Circuit analysis is the foundation of electrotechnological design. Understanding how to calculate voltage, current, and resistance in series and parallel circuits, apply Ohm’s law, and use Kirchhoff’s laws enables engineers to predict circuit behaviour, select components, and troubleshoot faults.
KEY TAKEAWAY: Series circuits share the same current; parallel circuits share the same voltage. These two rules, combined with Ohm’s law, solve most circuit problems.
Ohm’s Law states that the voltage across a conductor is directly proportional to the current flowing through it (at constant temperature):
$$V = IR$$
| Symbol | Quantity | Unit |
|---|---|---|
| $V$ | Voltage | Volt (V) |
| $I$ | Current | Ampere (A) |
| $R$ | Resistance | Ohm (Ω) |
Three forms:
$$V = IR \qquad I = \frac{V}{R} \qquad R = \frac{V}{I}$$
Worked example: A 12 V battery drives current through a 47 Ω resistor.
$$I = \frac{12}{47} = 0.255 \text{ A} = 255 \text{ mA}$$
EXAM TIP: Units must be consistent — convert mA to A and kΩ to Ω before substituting into Ohm’s law.
In a series circuit, components are connected end-to-end in a single path. The same current flows through every component.
Rules for series circuits:
Current is the same throughout:
$$I_{total} = I_1 = I_2 = I_3 = \ldots$$
Voltages add (voltage divides across components):
$$V_{total} = V_1 + V_2 + V_3 + \ldots$$
Resistances add:
$$R_{total} = R_1 + R_2 + R_3 + \ldots$$
Worked example: Three resistors 10 Ω, 20 Ω, and 30 Ω in series, connected to 12 V.
$$R_T = 10 + 20 + 30 = 60 \text{ Ω}$$
$$I = \frac{12}{60} = 0.2 \text{ A}$$
$$V_1 = 0.2 \times 10 = 2 \text{ V}, \quad V_2 = 4 \text{ V}, \quad V_3 = 6 \text{ V}$$
$$\text{Check: } 2 + 4 + 6 = 12 \text{ V} \checkmark$$
APPLICATION: Series circuits are used as voltage dividers — for example, a thermistor and fixed resistor in series create a voltage that changes with temperature, which a microcontroller can read as an analogue input.
In a parallel circuit, components are connected across the same two nodes — all have the same voltage.
Rules for parallel circuits:
Voltage is the same across each branch:
$$V_{total} = V_1 = V_2 = V_3 = \ldots$$
Currents add (total current splits between branches):
$$I_{total} = I_1 + I_2 + I_3 + \ldots$$
Reciprocal resistance rule:
$$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots$$
For two resistors only:
$$R_{total} = \frac{R_1 \times R_2}{R_1 + R_2}$$
Worked example: 30 Ω and 60 Ω in parallel, connected to 12 V.
$$R_T = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20 \text{ Ω}$$
$$I_{total} = \frac{12}{20} = 0.6 \text{ A}$$
$$I_1 = \frac{12}{30} = 0.4 \text{ A}, \quad I_2 = \frac{12}{60} = 0.2 \text{ A}$$
$$\text{Check: } 0.4 + 0.2 = 0.6 \text{ A} \checkmark$$
REMEMBER: The total resistance of a parallel combination is always less than the smallest individual resistor. Adding more parallel paths always reduces total resistance.
“The sum of currents entering a node equals the sum of currents leaving it.”
$$\sum I_{in} = \sum I_{out}$$
This is a consequence of conservation of charge — charge cannot accumulate at a node.
Example: At a junction, currents of 3 A and 2 A flow in. One branch carries 4 A out. The remaining branch carries:
$$I = 3 + 2 - 4 = 1 \text{ A out}$$
“The sum of all voltages around any closed loop in a circuit equals zero.”
$$\sum V = 0 \quad \text{(around any closed loop)}$$
This is a consequence of conservation of energy — energy supplied by sources equals energy consumed by components.
Sign convention: Voltage rises (going from − to + through a source) are positive; voltage drops across resistors are negative.
Worked example (single loop):
A 9 V battery, a 100 Ω resistor, and a 50 Ω resistor in series.
$\$9 - I(100) - I(50) = 0$$
$\$9 = 150I \quad \Rightarrow \quad I = 0.06 \text{ A}$$
$$V_{100\Omega} = 6 \text{ V}, \quad V_{50\Omega} = 3 \text{ V}$$
VCAA FOCUS: KVL is essential for circuits with multiple loops or multiple sources. Practice setting up loop equations systematically, assigning current directions before writing equations.
Many practical circuits combine series and parallel sections. Approach these by simplifying step by step:
Worked example: R1 = 10 Ω in series with the parallel combination of R2 = 20 Ω and R3 = 30 Ω. Supply = 15 V.
$$R_{23} = \frac{20 \times 30}{20 + 30} = 12 \text{ Ω}$$
$$R_{total} = 10 + 12 = 22 \text{ Ω}$$
$$I_{total} = \frac{15}{22} = 0.682 \text{ A}$$
$$V_{R1} = 0.682 \times 10 = 6.82 \text{ V}$$
$$V_{23} = 15 - 6.82 = 8.18 \text{ V (across both R2 and R3)}$$
$$I_2 = \frac{8.18}{20} = 0.409 \text{ A}, \quad I_3 = \frac{8.18}{30} = 0.273 \text{ A}$$
COMMON MISTAKE: In a series-parallel circuit, students sometimes try to use total current in a parallel branch. Remember: total current only flows through series components, not through individual parallel branches.
| Property | Series | Parallel |
|---|---|---|
| Current | Same everywhere | Splits between branches |
| Voltage | Divides (proportional to R) | Same across all branches |
| Total resistance | $R_T = R_1 + R_2 + \ldots$ | $1/R_T = 1/R_1 + 1/R_2 + \ldots$ |
| Effect of removing one component | All go off | Others unaffected |
| Example application | Voltage divider, LED + resistor | Mains household circuit, battery bank |
STUDY HINT: Always draw a clear circuit diagram before starting calculations. Label every component, junction, and current direction. Systematic labelling prevents most errors.