Skip to main content
WalesElectronicsSyllabus dot point

How do Ohm's law, Kirchhoff's laws and the potential divider let you analyse any DC circuit?

DC electrical circuits: charge, current, voltage and resistance, Ohm's law, series and parallel resistors, Kirchhoff's current and voltage laws, the potential divider, and power.

A focused answer to the WJEC A-Level Electronics core concept of DC electrical circuits, covering current, voltage and resistance, Ohm's law, series and parallel combinations, Kirchhoff's two laws, the potential divider equation, and electrical power.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

Every circuit you analyse in WJEC Electronics rests on the same handful of DC rules. The specification expects you to define current, voltage and resistance, apply Ohm's law, combine resistors in series and parallel, use Kirchhoff's two laws to find unknown currents and voltages, set an output with a potential divider, and work out power. The potential divider in particular is the single most reused circuit on the whole course, because it is how every sensor turns a resistance change into a voltage.

The answer

Current, voltage and resistance

Ohm's law states that the current through an ohmic conductor is proportional to the voltage across it at constant temperature, so V=IRV = IR with RR constant. Many electronic components (diodes, transistors, lamps) are non-ohmic, so Ohm's law is a tool for resistors and metal conductors, not a universal rule.

Combining resistors

In series the same current flows through every resistor and the voltages add. In parallel the same voltage appears across every resistor and the currents add. The parallel total is always smaller than the smallest individual resistor.

Kirchhoff's laws

These two laws let you find every current and voltage in a network, even one too tangled to reduce by simple series and parallel rules.

The potential divider

Two resistors in series across a supply split the supply voltage in the ratio of their resistances. The output is taken across R2R_2. This is the workhorse of input sub-systems: replace R2R_2 with a thermistor or LDR and the output voltage tracks temperature or light.

Power

Power is the rate of energy transfer, in watts. The three forms are interchangeable: pick the one that uses the quantities you already know. Power matters for choosing component ratings, since a resistor that dissipates more than its rating overheats.

Examples in context

Example 1. Setting a comparator reference
A fixed potential divider of two equal resistors gives exactly half the supply, a stable reference voltage for a comparator. Because the divider current is small, the reference barely changes when the comparator input draws a tiny current, which is why dividers are the standard way to set thresholds.
Example 2. Limiting current to an LED
An LED dropping 2.0V2.0\,\text{V} from a 5.0V5.0\,\text{V} supply needs a series resistor to take the remaining 3.0V3.0\,\text{V}. Ohm's law sizes it: for 15mA15\,\text{mA}, R=3.00.015=200ΩR = \frac{3.0}{0.015} = 200\,\Omega. Power in the resistor, P=I2R=0.045WP = I^2 R = 0.045\,\text{W}, confirms a standard quarter-watt resistor is fine.
Example 3. Sharing current in parallel loads
Two equal resistive heaters in parallel across a supply each see the full supply voltage, so each draws the same current and the total current doubles. Kirchhoff's first law accounts for it: the supply current equals the sum of the branch currents.

Try this

Q1. A 3.0kΩ3.0\,\text{k}\Omega and a 6.0kΩ6.0\,\text{k}\Omega resistor are in parallel. Find the combined resistance. [2 marks]

  • Cue. 1R=13.0+16.0=36.0kΩ1\frac{1}{R} = \frac{1}{3.0} + \frac{1}{6.0} = \frac{3}{6.0}\,\text{k}\Omega^{-1}, so R=2.0kΩR = 2.0\,\text{k}\Omega.

Q2. A resistor carries 20mA20\,\text{mA} with 5.0V5.0\,\text{V} across it. Find its resistance and the power it dissipates. [3 marks]

  • Cue. R=VI=5.00.020=250ΩR = \frac{V}{I} = \frac{5.0}{0.020} = 250\,\Omega; P=VI=5.0×0.020=0.10WP = VI = 5.0 \times 0.020 = 0.10\,\text{W}.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC Eduqas 20194 marksA potential divider is made from a 10kΩ10\,\text{k}\Omega fixed resistor in series with a thermistor across a 9.0V9.0\,\text{V} supply. The output is taken across the thermistor. When the thermistor resistance is 5.0kΩ5.0\,\text{k}\Omega, calculate the output voltage and the power dissipated in the fixed resistor.
Show worked answer →

Use the potential divider equation, then power in the fixed resistor.

Output across the thermistor: Vout=VSRthRth+R1=9.0×5.05.0+10=9.0×13=3.0VV_{out} = V_S \dfrac{R_{th}}{R_{th} + R_1} = 9.0 \times \dfrac{5.0}{5.0 + 10} = 9.0 \times \dfrac{1}{3} = 3.0\,\text{V}.

The current is the same everywhere in the series chain: I=VSRth+R1=9.015×103=0.60mAI = \dfrac{V_S}{R_{th} + R_1} = \dfrac{9.0}{15 \times 10^3} = 0.60\,\text{mA}.

Power in the fixed resistor: P=I2R1=(0.60×103)2×10×103=3.6mWP = I^2 R_1 = (0.60 \times 10^{-3})^2 \times 10 \times 10^3 = 3.6\,\text{mW}.

Markers reward the divider ratio with the thermistor on top, the series current, and the power from P=I2RP = I^2 R.

WJEC Eduqas 20223 marksAt a junction, 12mA12\,\text{mA} flows in along one wire and 5mA5\,\text{mA} flows out along a second wire. State Kirchhoff's first law and use it to find the current in the third wire and its direction.
Show worked answer →

Kirchhoff's first law: the total current flowing into a junction equals the total current flowing out, because charge is conserved.

In: 12mA12\,\text{mA}. Out so far: 5mA5\,\text{mA}. The remaining current must leave the junction along the third wire.

Current in the third wire: 125=7mA12 - 5 = 7\,\text{mA}, flowing out of the junction.

Markers reward the conservation-of-charge statement, the value 7mA7\,\text{mA}, and the direction (out of the junction).

Related dot points

Sources & how we know this