StudyVault
Home Unit Overview Exam Technique Revision Techniques ← Previous Lesson Next Lesson →

← Back to Dashboard

Overlay:
0:00 / 0:00 Narration generated using AI text-to-speech.

Series and Parallel Circuits

Series Circuits

In a series circuit, all components are connected in a single loop. There is only one path for current to flow, so the current is the same at every point in the circuit.

Key Fact

In a series circuit: current is the same everywhere, potential differences add up to equal the supply voltage, and total resistance equals the sum of all individual resistances: \(R_{\text{total}} = R_1 + R_2 + R_3\).

The potential difference is shared between components in series. The larger the resistance of a component, the greater its share of the total voltage. If two identical resistors are in series, each gets half the supply voltage.

A 12 V battery is connected in series with a 4 Ω resistor and an 8 Ω resistor. Calculate the current and the voltage across each resistor.

Step 1: Total resistance \(R_{\text{total}} = 4 + 8 = 12 \; \Omega\)

Step 2: Current \(I = \frac{V}{R} = \frac{12}{12} = 1\) A (same throughout)

Step 3: \(V_1 = IR_1 = 1 \times 4 = 4\) V across the 4 Ω resistor

Step 4: \(V_2 = IR_2 = 1 \times 8 = 8\) V across the 8 Ω resistor

Check: \(4 + 8 = 12\) V, which equals the supply voltage.

Parallel Circuits

In a parallel circuit, components are connected on separate branches. Each branch provides a different path for the current to flow through.

Key Fact

In a parallel circuit: potential difference is the same across each branch, currents in the branches add up to give the total current from the supply, and total resistance is less than the smallest individual resistance.

Adding more resistors in parallel decreases the total resistance, because there are more paths for the current. This means the total current from the battery increases.

Two resistors (6 Ω and 3 Ω) are connected in parallel across a 12 V supply. Calculate the current in each branch and the total current.

Branch 1: \(I_1 = \frac{V}{R_1} = \frac{12}{6} = 2\) A

Branch 2: \(I_2 = \frac{V}{R_2} = \frac{12}{3} = 4\) A

Total current: \(I_{\text{total}} = 2 + 4 = 6\) A

Investigating Components

You need to know the I-V characteristics of three components: fixed resistors (ohmic conductors), filament lamps, and diodes.

A fixed resistor gives a straight line through the origin on an I-V graph — current is directly proportional to voltage (Ohm’s law). A filament lamp gives a curve: as current increases, the wire heats up and resistance increases. A diode only conducts in one direction — it has very high resistance in reverse bias.

To investigate the I-V characteristics of a component, connect it in series with a variable resistor, ammeter, and power supply. Connect a voltmeter in parallel across the component. Vary the resistance using the variable resistor to change the current and voltage. Record pairs of I and V values, then swap the power supply connections to get negative values. Plot V on the x-axis and I on the y-axis.

Two other important components are the thermistor and the LDR (light-dependent resistor). A thermistor’s resistance drops as temperature rises, making it useful in temperature sensors. An LDR’s resistance drops as light intensity increases, making it useful in automatic lighting circuits.

Practice Questions

Test yourself on this lesson. Pick a question, write your answer, then get it marked.

AI feedback
AI feedback is guidance, not a final exam mark. If you disagree, your teacher's view is always the tie-breaker.
← Back to unit
Video

© 2026 StudyVault · About · FAQ · Copyright & IP · Privacy · Disclaimers

StudyVault Education Ltd · registered in England and Wales, company no. 17490559 · 71–75 Shelton Street, London WC2H 9JQ