wire and some reference", usually the negative terminal or ground.

**Current (I, amperes or amps)** is the rate at which charge flows past a point. One amp is one coulomb of charge per second. Current is measured *through* a part of a circuit.

**Resistance (R, ohms, Ω)** is how strongly a material or component opposes current. Copper wire has very low resistance; a resistor is made to have a known resistance; a lamp filament has a resistance that rises as it heats up.

**Power (P, watts)** is the rate at which energy is converted, for example into heat or light. One watt is one joule per second.

A water analogy helps at first: voltage is like pressure, current is like flow rate, and resistance is like a narrow pipe. Do not push the analogy too far, but it explains why more push gives more flow and a narrower pipe gives less.

## Ohm's law

For a resistor, current is proportional to voltage:

```text
V = I × R      I = V / R      R = V / I
```

Units must match: volts, amps and ohms. Convert milliamps (mA) to amps by dividing by 1,000, and kilohms (kΩ) to ohms by multiplying by 1,000.

**Worked example 1.** A 9 V battery is connected across a 470 Ω resistor. What current flows?

```text
I = V / R = 9 / 470 = 0.0191 A ≈ 19 mA
```

**Worked example 2.** You measure 2.5 V across a resistor and 5 mA through it. What is its resistance?

```text
R = V / I = 2.5 / 0.005 = 500 Ω
```

## Power

Power is voltage times current:

```text
P = V × I
```

Substituting Ohm's law gives two more forms that are handy when you know only two quantities:

```text
P = I² × R        P = V² / R
```

**Worked example 3.** In example 1, how much power does the resistor turn into heat?

```text
P = V × I = 9 × 0.0191 = 0.172 W
```

A standard small resistor is rated 0.25 W, so this one is fine. If the answer had been 0.4 W, you would need a 0.5 W or larger resistor, or it would overheat.

**Worked example 4.** A 12 V, 36 W lamp. What current does it draw, and what is its hot resistance?

```text
I = P / V = 36 / 12 = 3 A
R = V / I = 12 / 3 = 4 Ω
```

## Energy

Energy is power multiplied by time. Utilities bill in kilowatt-hours (kWh): a 1,000 W heater running for one hour uses 1 kWh. Batteries are rated in amp-hours (Ah) or milliamp-hours (mAh): a 2,000 mAh battery could, in theory, supply 20 mA for 100 hours. Real run times are shorter, because capacity falls at higher currents and the voltage sags as the battery empties.

## Common mistakes

- **Mixing units.** 19 mA is 0.019 A, not 19 A. Write units on every line of a calculation.
- **Forgetting that voltage is between two points.** A meter has two leads for a reason.
- **Using Ohm's law on parts that are not resistors.** LEDs, diodes and motors do not have a fixed resistance; you will treat LEDs differently in the breadboard lesson.
- **Ignoring power ratings.** A calculation can be right and the part can still burn out because its wattage rating is too low.
- **Trusting a result that fails a sanity check.** A phone charger drawing 300 A is a sign you misplaced a decimal point.

## Safety note

The formulas are the same at 9 V and at 480 V, but the danger is not. In this course you practise only on batteries of 9 V or less and in simulators. Work on mains or industrial circuits requires employer or school training, supervision by a qualified person and the right protective equipment.

## Recap

- Voltage is push (V), current is flow (A), resistance opposes flow (Ω), power is the rate of energy use (W).
- Ohm's law: V = I × R.
- Power: P = V × I = I² × R = V² / R.
- Convert mA and kΩ before calculating, and check the answer makes sense.

## Try it

Open the **Do** tab and finish a small Ohm's law and power calculator by writing the power formula.


Relate voltage, current, resistance and power with Ohm's law and the power formula.

Voltage, current, resistance, power

Voltage Current and Resistance

Electrical Formulas - Basic Electricity For Beginners

## Why it matters

Real circuits are rarely one battery and one resistor. Lights on a string, outlets on a branch circuit, sensors on a control panel and the resistors on a circuit board are all connected in series, in parallel or in a mix of both. Reducing a network to a single equivalent resistance lets you find the total current, then work back to the voltage and current at every part. It is the most common hand calculation in circuit work.

## Series circuits

Components are in **series** when they sit one after another on a single path, so the **same current** flows through each one.

- Total resistance is the sum: `R_total = R1 + R2 + R3 + ...`
- The source voltage is shared out: each resistor's voltage is `I × R`, and the drops add up to the source voltage.
- If one part opens (a lamp filament breaks), current stops everywhere on that path.

**Worked example 1.** A 12 V battery drives 100 Ω, 200 Ω and 300 Ω in series.

```text
R_total = 100 + 200 + 300 = 600 Ω
I = 12 / 600 = 0.02 A (20 mA)
V1 = 0.02 × 100 = 2 V,  V2 = 4 V,  V3 = 6 V   (2 + 4 + 6 = 12 V)
```

The biggest resistor takes the biggest share of the voltage. This is the idea behind a **voltage divider**: two resistors in series give an output of `Vout = Vin × R2 / (R1 + R2)`.

**Worked example (divider).** A sensor input needs about 3 V from a 9 V battery. With R1 = 20 kΩ on top and R2 = 10 kΩ below:

```text
Vout = 9 × 10,000 / (20,000 + 10,000) = 3 V
```

Dividers are used for reference voltages and for reading sensors, but they are not power supplies: connecting a load across R2 puts it in parallel with R2 and pulls the output voltage down.

## Spotting series and parallel on a drawing

Schematics are not always drawn neatly, so look at the **nodes** (the connection points), not at the shape of the drawing. Trace each wire with a finger or a coloured pen:

- Two parts are in series if the only thing joined at the node between them is those two parts.
- Two parts are in parallel if both of their ends land on the same pair of nodes, even if one is drawn at the top of the page and the other at the bottom.

Colouring every node a different colour first is a habit that prevents most reduction mistakes.

## Parallel circuits

Components are in **parallel** when they connect across the same two points, so they all have the **same voltage**.

- Currents add: the total current is the sum of the branch currents.
- The equivalent resistance is found from reciprocals:

```text
1 / R_total = 1/R1 + 1/R2 + 1/R3 + ...
```

- For exactly two resistors there is a shortcut, "product over sum": `R_total = (R1 × R2) / (R1 + R2)`.
- The equivalent resistance is always **smaller than the smallest branch**. Adding a branch gives current another path, so total current goes up.
- If one branch opens, the others keep working. That is why outlets and lights in a building are wired in parallel.

**Worked example 2.** 100 Ω and 300 Ω in parallel across 12 V.

```text
R_total = (100 × 300) / (100 + 300) = 75 Ω
I_total = 12 / 75 = 0.16 A
I1 = 12 / 100 = 0.12 A,  I2 = 12 / 300 = 0.04 A   (0.12 + 0.04 = 0.16 A)
```

Two equal resistors in parallel give half the value: 220 Ω ∥ 220 Ω = 110 Ω.

## Series-parallel circuits

Reduce the network step by step, from the part furthest from the source back toward it:

1. Find groups that are clearly in series or clearly in parallel.
2. Replace each group with its equivalent resistance and redraw.
3. Repeat until one resistance is left.
4. Find the total current, then expand the drawing again in reverse to find each voltage and current.

**Worked example 3.** A 10 V source feeds R1 = 40 Ω in series with a parallel pair R2 = 60 Ω and R3 = 30 Ω.

```text
R2 ∥ R3 = (60 × 30) / (60 + 30) = 20 Ω
R_total = 40 + 20 = 60 Ω
I_total = 10 / 60 = 0.1667 A
V across R1 = 0.1667 × 40 = 6.67 V
V across the pair = 0.1667 × 20 = 3.33 V
I2 = 3.33 / 60 = 0.0556 A,  I3 = 3.33 / 30 = 0.111 A
```

Check: 6.67 + 3.33 = 10 V, and 0.0556 + 0.111 = 0.1667 A. Always close the loop with a check like this.

## Common mistakes

- **Adding parallel resistances.** If your parallel answer is bigger than the smallest branch, you added instead of using reciprocals.
- **Forgetting the final reciprocal.** `1/100 + 1/300 = 0.01333`, but that is 1/R. The resistance is 1 / 0.01333 = 75 Ω.
- **Calling resistors "in series" because they are drawn in a line.** Series means the same current with no branch point between them; parallel means the same two end nodes.
- **Skipping the expand-back step.** The equivalent resistance gives the total current, not the current in each branch.

## Recap

- Series: same current, resistances add, voltages share out.
- Parallel: same voltage, currents add, `1/R_total = Σ 1/R`, result smaller than the smallest branch.
- Reduce series-parallel networks step by step, then expand back and check with the totals.

## Try it

Open the **Do** tab and write the parallel-resistance formula in a small network calculator.


Solve series, parallel and series-parallel resistor networks for total resistance, current and voltage.

Series and parallel

Series and Parallel Circuits

Equivalent Resistance of Complex Circuits - Resistors In Series and Parallel Combinations

## Why it matters

Series and parallel reduction stops working as soon as a circuit has two sources or a resistor shared between two loops. Kirchhoff's two laws work on any circuit, however tangled. Engineers use them to write the equations that simulators solve, technicians use them to check whether a set of measurements adds up, and they are the foundation of mesh and nodal analysis that you will meet in power and electronics courses.

## Kirchhoff's current law (KCL)

**The total current flowing into a node equals the total current flowing out.** A node is any point where two or more components meet. Charge does not pile up in a wire, so what goes in must come out.

```text
If 3 A and 2 A flow into a node, and one wire leaves it,
that wire carries 3 + 2 = 5 A out.
```

A common way to write it: the currents into a node, counted positive in and negative out, sum to zero.

## Kirchhoff's voltage law (KVL)

**Around any closed loop, the voltage rises equal the voltage drops.** Equivalently, the signed voltages around a loop sum to zero. Energy given to charge by sources is all used by the loads before the charge returns to where it started.

To apply it, pick a direction to walk around the loop and stick to it:

- Walking through a battery from − to +: a **rise** of its voltage.
- Walking through a resistor in the same direction as the current you assumed: a **drop** of `I × R`.
- Walking against the assumed current: a rise of `I × R`.

You do not need to guess current directions correctly. If you guess wrong, the answer simply comes out negative, which tells you the current flows the other way.

## Mesh analysis: a two-loop example

Mesh analysis applies KVL with a loop current for each "window" of the circuit.

The circuit: a 12 V battery V1 on the left side and a 6 V battery V2 on the right side, both with their + terminal at the top. Along the top, R1 = 2 Ω sits on the left and R2 = 4 Ω on the right. A shared resistor R3 = 6 Ω runs vertically down the middle.

```text
   +--- R1 (2 Ω) ---+--- R2 (4 Ω) ---+
   |                |                |
  V1 12 V       R3 (6 Ω)          V2 6 V
   |                |                |
   +----------------+----------------+
```

Assign a clockwise mesh current **I1** to the left window and **I2** to the right window. R3 carries both: its current is `I1 − I2` (downward).

**Loop 1 (left, clockwise):** rise through V1, drop across R1, drop across R3.

```text
V1 = I1·R1 + (I1 − I2)·R3
12 = 8·I1 − 6·I2
```

**Loop 2 (right, clockwise):** up through R3, across R2, then down through V2 from + to −, which is a drop of 6 V.

```text
0 = (I2 − I1)·R3 + I2·R2 + V2
−6 = −6·I1 + 10·I2
```

Two equations, two unknowns. Solve by substitution or Cramer's rule:

```text
det = (8)(10) − (−6)(−6) = 80 − 36 = 44
I1 = (12·10 − (−6)(−6)) / 44 = 84 / 44 = 1.909 A
I2 = (8·(−6) − (−6)(12)) / 44 = 24 / 44 = 0.545 A
I_R3 = I1 − I2 = 1.364 A (downward)
```

**Check with KCL and Ohm's law.** The voltage at the top middle node is `12 − 1.909 × 2 = 8.18 V`. Then R3 carries `8.18 / 6 = 1.364 A` and R2 carries `(8.18 − 6) / 4 = 0.545 A`. At the top node, 1.909 A comes in and 1.364 + 0.545 = 1.909 A goes out. KCL holds.

Notice what I2 being positive means: current flows *into* the + terminal of V2. The 12 V battery is charging the 6 V one. Real batteries do not like that, which is one reason chargers include current limiting.

## A general recipe

1. Label every component and value; draw a mesh current in each window, all clockwise.
2. Write one KVL equation per mesh. Shared resistors use the difference of the two mesh currents.
3. Collect terms into the form `a·I1 + b·I2 = c`.
4. Solve the system (by hand for two loops; with software for more).
5. Find branch currents from the mesh currents, then check with KCL at a node.

## Common mistakes

- **Changing direction partway round a loop.** Pick one direction and follow it all the way.
- **Sign errors on sources.** Ask: am I going from − to + (rise) or + to − (drop)?
- **Using I1 alone for a shared resistor.** A resistor between two meshes carries `I1 − I2`.
- **Panicking at a negative answer.** It only means the real current is opposite to your arrow.
- **Not checking.** A KCL check at one node catches most arithmetic slips.

## Recap

- KCL: current in = current out at every node.
- KVL: rises = drops around every closed loop.
- Mesh analysis writes one KVL equation per window and solves the system.

## Try it

Open the **Do** tab and write the loop 1 equation for a two-mesh solver in Python.


Use Kirchhoff's current and voltage laws and mesh analysis to solve a two-loop circuit.

Kirchhoff's laws

Kirchhoff's Voltage Law - KVL Circuits, Loop Rule & Ohm's Law - Series Circuits, Physics

Mesh Analysis for Circuits Explained

Voltage, current, resistance and power; series and parallel networks; and Kirchhoff's laws for multi-loop circuits.