Circuits

Series and parallel circuits, Kirchhoff's laws, and resistor combinations

SPEC 3.5.1.4 • ELECTRICITY

Circuit Symbols

You must recognise and draw standard circuit symbols accurately in exams.

LED

Series Circuits

In a series circuit, components are connected one after another in a single loop.

Series Circuit Rules

Current: The same through all components: I₁ = I₂ = I₃ = I

Voltage: Shared between components: V = V₁ + V₂ + V₃

Resistance: Adds up: R_total = R₁ + R₂ + R₃

Formula

R_{t o t a l} = R_{1} + R_{2} + R_{3} + ...

01Worked Example

Two resistors of 6 Ω and 12 Ω are connected in series to a 9 V battery. Calculate:

(a) the total resistance

(b) the current in the circuit

(c) the p.d. across each resistor

a)

For series: R_total = R₁ + R₂ = 6 + 12 = 18 Ω

b)

I = V / R_total = 9 / 18 = 0.5 A

c)

V₁ = IR₁ = 0.5 × 6 = 3 V

V₂ = IR₂ = 0.5 × 12 = 6 V

Check: V₁ + V₂ = 3 + 6 = 9 V ✓

Parallel Circuits

In a parallel circuit, components are connected across the same two points, providing multiple paths for current.

Parallel Circuit Rules

Voltage: The same across all branches: V₁ = V₂ = V₃ = V

Current: Splits between branches: I = I₁ + I₂ + I₃

Resistance: Combined using reciprocals

Formula

\frac{1}{R _{t o t a l}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}} + ...

Worked Example02

Two resistors of 6 Ω and 12 Ω are connected in parallel. Calculate the total resistance.

1

Use the parallel resistance formula:

\frac{1}{R _{t o t a l}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

2

Substitute values:

\frac{1}{R _{t o t a l}} = \frac{1}{6} + \frac{1}{12}

3

Find common denominator:

\frac{1}{R _{t o t a l}} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}

4

Take reciprocal: R_total = 4 Ω

Note: Parallel resistance is always less than the smallest individual resistor.

03Worked Example

Three resistors of 4 Ω, 6 Ω, and 12 Ω are connected in parallel to a 12 V supply. Calculate:

(a) the total resistance

(b) the total current from the supply

(c) the current through each resistor

a)

\frac{1}{R _{T}} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} = \frac{3}{12} + \frac{2}{12} + \frac{1}{12} = \frac{6}{12} = \frac{1}{2}

R_total = 2 Ω

b)

I_total = V / R_total = 12 / 2 = 6 A

c)

Each branch has 12 V across it:

I₁ = 12/4 = 3 A

I₂ = 12/6 = 2 A

I₃ = 12/12 = 1 A

Check: 3 + 2 + 1 = 6 A ✓

Cells in Series and Parallel

Total: 6V

Voltages ADD:

V_total = V₁ + V₂ + V₃

Total: 2V

Voltage SAME as one cell

But battery lasts longer!

Why Use Parallel Cells?

Parallel cells provide the same voltage but can supply current for longer — useful when you need extended battery life without increasing voltage.

04Worked Example

Four identical cells, each with EMF 1.5 V, are connected. Calculate the total EMF when they are connected:

(a) all in series

(b) all in parallel

(c) two pairs in series, with the pairs connected in parallel

a)

In series, EMFs add up:

Total EMF = 1.5 + 1.5 + 1.5 + 1.5 = 6.0 V

b)

In parallel, EMF stays the same as one cell:

Total EMF = 1.5 V

(But the combination can supply current for 4× longer)

c)

Each series pair: 1.5 + 1.5 = 3.0 V

Two 3.0 V pairs in parallel: Total EMF = 3.0 V

(This gives higher voltage than parallel alone, and longer life than series alone)

Kirchhoff's First Law

The sum of currents entering a junction equals the sum of currents leaving the junction.

Σ I_{in} = Σ I_{o u t}

This law is based on conservation of charge — charge cannot be created or destroyed at a junction.

At the junction: I₁ = I₂ + I₃

Exam Tip

Kirchhoff's First Law applies at any junction in a circuit. Use it to find unknown currents when you know other branch currents.

Worked Example05

At a junction in a circuit, three wires meet. Wire A carries 2.5 A into the junction. Wire B carries 1.0 A out of the junction. Calculate the current in wire C and state its direction.

1

Apply Kirchhoff's First Law:

Σ I_{in} = Σ I_{o u t}

2

Current in = Current out:

2.5 = 1.0 + I_{C}

3

Solve for I_C:

I_{C} = 2.5 - 1.0 = 1.5 A

Direction: Out of the junction (since total in must equal total out)

Worked Example06

In a circuit, four wires meet at junction P. Wire A carries 3.2 A into P. Wire B carries 1.8 A into P. Wire C carries 2.7 A out of P. Calculate the current in wire D, stating whether it flows into or out of the junction.

1

Apply Kirchhoff's First Law: Total current in = Total current out

2

Total current into junction:

I_{in} = 3.2 + 1.8 = 5.0 A

3

We know 2.7 A flows out through C. Let I_D be positive if flowing out:

5.0 = 2.7 + I_{D}

4

Solve:

I_{D} = 5.0 - 2.7 = 2.3 A out of the junction

Kirchhoff's Second Law

The sum of EMFs in any closed loop equals the sum of potential differences around that loop.

Σ ε = Σ I R

This law is based on conservation of energy — energy gained from sources equals energy dissipated in components.

Around the loop: ε = I₁R₁ + I₂R₂ + I₃R₃

Applying the Law

Choose a direction around the loop. Add EMFs if you go from − to + through them; subtract if + to −. Add IR drops in the direction of current flow.

Worked Example07

A 12 V battery is connected to two resistors in series: R₁ = 8 Ω and R₂ = 4 Ω. Use Kirchhoff's Second Law to find the current in the circuit.

1

Apply Kirchhoff's Second Law around the loop:

Σ ε = Σ I R

2

The EMF = 12 V. The p.d.s across resistors are IR₁ and IR₂:

12 = I \times 8 + I \times 4

3

Simplify:

12 = 12 I

4

Solve: I = 12/12 = 1 A

Worked Example08

A circuit contains two cells and two resistors in series. Cell 1 has EMF 9 V and cell 2 has EMF 3 V (connected in opposition). R₁ = 4 Ω and R₂ = 2 Ω. Use Kirchhoff's Second Law to find the current and the p.d. across each resistor.

1

Cells in opposition means EMFs subtract:

Σ ε = 9 - 3 = 6 V

2

Apply K2: Sum of EMFs = Sum of p.d.s

6 = I \times 4 + I \times 2 = 6 I

3

Current: I = 6/6 = 1 A

4

P.d. across R₁: V₁ = IR₁ = 1 × 4 = 4 V

P.d. across R₂: V₂ = IR₂ = 1 × 2 = 2 V

Check: 4 + 2 = 6 V ✓

Energy and Power in Circuits

When charge flows through a component, energy is transferred. The rate of energy transfer is power.

Energy Transferred

E = I V t

Energy (J) = Current (A) × Voltage (V) × Time (s)

Power

P = I V

Power (W) = Current (A) × Voltage (V)

Using Ohm's law (V = IR), we can derive two more useful forms:

Substitute V = IR

P = I^{2} R

Use when you know I and R

Substitute I = V/R

P = \frac{V ^{2}}{R}

Use when you know V and R

Which Formula?

Choose the formula based on what quantities you're given. All three give the same answer — pick the one that avoids extra calculation steps.

09Worked Example

A 12 V supply is connected to a 6 Ω resistor. Calculate:

(a) the current through the resistor

(b) the power dissipated in the resistor

(c) the energy transferred in 5 minutes

a)

Current: I = V/R = 12/6 = 2 A

b)

Power (using any formula):

P = IV = 2 × 12 = 24 W

Or: P = I²R = 2² × 6 = 24 W ✓

Or: P = V²/R = 12²/6 = 144/6 = 24 W ✓

c)

Time = 5 min = 5 × 60 = 300 s

Energy: E = Pt = 24 × 300 = 7200 J (or 7.2 kJ)

10

Worked Example

Two resistors, 4 Ω and 8 Ω, are connected in series to a 24 V supply. Calculate the power dissipated in each resistor.

1

Total resistance: R_total = 4 + 8 = 12 Ω

2

Current (same through both): I = V/R = 24/12 = 2 A

3

Power in 4 Ω resistor: P₁ = I²R₁ = 2² × 4 = 16 W

4

Power in 8 Ω resistor: P₂ = I²R₂ = 2² × 8 = 32 W

Note: The larger resistor dissipates more power (in series, where current is the same). Total power = 16 + 32 = 48 W = IV = 2 × 24 ✓

Combined Series-Parallel Circuits

Many circuits contain both series and parallel combinations. Solve these step by step.

Strategy

Identify which resistors are in series and which are in parallel

Simplify parallel combinations first to get equivalent resistances

Add series resistances to get total resistance

Calculate currents and voltages as needed

11

Worked Example

In the circuit shown, R₁ = 4 Ω is in series with a parallel combination of R₂ = 6 Ω and R₃ = 12 Ω. The supply voltage is 12 V. Calculate:

(a) the total resistance

(b) the current from the supply

1

First find the parallel combination of R₂ and R₃:

\frac{1}{R _{P}} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}

So R_P = 4 Ω

2

Now add R₁ in series: R_total = R₁ + R_P = 4 + 4 = 8 Ω

3

Calculate current: I = V / R_total = 12 / 8 = 1.5 A

12

Worked Example

A 24 V supply is connected to: R₁ = 6 Ω in series with a parallel combination of R₂ = 8 Ω and R₃ = 8 Ω. Calculate:

(a) the total resistance

(b) the current through R₁

(c) the current through R₂

a)

Parallel combination:

\frac{1}{R _{P}} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}

→ R_P = 4 Ω

Total: R_total = 6 + 4 = 10 Ω

b)

Total current (through R₁): I = V/R = 24/10 = 2.4 A

c)

P.d. across parallel section: V_P = IR_P = 2.4 × 4 = 9.6 V

Current through R₂: I₂ = V_P/R₂ = 9.6/8 = 1.2 A

Note: R₂ and R₃ are equal, so current splits equally (1.2 A each)

13

Worked Example

Three identical resistors, each of resistance R, are connected in parallel. The total resistance is 5 Ω. Find the value of R.

1

For three identical resistors R in parallel:

\frac{1}{R _{t o t a l}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}

2

Therefore:

R_{t o t a l} = \frac{R}{3}

3

Given R_total = 5 Ω:

5 = \frac{R}{3}

4

Solve: R = 5 × 3 = 15 Ω

Common Mistakes to Avoid

Warning 1

Forgetting the final reciprocal — When using the parallel formula, remember to take the reciprocal at the end!

If 1/R = 1/4, then R = 4 Ω, not 0.25 Ω

Warning 2

Misidentifying series and parallel — Components are in series if the same current flows through them. They're in parallel if they have the same voltage across them.

Warning 3

Thinking current is "used up" — Current is the same throughout a series circuit. It doesn't get smaller as it passes through resistors.

Quick Check

Parallel resistance is always less than the smallest individual resistor. If your answer is larger, you've made an error!

Practice

Question 1: Two resistors of 10 Ω and 15 Ω are connected in series with a 20 Ω resistor in parallel with them. Calculate the total resistance of the circuit.

1

First, the 10 Ω and 15 Ω in series: R_series = 10 + 15 = 25 Ω

2

This 25 Ω is in parallel with 20 Ω:

\frac{1}{R _{T}} = \frac{1}{25} + \frac{1}{20} = \frac{4}{100} + \frac{5}{100} = \frac{9}{100}

3

R_total = 100/9 = 11.1 Ω (3 s.f.)

Practice

Question 2: A current of 3.0 A flows into a junction where it splits into three branches. The currents in two branches are 1.2 A and 0.8 A. What is the current in the third branch?

1

Apply Kirchhoff's First Law: ΣIᵢₙ = ΣI_out

2

3.0 = 1.2 + 0.8 + I₃

3

I₃ = 3.0 − 1.2 − 0.8 = 1.0 A

Practice

Question 3: Four identical 100 Ω resistors are available. What arrangements give: (a) maximum resistance, (b) minimum resistance?

a)

Maximum: All four in series

R_max = 100 + 100 + 100 + 100 = 400 Ω

b)

Minimum: All four in parallel

\frac{1}{R _{min}} = \frac{1}{100} + \frac{1}{100} + \frac{1}{100} + \frac{1}{100} = \frac{4}{100}

R_min = 100/4 = 25 Ω

Summary

Key Formulas

SeriesParallel

R_total = R₁ + R₂ + R₃1/R_total = 1/R₁ + 1/R₂ + 1/R₃

Same current: I₁ = I₂ = I₃Same voltage: V₁ = V₂ = V₃

Voltages add: V = V₁ + V₂ + V₃Currents add: I = I₁ + I₂ + I₃

Kirchhoff's Laws

K1: ΣIᵢₙ = ΣI_out (charge conservation)

K2: Σε = ΣIR (energy conservation)

Power and Energy

Power: P = IV = I²R = V²/R

Energy: E = IVt = Pt

Exam Skills

Always identify series and parallel sections first

Simplify parallel combinations before adding series resistances

Check: parallel R < smallest individual R

Use Kirchhoff's laws for complex circuits

Choose the power formula that matches given quantities