PhysicsICSE Class 10 · Ohm's Law and Resistances

Resistances in Series and Parallel: Derivations and Network Problems (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

In series the same current flows through every resistor and the pds add, so R = R₁ + R₂ + R₃. In parallel every resistor has the same pd and the currents add, so 1/R = 1/R₁ + 1/R₂ + 1/R₃. Series raises the total resistance; parallel lowers it below the smallest resistor.

Understand Resistances in Series and Parallel

Series connection. Resistors are in series when they are joined end to end, so that there is only one path for the current.

The same current $I$ flows through each resistor.
The pd divides: $V = V_{1} + V_{2} + V_{3}$ .

Derivation (series). By Ohm's law $V_{1} = I R_{1}$ , $V_{2} = I R_{2}$ , $V_{3} = I R_{3}$ . If $R_{s}$ is the equivalent resistance, $V = I R_{s}$ . Then

$I R_{s} = I R_{1} + I R_{2} + I R_{3} \Rightarrow R_{s} = R_{1} + R_{2} + R_{3} .$

The equivalent resistance is greater than the largest individual resistance. The pd across each resistor is in proportion to its resistance: the larger resistor gets the larger share.

Parallel connection. Resistors are in parallel when each is connected across the same two points, giving separate paths for the current.

The same pd $V$ is across each resistor.
The current divides: $I = I_{1} + I_{2} + I_{3}$ .

Derivation (parallel). $I_{1} = \frac{V}{R _{1}}$ , $I_{2} = \frac{V}{R _{2}}$ , $I_{3} = \frac{V}{R _{3}}$ and $I = \frac{V}{R _{p}}$ . So

$\frac{V}{R _{p}} = \frac{V}{R _{1}} + \frac{V}{R _{2}} + \frac{V}{R _{3}} \Rightarrow \frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}} .$

The equivalent resistance is less than the smallest individual resistance. For two resistors, $R_{p} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$ (product over sum). For $n$ equal resistors $R$ : series gives $n R$ , parallel gives $\frac{R}{n}$ .

Current division in parallel. The pd is common, so the current in each branch is inversely proportional to its resistance: the smaller resistance takes the larger current. For two branches, $I_{1} = I \times \frac{R _{2}}{R _{1} + R _{2}}$ . If two branches have equal resistance, the current simply splits equally; no long calculation is needed.

Comparing series and parallel.

Series	Parallel
One path for current	Several paths
Same current in each resistor	Same pd across each resistor
pds add	Currents add
$R_{s} = R_{1} + R_{2} + \dots$ (larger)	$\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots$ (smaller)
If one breaks, all stop	Others keep working

Household connection. Home appliances are connected in parallel: each gets the full mains voltage, each can be switched independently, and a fault in one does not switch off the others. Old decorative bulb strings were in series: one fused bulb put out the whole string.

Solving a simple network.
1. Spot groups that are clearly in series (one path) or in parallel (same two end points).
2. Replace each group by its equivalent, working from the innermost group outwards.
3. Find the total current from the supply: $I = \frac{V}{R _{total}}$ (add internal resistance if given).
4. Work back: find the pd across each group, then the current in each branch.
Redraw an unusual diagram if needed: resistors are in parallel only if both their ends join the same two points.

Resistances in Series and Parallel: formulas and key facts

Series: $R_{s} = R_{1} + R_{2} + R_{3}$; Same current; pds add.
Parallel: $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$; Same pd; currents add.
Two in parallel: $R_{p} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$; Product over sum.
Current division (two branches): $I_{1} = I \times \frac{R _{2}}{R _{1} + R _{2}}$; Smaller resistance takes the larger current.
n equal resistors R: $R_{series} = n R, R_{parallel} = \frac{R}{n}$; Ratio series : parallel = n² : 1.

How to solve Resistances in Series and Parallel questions

Reduce the network step by step, then work back to individual currents and pds.

1Mark the two supply terminals and trace each path; identify series and parallel groups.
2Replace parallel groups with $\frac{R _{1} R _{2}}{R _{1} + R _{2}}$ (or the reciprocal sum) and add series resistances.
3Find the total current $I = \frac{V}{R _{total}}$ , adding the cell's internal resistance to $R_{total}$ if given.
4Find the pd across a series part ( $I R$ ) or across a parallel group ( $I \times R_{group}$ ), then each branch current ( $\frac{V}{R}$ ).
5Check: currents into a junction equal currents out; pds round the circuit add up to the supply voltage. Give units (Ω, A, V).

Resistances in Series and Parallel in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Why homes are wired in parallel

Suppose your room has a 220 V supply feeding a lamp of resistance 1100 Ω and a fan of resistance 550 Ω.

Find the current in each and the total current if they are in parallel, and say what would go wrong if they were in series.

Try it first, then show the solutionSolution

Parallel: lamp $\frac{220}{1100} = 0.2$ A; fan $\frac{220}{550} = 0.4$ A; total $0.6$ A. Each gets the full 220 V.
Series: they would share 220 V (the lamp about 147 V, the fan about 73 V), neither would work properly, and switching one off would switch off both.

Answer: 0.2 A and 0.4 A, total 0.6 A in parallel; in series both would be under-run and could not be switched separately.

Memory hook: Parallel: full voltage for everyone.

Making a resistor you do not have

Imagine a robotics club needs a 4 Ω resistor for a sensor circuit, but the kit has only 6 Ω resistors.

How can they make 4 Ω from three 6 Ω resistors?

Try it first, then show the solutionSolution

Two 6 Ω in series give 12 Ω.
12 Ω in parallel with the third 6 Ω: $\frac{12 \times 6}{12 + 6} = 4$ Ω.

Answer: Join two in series and put the third in parallel with them.

Memory hook: Series to grow, parallel to shrink.

Tricks to remember Resistances in Series and Parallel

Short memory aids. Say each one aloud once, then use it on a question the same day.

Series: Same current; Parallel: Pd same
S-S and P-P: Series has the Same current; Parallel has the same Pd.
Use it when: Knowing what is shared.
Product over sum
For two in parallel, multiply on top, add below.
Use it when: Quick two-resistor parallel.
Below the smallest
A parallel combination is always less than its smallest member: a quick answer check.
Use it when: Spotting errors such as 1/16 + 1/16 = 8 Ω written as 1/8 = 8.
Lazy current
Current prefers the easier path: more current through the smaller resistance.
Use it when: Comparing branch currents.

Resistances in Series and Parallel: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Three equal resistors in parallel · 1 mark · MCQBoard level

Three 9 Ω resistors are joined side by side between the same two points. Together they behave like a single resistor of:
(a) 27 Ω
(b) 9 Ω
(c) 3 Ω
(d) 1 Ω

Try it first, then show the solutionSolution

$n$ equal resistors in parallel: $\frac{R}{n} = \frac{9}{3} = 3$ Ω.

Answer: (c) 3 Ω

Example 2: Deriving the series formula · 3 marks · Short answerBoard level

Three resistors $R_{1}$ , $R_{2}$ and $R_{3}$ are connected in series to a cell of pd $V$ . Derive an expression for their equivalent resistance.

Try it first, then show the solutionSolution

In series the same current $I$ flows through each resistor, and the total pd is the sum: $V = V_{1} + V_{2} + V_{3}$ . (1)
By Ohm's law, $V_{1} = I R_{1}$ , $V_{2} = I R_{2}$ , $V_{3} = I R_{3}$ , and $V = I R_{s}$ . (1)
So $I R_{s} = I (R_{1} + R_{2} + R_{3})$ , giving $R_{s} = R_{1} + R_{2} + R_{3}$ . (1)

Answer: $R_{s} = R_{1} + R_{2} + R_{3}$

Example 3: Three resistors in series · 3 marks · Short answerHarder board level

Resistors of 5 Ω, 10 Ω and 15 Ω are joined end to end and connected to a 12 V supply. Calculate (a) their total resistance, (b) the circuit current, (c) the pd across each resistor.

Try it first, then show the solutionSolution

(a) $R_{s} = 5 + 10 + 15 = 30$ Ω. (1)
(b) $I = \frac{12}{30} = 0.4$ A. (1)
(c) $V_{5} = 0.4 \times 5 = 2$ V; $V_{10} = 4$ V; $V_{15} = 6$ V (check: $2 + 4 + 6 = 12$ V). (1)

Answer: (a) 30 Ω (b) 0.4 A (c) 2 V, 4 V and 6 V

Example 4: Two in parallel · 3 marks · Short answerHarder board level

A 12 Ω resistor and a 4 Ω resistor are each connected directly across the terminals of a 9 V battery of negligible internal resistance. Calculate (a) their combined resistance, (b) the battery current, (c) the current through each resistor.

Try it first, then show the solutionSolution

(a) $R_{p} = \frac{12 \times 4}{12 + 4} = 3$ Ω. (1)
(b) $I = \frac{9}{3} = 3$ A. (1)
(c) $I_{12} = \frac{9}{12} = 0.75$ A; $I_{4} = \frac{9}{4} = 2.25$ A (check: $0.75 + 2.25 = 3$ A). (1)

Answer: (a) 3 Ω (b) 3 A (c) 0.75 A and 2.25 A

Example 5: A simple network · 4 marks · Short answerHarder board level

An 8 V supply drives current through a 2 Ω resistor and then through a pair of resistors, 10 Ω and 15 Ω, joined side by side. Find (a) the total resistance, (b) the current from the supply, (c) the pd across the parallel part, (d) the current in the 10 Ω resistor.

Try it first, then show the solutionSolution

(a) Parallel part: $\frac{10 \times 15}{10 + 15} = 6$ Ω; total $= 2 + 6 = 8$ Ω. (1)
(b) $I = \frac{8}{8} = 1$ A. (1)
(c) $V = 1 \times 6 = 6$ V (the other 2 V is across the 2 Ω). (1)
(d) $I_{10} = \frac{6}{10} = 0.6$ A (and 0.4 A in the 15 Ω). (1)

Answer: (a) 8 Ω (b) 1 A (c) 6 V (d) 0.6 A

Example 6: Four values from three resistors · 4 marks · Short answerHOTS / competency

You have three 6 Ω resistors. Showing how they are connected, obtain an equivalent resistance of (a) 18 Ω, (b) 2 Ω, (c) 9 Ω, (d) 4 Ω.

Try it first, then show the solutionSolution

(a) All three in series: $6 + 6 + 6 = 18$ Ω. (1)
(b) All three in parallel: $\frac{6}{3} = 2$ Ω. (1)
(c) Two in parallel (3 Ω) in series with the third: $3 + 6 = 9$ Ω. (1)
(d) Two in series (12 Ω) in parallel with the third: $\frac{12 \times 6}{12 + 6} = 4$ Ω. (1)

Answer: (a) Series (b) Parallel (c) (6 ∥ 6) + 6 (d) (6 + 6) ∥ 6

Example 7: Case study: cutting a wire · 4 marks · Case-basedHOTS / competency

A uniform wire of resistance 24 Ω is cut into four equal pieces.
(a) What is the resistance of each piece?
(b) Find the equivalent resistance when all four pieces are joined in parallel.
(c) The pieces are instead joined in two pairs, each pair in series, and the two pairs are connected in parallel. Find the equivalent resistance.
(d) The all-parallel arrangement is connected to a 3 V cell of negligible internal resistance. Find the total current.

Try it first, then show the solutionSolution

(a) Resistance is proportional to length: $\frac{24}{4} = 6$ Ω each. (1)
(b) Four equal resistors in parallel: $\frac{6}{4} = 1.5$ Ω. (1)
(c) Each pair: $6 + 6 = 12$ Ω; two 12 Ω in parallel: $6$ Ω. (1)
(d) $I = \frac{3}{1.5} = 2$ A. (1)

Answer: (a) 6 Ω (b) 1.5 Ω (c) 6 Ω (d) 2 A

Where students lose marks in Resistances in Series and Parallel

The slips examiners see most often, and what to write instead.

Mistake: Mixing up R and 1/R in the parallel formula
The parallel formula gives $\frac{1}{R}$ , not $R$ .
Wrong working
$\frac{1}{20} + \frac{1}{20} = \frac{1}{10} = 10$ Ω (a false step, since $\frac{1}{10} \neq = 10$ ), or forgetting to invert and giving $R = 0.1$ Ω.
Right working
$\frac{1}{R} = \frac{1}{20} + \frac{1}{20} = \frac{1}{10}$ , so $R = 10$ Ω.
Do this: Write ' $\frac{1}{R} =$ ' on the left and invert at the end; check the result is below the smallest resistor.
Usually costs: CISCE examiners reported steps such as 1/16 + 1/16 = 1/8 = 8 Ω and missing units in 2024 Q8(iii).
Mistake: Long calculations for equal branches
If two parallel branches have equal resistance, the current divides equally.
Wrong working
Working out each branch current with fresh pds and resistances.
Right working
'Both branches are 16 Ω, so the current divides equally.'
Do this: Look for symmetry before calculating.
Usually costs: Reported by CISCE examiners for 2024 Q8(iii)(c).
Mistake: Misreading an unusual diagram
Resistors are in parallel only if both their ends join the same two points, whatever way they are drawn.
Wrong working
Calling two resistors drawn one above the other 'series'.
Right working
Trace each end: same two junctions means parallel.
Do this: Redraw the circuit in a standard layout.
Usually costs: Reported by CISCE examiners for 2025 Q1(xii).
Mistake: Vague answers to 'increases, decreases or remains same'
Removing one of several series bulbs lowers the total resistance, so current and brightness increase.
Wrong working
'It changes' or 'it becomes more bright and less resisted'.
Right working
'Resistance decreases; intensity of light increases.'
Do this: Use exactly the words offered in the question.
Usually costs: Reported by CISCE examiners for 2024 Q2(vi).

Exam tips

When asked to derive, start from what is common (current in series, pd in parallel) and show the Ohm's-law step.
Give every resistance in Ω, current in A and pd in V; units carry marks.
After finding a parallel equivalent, check it is smaller than the smallest resistor in the group.
For networks, write the order of reduction in words: 'the 10 Ω and 15 Ω are in parallel, giving 6 Ω, which is in series with 2 Ω'.

Practice questions on Resistances in Series and Parallel

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In a series circuit, the quantity that is the same for every resistor is the:
(a) pd
(b) current
(c) power
(d) resistance
Hint
One path only.
Answer
(b) current
Q2 · 1 mark · Very short answerBoard level
Find the equivalent resistance of 20 Ω and 30 Ω in parallel.
Hint
Product over sum.
Answer
$\frac{20 \times 30}{20 + 30} = 12$ Ω.
Q3 · 2 marks · Short answerHarder board level
Resistors of 2 Ω, 4 Ω and 6 Ω are in series with a 6 V battery. Find the current and the pd across the 4 Ω resistor.
Hint
Total first.
Answer
$R = 12$ Ω, $I = \frac{6}{12} = 0.5$ A; $V_{4} = 0.5 \times 4 = 2$ V.
Q4 · 2 marks · Short answerHarder board level
Two resistors in parallel have an equivalent resistance of 2.4 Ω. One of them is 4 Ω. Find the other.
Hint
$\frac{1}{R _{2}} = \frac{1}{2.4} - \frac{1}{4}$ .
Answer
$\frac{1}{R _{2}} = \frac{1}{2.4} - \frac{1}{4} = \frac{1}{6}$ , so $R_{2} = 6$ Ω.
Q5 · 3 marks · Short answerBoard level
Derive the expression for the equivalent resistance of three resistors connected in parallel.
Hint
Same pd; currents add.
Answer
Same pd $V$ across each; $I = I_{1} + I_{2} + I_{3}$ . With $I_{1} = \frac{V}{R _{1}}$ etc. and $I = \frac{V}{R _{p}}$ : $\frac{V}{R _{p}} = \frac{V}{R _{1}} + \frac{V}{R _{2}} + \frac{V}{R _{3}}$ , so $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ .
Q6 · 3 marks · Short answerHarder board level
A 3 Ω resistor is in series with two 6 Ω resistors that are in parallel with each other. The combination is connected to a 12 V supply. Find (a) the total resistance, (b) the supply current, (c) the current in each 6 Ω resistor.
Hint
Equal branches split the current equally.
Answer
(a) $3 + \frac{6}{2} = 6$ Ω. (b) $\frac{12}{6} = 2$ A. (c) The branches are equal, so 1 A each.
Q7 · 2 marks · Short answerHarder board level
Two identical bulbs are connected (i) in series, (ii) in parallel, to the same battery. In which case does the battery supply more current, and in which case does each bulb glow brighter? Give reasons.
Hint
Compare total resistances and pd across each bulb.
Answer
Parallel: the total resistance is a quarter of the series value, so the battery supplies more current; each bulb gets the full battery pd, so each glows brighter than in series, where each gets only half.
Q8 · 4 marks · Long answerHOTS / competency
A 16 V supply is connected to a 5 Ω resistor followed by a 4 Ω and a 12 Ω resistor joined side by side. Find (a) the total resistance, (b) the supply current, (c) the pd across the 5 Ω resistor and across the parallel part, (d) the currents in the 4 Ω and 12 Ω resistors.
Hint
Reduce, then work back.
Answer
(a) $5 + \frac{4 \times 12}{16} = 5 + 3 = 8$ Ω. (b) $\frac{16}{8} = 2$ A. (c) $V_{5} = 10$ V; parallel part $2 \times 3 = 6$ V. (d) $I_{4} = \frac{6}{4} = 1.5$ A; $I_{12} = \frac{6}{12} = 0.5$ A.
Q9 · 1 mark · Very short answerHarder board level
How many 8 Ω resistors must be connected in parallel to give an equivalent resistance of 2 Ω?
Hint
$\frac{R}{n}$ .
Answer
$\frac{8}{n} = 2$ , so $n = 4$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Equivalent of R₁ and R₂ in series?Show answer
$R_{1} + R_{2}$ .
2.Equivalent of R₁ and R₂ in parallel?Show answer
$\frac{R _{1} R _{2}}{R _{1} + R _{2}}$ .
3.What is common in a parallel combination?Show answer
The pd.
4.n equal resistors R in parallel?Show answer
$\frac{R}{n}$ .
5.How are household appliances connected?Show answer
In parallel.

Resistances in Series and Parallel: frequently asked questions

What is the formula for resistances in series and in parallel?

Series: R = R₁ + R₂ + R₃. Parallel: 1/R = 1/R₁ + 1/R₂ + 1/R₃; for two resistors this gives R = R₁R₂/(R₁ + R₂).

Why is the equivalent resistance in parallel less than each resistor?

Each extra parallel branch provides another path for current, so for the same pd more total current flows. More current for the same pd means a smaller equivalent resistance.

Why are household appliances connected in parallel?

So each gets the full mains voltage, each can be switched on or off independently, and a fault in one does not cut off the others.