SEAB Original-style practice: Two resistors of 200\ and 300\ are connected in series across a 10\ V supply. Calculate (a) the total resistance, (b) the current from the supply, and (c) the voltage across the 300\ resistor.

(a) In series, resistances add: R_T = 200 + 300 = 500\. (b) The current from the supply is I = VR_T = 10500 = 0.020\ A = 20\ mA. (c) The same current flows through both resistors, so the voltage across the 300\ resistor is V = IR = 0.020 × 300 = 6.0\ V. What markers reward: adding resistances in series, using the total to find one common current, and applying V = IR to the chosen resistor. The two voltages must sum to 10\ V, a useful check.

SingaporeElectronicsSyllabus dot point

How do current, voltage and resistance behave differently when components are wired in series compared with in parallel?

Apply the rules for current, voltage and resistance in series and parallel circuits, and calculate combined resistance

A focused answer to the O-Level Electronics outcome on series and parallel circuits. The current, voltage and resistance rules for each, and how to calculate the combined resistance.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

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

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to apply the rules for current, voltage and resistance in series and in parallel circuits, and to calculate the combined resistance of resistors in each arrangement. The central insight is a clean pair of patterns: in series the current is shared in the sense of being common and the voltage divides, while in parallel the voltage is common and the current divides. Getting these the right way round is the foundation of every analogue circuit.

The answer

Series circuits

Components in series are connected end to end so there is only one path for the current. The rules are:

Current is the same at every point: $I_1 = I_2 = I_{total}$ .
Voltage is shared: the supply voltage equals the sum of the voltages across the components, $V_{total} = V_1 + V_2 + \dots$
Resistance adds: $R_T = R_1 + R_2 + \dots$

The total resistance of a series chain is always larger than the largest single resistor.

Parallel circuits

Components in parallel are connected across the same two points, so the current has more than one path. The rules are:

Voltage is the same across each branch: $V_1 = V_2 = V_{total}$ .
Current divides between the branches and recombines, so the supply current equals the sum of the branch currents, $I_{total} = I_1 + I_2 + \dots$
Resistance combines by reciprocals: $\dfrac{1}{R_T} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$

The total resistance of a parallel combination is always smaller than the smallest single resistor, because adding another path makes it easier for current to flow.

The two-resistor parallel shortcut

For exactly two resistors in parallel, the reciprocal rule simplifies to a product-over-sum form:

R_T = \frac{R_1 R_2}{R_1 + R_2}

This is quick and safe for the two-component cases that dominate the syllabus.

How current divides in parallel

Because each branch has the same voltage, $I = V/R$ tells you that the branch with the smaller resistance carries the larger current. The currents are in inverse proportion to the resistances, so a $3\ \Omega$ branch carries twice the current of a $6\ \Omega$ branch across the same voltage.

Worked example

A $4.0\ \Omega$ resistor and a $12\ \Omega$ resistor are connected in parallel, and this combination is in series with a $5.0\ \Omega$ resistor across a $12\ \text{V}$ supply. Find the total resistance and the current from the supply.

Step 1: Combine the parallel pair

Using the product-over-sum shortcut: $R_P = \dfrac{4.0 \times 12}{4.0 + 12} = \dfrac{48}{16} = 3.0\ \Omega$ .

Step 2: Check the parallel value is sensible

$3.0\ \Omega$ is less than the smaller branch ( $4.0\ \Omega$ ), which is exactly what a parallel combination must give.

Step 3: Add the series resistor

The parallel block is in series with the $5.0\ \Omega$ resistor, so the total resistance is $R_T = R_P + 5.0 = 3.0 + 5.0 = 8.0\ \Omega$ .

Step 4: Find the supply current

Apply Ohm's law to the whole circuit: $I = \dfrac{V}{R_T} = \dfrac{12}{8.0} = 1.5\ \text{A}$ .

Examples in context

Example 1. Christmas lights. Old fairy lights wired in series all go out if one bulb fails, because breaking the single path stops the common current. Modern lights wire bulbs in parallel so each has the full supply voltage and an independent path, meaning one failed bulb leaves the rest glowing. This is the series-versus-parallel choice made visible.

Example 2. Home wiring. Sockets and lights in a house are wired in parallel so that each appliance gets the full mains voltage and can be switched on or off without affecting the others. The total current drawn rises as more appliances are switched on, which is why circuits are protected by a fuse rated for the maximum safe current.

Try this

Cue. Three $90\ \Omega$ resistors are connected in series. Find the total resistance. In series they add: $R_T = 90 + 90 + 90 = 270\ \Omega$ .
Cue. Two $10\ \Omega$ resistors are connected in parallel. Find the combined resistance. Identical parallel resistors halve: $R_T = 10/2 = 5.0\ \Omega$ .
Cue. Explain why the total current rises when an extra lamp is added in parallel to a supply. Each added parallel branch lowers the total resistance, so by $I = V/R$ the supply delivers more total current.

Exam-style practice questions

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

Original4 marksTwo resistors of

200\ \Omega

and

300\ \Omega

are connected in series across a

10\ \text{V}

supply. Calculate (a) the total resistance, (b) the current from the supply, and (c) the voltage across the

300\ \Omega

resistor.

Show worked answer →

(a) In series, resistances add: $R_T = 200 + 300 = 500\ \Omega$ .

(b) The current from the supply is $I = \dfrac{V}{R_T} = \dfrac{10}{500} = 0.020\ \text{A} = 20\ \text{mA}$ .

(c) The same current flows through both resistors, so the voltage across the $300\ \Omega$ resistor is $V = IR = 0.020 \times 300 = 6.0\ \text{V}$ .

What markers reward: adding resistances in series, using the total to find one common current, and applying $V = IR$ to the chosen resistor. The two voltages must sum to $10\ \text{V}$ , a useful check.

Original3 marksTwo resistors of

6.0\ \Omega

and

3.0\ \Omega

are connected in parallel. Calculate their combined resistance and state how the supply current divides between them.