ScienceCBSE Class 10 · Electricity

Resistors in Series and Parallel: Class 10 Explained

Updated 14 October 2026 · Fact-checked

Resistors joined end to end are in series: the same current flows through each and the total resistance is the sum. Resistors joined between the same two points are in parallel: each gets the same voltage and the reciprocals add. Expect numericals on mixed circuits, reasons for parallel home wiring and the practical.

In series one current passes through every resistor and resistances add; in parallel each branch has the same voltage and the current splits.

Illustration drawn for QuestionBuddy.

Understand Resistors in Series and Parallel

Most gadgets use more than one resistor, so you need to know how resistors behave together. There are only two basic ways to join them. In series, resistors are joined end to end, so the charge has just one path. In parallel, each resistor is connected between the same two points, so the current has several separate paths.

In series, there is nowhere else for charge to go, so the current is the same through every resistor (Activity 11.4: the ammeter reads the same wherever you put it). The battery's potential difference is shared out: $V = V_{1} + V_{2} + V_{3}$ (Activity 11.5). Using Ohm's law on each resistor, $I R_{s} = I R_{1} + I R_{2} + I R_{3}$ , and cancelling $I$ gives $R_{s} = R_{1} + R_{2} + R_{3}$ . Charge must cross every obstacle one after another, so $R_{s}$ is bigger than the largest resistor. The larger resistor also takes the larger share of voltage, because $V = I R$ with the same $I$ .

In parallel, every branch is joined across the same two points, so the potential difference is the same across each (Activity 11.6). The total current splits: $I = I_{1} + I_{2} + I_{3}$ . Using $I = V / R$ for each branch and for the whole, then cancelling $V$ , gives $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ . Each extra branch opens a new path, so more current flows for the same voltage, and $R_{p}$ is smaller than the smallest resistor.

This topic is really Ohm's law applied twice: once to each resistor and once to the single equivalent resistor that could replace the group without changing $V$ or $I$ . For a mixed circuit, reduce it step by step: combine each parallel group, then add the series parts.

It explains daily life too. A series circuit gives every device the same current. This is a poor choice for a bulb and a heater together, because they need very different currents, and if one part fails the whole circuit stops, as with a string of fairy lights. Parallel wiring gives each device the full supply voltage, lets each draw the current it needs, lowers total resistance and keeps the others working if one fails. That is why homes are wired in parallel.

Resistors in Series and Parallel: formulas and key facts

Series: equivalent resistance: $R_{s} = R_{1} + R_{2} + R_{3}$; $R_{s}$ is greater than the largest individual resistor.
Series: current and voltage: $I same in all; V = V_{1} + V_{2} + V_{3}$; Each share is $V_{1} = I R_{1}$ , so voltage divides in proportion to resistance.
Parallel: equivalent resistance: $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$; Invert at the end. $R_{p}$ is less than the smallest resistor.
Two resistors in parallel (shortcut): $R_{p} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$; Follows from the reciprocal rule; works for two resistors only.
Parallel: current and voltage: $V same across all; I = I_{1} + I_{2} + I_{3}$; Each branch current is $I_{1} = V / R_{1}$ ; the smaller resistor takes the bigger current.
n identical resistors: $R_{s} = n R; R_{p} = \frac{R}{n}$; So $R_{s} / R_{p} = n^{2}$ . Useful for 'wire cut into n equal parts' questions.

How to solve Resistors in Series and Parallel questions

Use this method for any series, parallel or mixed circuit numerical.

1Read the circuit in words and mark which resistors share the same current (series) and which share the same two end points (parallel).
2Combine each parallel group into one resistor using the reciprocal rule, then invert.
3Add the series parts to get the total equivalent resistance of the circuit.
4Find the battery current with $I = V / R_{t o t a l}$ .
5Work back: p.d. across a series part is $I R$ ; p.d. across a parallel group is $I \times R_{p}$ ; each branch current is that p.d. divided by the branch resistance.
6Check: branch currents add to the total, and voltages round the series path add to the battery voltage. Write units on every answer.

Resistors 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.

Resistors in parallel and their use in daily life

Imagine your school science lab has one 220 V supply line protected for a maximum current of 5 A. The teacher wants to run several identical 220 V, 44 W lamps on it, all in parallel, for a night exhibition.

Find the resistance of one lamp, the current each lamp draws, the most lamps that can be run together, and the equivalent resistance of that many lamps in parallel.

Try it first, then show the solutionSolution

Current per lamp: $I = P / V = 44 W /220 V = 0.2 A$ .
Resistance of one lamp: $R = V / I = 220 V /0.2 A = 1100 Ω$ .
In parallel, currents add: number of lamps $= 5 A /0.2 A = 25$ .
Equivalent resistance: $R_{p} = R / n = 1100 Ω/25 = 44 Ω$ .
Check: $I = 220 V /44 Ω = 5 A$ , matching the limit.

Answer: Each lamp is $1100 Ω$ and draws 0.2 A; at most 25 lamps; their equivalent resistance is $44 Ω$ .

Memory hook: Every lamp added in parallel adds its own current to the line.

Resistors in series

Suppose a toy shop sells a small 3 V torch bulb that needs 0.25 A, but you only have a 9 V battery at home. You plan to put a resistor in series with the bulb so it does not burn out.

Find the resistance of the bulb, the resistor needed in series, and the p.d. across the resistor.

Try it first, then show the solutionSolution

Bulb resistance: $R_{b} = 3 V /0.25 A = 12 Ω$ .
Total resistance needed for 0.25 A from 9 V: $R_{s} = 9 V /0.25 A = 36 Ω$ .
Series resistor: $R = 36 Ω - 12 Ω = 24 Ω$ .
P.d. across resistor: $V = 0.25 A \times 24 Ω = 6 V$ ; check $3 + 6 = 9 V$ .

Answer: Bulb $12 Ω$ ; series resistor $24 Ω$ ; p.d. across it 6 V.

Memory hook: In series the extra resistor soaks up the extra voltage.

Tricks to remember Resistors in Series and Parallel

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

Series Shares Voltage, Parallel Shares Current
S-S-V: Series Shares Voltage (current same). P-S-C: Parallel Shares Current (voltage same).
Use it when: Decide instantly which quantity is common before you write any equation.
Bigger than biggest, smaller than smallest
Series total is bigger than the biggest; parallel total is smaller than the smallest.
Use it when: Sanity-check every answer and eliminate MCQ options fast.
Product over sum
For two in parallel: product over sum, $\frac{R _{1} R _{2}}{R _{1} + R _{2}}$ .
Use it when: Quick parallel pairs like 6 and 3 giving 2.
n squared
n equal pieces: series to parallel divides resistance by $n^{2}$ .
Use it when: Wire-cut questions: cut into 5 and join in parallel gives R/25.

Resistors in Series and Parallel: solved examples

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

Example 1: Wire cut and joined in parallel · 1 mark · MCQBoard level

A wire of resistance $20 Ω$ is cut into four equal parts, and the parts are joined in parallel. The equivalent resistance is:
(a) $80 Ω$
(b) $5 Ω$
(c) $1.25 Ω$
(d) $0.8 Ω$

Try it first, then show the solutionSolution

Each part has a quarter of the length, so $R = 20/4 = 5 Ω$ ; four equal parts in parallel: $R_{p} = 5/4 = 1.25 Ω$ (1 mark).

Answer: (c) $1.25 Ω$

Example 2: Arranging three resistors · 3 marks · Short answerBoard level

You are given three resistors of $4 Ω$ , $12 Ω$ and $6 Ω$ . Using all three each time, show how you would connect them to get a total resistance of (a) $8 Ω$ and (b) $2 Ω$ . Show the working.

Try it first, then show the solutionSolution

(a) $12 Ω$ and $6 Ω$ in parallel: $R_{p} = \frac{12 \times 6}{12 + 6} = 4 Ω$ (1 mark).
This $4 Ω$ in series with the $4 Ω$ resistor: $4 + 4 = 8 Ω$ (½ mark).
(b) All three in parallel: $\frac{1}{R _{p}} = \frac{1}{4} + \frac{1}{12} + \frac{1}{6} = \frac{3 + 1 + 2}{12} = \frac{6}{12}$ (1 mark).
So $R_{p} = 2 Ω$ (½ mark).

Answer: (a) 12 Ω and 6 Ω in parallel, that pair in series with 4 Ω. (b) All three in parallel.

Example 3: Mixed circuit: currents and voltages · 5 marks · Long answerHarder board level

A $4 Ω$ resistor is connected in series with a parallel combination of a $12 Ω$ and a $6 Ω$ resistor. The arrangement is connected to a 12 V battery with a plug key. Find (a) the equivalent resistance, (b) the current from the battery, (c) the p.d. across the $4 Ω$ resistor and across the parallel part, (d) the current through the $12 Ω$ and $6 Ω$ resistors.

Try it first, then show the solutionSolution

Parallel part: $R_{p} = \frac{12 \times 6}{12 + 6} = 4 Ω$ (1 mark).
Total: $R = 4 + 4 = 8 Ω$ (½ mark).
Battery current: $I = 12 V /8 Ω = 1.5 A$ (1 mark).
P.d. across $4 Ω$ : $1.5 \times 4 = 6 V$ ; across parallel part: $1.5 \times 4 = 6 V$ (1 mark).
Current in $12 Ω$ : $6/12 = 0.5 A$ ; in $6 Ω$ : $6/6 = 1 A$ (1 mark).
Check: $0.5 + 1 = 1.5 A$ and $6 + 6 = 12 V$ (½ mark).

Answer: (a) $8 Ω$ (b) 1.5 A (c) 6 V and 6 V (d) 0.5 A and 1 A.

Example 4: Network of five resistors: current and p.d. between points · 5 marks · Long answerHarder board level

The circuit in words: between points A and B, resistors of $6 Ω$ and $3 Ω$ are joined in parallel. Between points B and C, three resistors of $12 Ω$ , $6 Ω$ and $12 Ω$ are joined in parallel. A 15 V battery and a plug key are connected across A and C. Find (a) the equivalent resistance between A and B and between B and C, (b) the total current from the battery, (c) the p.d. between A and B and between B and C, (d) the current through the $6 Ω$ resistor between B and C.

Try it first, then show the solutionSolution

$R_{A B} = \frac{6 \times 3}{6 + 3} = 2 Ω$ (1 mark).
$\frac{1}{R _{B C}} = \frac{1}{12} + \frac{1}{6} + \frac{1}{12} = \frac{1 + 2 + 1}{12} = \frac{4}{12}$ , so $R_{B C} = 3 Ω$ (1 mark).
The two groups are in series: $R = 2 + 3 = 5 Ω$ ; $I = 15 V /5 Ω = 3 A$ (1 mark).
$V_{A B} = 3 A \times 2 Ω = 6 V$ ; $V_{B C} = 3 A \times 3 Ω = 9 V$ ; check $6 + 9 = 15 V$ (1 mark).
Current in the $6 Ω$ between B and C: $9/6 = 1.5 A$ ; check: each $12 Ω$ takes $9/12 = 0.75 A$ , and $0.75 + 1.5 + 0.75 = 3 A$ (1 mark).

Answer: (a) $2 Ω$ and $3 Ω$ (b) 3 A (c) 6 V and 9 V (d) 1.5 A.

Example 5: Derive series combination formula · 5 marks · Long answerHarder board level

(a) Draw a labelled circuit diagram of three resistors $R_{1}$ , $R_{2}$ and $R_{3}$ connected in series with a battery, a plug key and an ammeter, with a voltmeter to measure the p.d. across the combination. Use it to derive the expression for the equivalent resistance. (b) Three resistors of $2 Ω$ , $4 Ω$ and $6 Ω$ are joined in series to a 6 V battery. Find the current and the p.d. across the $6 Ω$ resistor.

Try it first, then show the solutionSolution

Draw and label: battery and plug key; ammeter in series; $R_{1}$ , $R_{2}$ , $R_{3}$ joined end to end between points X and Y; voltmeter across X and Y; arrow for current I from the positive terminal through the resistors to the negative terminal; mark $V_{1}$ , $V_{2}$ , $V_{3}$ across each resistor. Slip to avoid: drawing the voltmeter in series or drawing the resistors side by side between the same two points (1 mark).
There is only one path, so the same current I flows through each resistor; the total p.d. is shared: $V = V_{1} + V_{2} + V_{3}$ (1 mark).
Ohm's law on each resistor: $V_{1} = I R_{1}$ , $V_{2} = I R_{2}$ , $V_{3} = I R_{3}$ ; for the equivalent resistor $V = I R_{s}$ (1 mark).
Substitute: $I R_{s} = I R_{1} + I R_{2} + I R_{3}$ ; cancel I: $R_{s} = R_{1} + R_{2} + R_{3}$ (1 mark).
(b) $R_{s} = 2 + 4 + 6 = 12 Ω$ ; $I = 6 V /12 Ω = 0.5 A$ ; p.d. across $6 Ω$ $= 0.5 A \times 6 Ω = 3 V$ (1 mark).

Answer: $R_{s} = R_{1} + R_{2} + R_{3}$ ; in (b) the current is 0.5 A and the p.d. across the $6 Ω$ resistor is 3 V.

Example 6: Derive parallel combination formula · 5 marks · Long answerHarder board level

(a) With a labelled circuit diagram, derive the expression for the equivalent resistance of three resistors connected in parallel. (b) State two advantages of connecting household appliances in parallel rather than in series.

Try it first, then show the solutionSolution

Draw and label: battery, plug key and ammeter in series; three resistors $R_{1}$ , $R_{2}$ , $R_{3}$ each joined between common points X and Y; voltmeter across X and Y; arrow for current from the positive terminal; currents $I_{1}$ , $I_{2}$ , $I_{3}$ marked in branches. Slip to avoid: drawing the voltmeter in series (1 mark).
Same p.d. V across each resistor; total current divides: $I = I_{1} + I_{2} + I_{3}$ (1 mark).
Ohm's law on each: $I_{1} = V / R_{1}$ , $I_{2} = V / R_{2}$ , $I_{3} = V / R_{3}$ , and for the whole $I = V / R_{p}$ (1 mark).
Substitute: $V / R_{p} = V / R_{1} + V / R_{2} + V / R_{3}$ , cancel V: $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ (1 mark).
Advantages: each appliance gets full supply voltage and draws the current it needs; if one fails or is switched off, others keep working (1 mark).

Answer: $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ ; parallel wiring gives full voltage to each device, lets each take its own current and work independently.

Example 7: Practical 5: equivalent resistance in series and parallel · 3 marks · Short answerBoard level

In the practical to find the equivalent resistance of two resistors, a student uses $R_{1} = 4 Ω$ and $R_{2} = 12 Ω$ . Her observation table reads:

Series: V = 3.2 V, I = 0.20 A
Parallel: V = 3.0 V, I = 1.0 A

(a) Outline how the two circuits are set up and the readings taken. (b) Find the experimental resistance in each case and compare it with the value from the formulas. (c) State two precautions.

Try it first, then show the solutionSolution

Set-up: join $R_{1}$ and $R_{2}$ end to end (series), then between the same two points (parallel); each time connect a battery, plug key and ammeter in series and a voltmeter across the combination; plug the key and note V and I (1 mark).
Experimental: series $R = 3.2 V /0.20 A = 16 Ω$ ; parallel $R = 3.0 V /1.0 A = 3 Ω$ (½ mark).
Formulas: $R_{s} = 4 + 12 = 16 Ω$ ; $R_{p} = \frac{4 \times 12}{4 + 12} = 3 Ω$ ; the measured values agree with these (½ mark).
Precautions: keep connections tight and clean; take the key out after each reading so the resistors do not heat up and change their resistance (1 mark).

Answer: Series 16 Ω and parallel 3 Ω, both matching the values from the formulas.

Example 8: Practical data: find the error · 4 marks · Case-basedHOTS / competency

In the practical on equivalent resistance, a student uses resistors $R_{1} = 3 Ω$ and $R_{2} = 6 Ω$ with a 6 V battery. For the series set-up she records V = 6 V and I = 0.67 A. For the parallel set-up she records V = 6 V and I = 1.0 A. She concludes the parallel combination has the larger resistance.
(a) Calculate the expected readings for both cases.
(b) Which recorded value is wrong and why can you be sure?
(c) State one precaution for this experiment.

Try it first, then show the solutionSolution

Series: $R_{s} = 9 Ω$ , $I = 6/9 = 0.67 A$ , so this reading fits (1 mark).
Parallel: $R_{p} = \frac{3 \times 6}{9} = 2 Ω$ , $I = 6/2 = 3 A$ (1 mark).
The 1.0 A parallel reading is wrong; parallel resistance must be less than $3 Ω$ , so current must exceed $6/3 = 2 A$ . Her conclusion is wrong (1 mark).
Precaution: keep connections tight and take the key out after each reading so resistors do not heat up; ammeter in series, voltmeter in parallel (1 mark).

Answer: Expected: series 0.67 A ( $9 Ω$ ), parallel 3 A ( $2 Ω$ ). The parallel reading of 1.0 A is wrong; parallel resistance is smaller.

Where students lose marks in Resistors in Series and Parallel

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

Mistake: Forgetting to invert in the parallel formula
Students stop after adding reciprocals.
Wrong working
$\frac{1}{R _{p}} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} = \frac{1}{3}$ , so $R_{p} = \frac{1}{3} Ω$
Right working
$\frac{1}{R _{p}} = \frac{1}{3}$ , so $R_{p} = 3 Ω$
Do this: Always write the last line as 'R_p = ...' by flipping the fraction.
Usually costs: The answer mark and every later step that uses it.
Mistake: Using full battery voltage across one series resistor
Confusing series with parallel.
Wrong working
$20 Ω$ and $4 Ω$ in series on 6 V: $I = 6/4 = 1.5 A$ in the $4 Ω$
Right working
$R_{s} = 24 Ω$ , $I = 6/24 = 0.25 A$ through both
Do this: In series, find the one common current from the total resistance first.
Usually costs: 1-2 marks in a 3-mark numerical.
Mistake: Adding reciprocals of only some resistors in a mixed circuit
Not identifying groups before calculating.
Wrong working
Treating $4 Ω$ in series with ( $12 Ω ∥ 6 Ω$ ) as all three in parallel
Right working
$12 ∥ 6 = 4 Ω$ , then $4 + 4 = 8 Ω$
Do this: Label each group as series or parallel in words before writing a formula.
Usually costs: Most of a 5-mark question.
Mistake: Product-over-sum used for three resistors
Overextending the shortcut.
Wrong working
$R_{p} = \frac{2 \times 3 \times 6}{2 + 3 + 6} = \frac{36}{11} Ω$
Right working
$\frac{1}{R _{p}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$ , $R_{p} = 1 Ω$
Do this: Use product over sum for two resistors only; combine pairwise otherwise.
Usually costs: The calculation and answer marks.

Exam tips

In derivations, state the key fact first (same current in series, same p.d. in parallel), then Ohm's law for each resistor and for the equivalent one; each line earns a mark.
Check your answer against 'bigger than biggest' for series and 'smaller than smallest' for parallel before moving on.
For circuit-drawing questions, draw the ammeter in series and the voltmeter in parallel, show the plug key and mark current direction from the positive terminal.
In 'give reasons' questions on home wiring, mention at least two points: full voltage to each device, different currents, independent switching and failure.

Practice questions on Resistors 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 · 3 marks · Short answerHarder board level
A $3 Ω$ resistor is connected in series with a parallel combination of a $4 Ω$ and a $12 Ω$ resistor, across a 12 V battery. Find the power used in each of the three resistors. Which one produces the most heat per second, and what is the ratio of the power in the $3 Ω$ resistor to that in the $4 Ω$ resistor?
Hint
Find the battery current first; use $P = I^{2} R$ for the series resistor and $P = V^{2} / R$ for the parallel ones.
Answer
$R_{p} = \frac{4 \times 12}{16} = 3 Ω$ , total $6 Ω$ , $I = 12/6 = 2 A$ . $P_{3} = 2^{2} \times 3 = 12 W$ . P.d. across the parallel part $= 2 \times 3 = 6 V$ , so $P_{4} = 36/4 = 9 W$ and $P_{12} = 36/12 = 3 W$ . The $3 Ω$ resistor produces most heat; ratio $12 : 9 = 4 : 3$ . Check: $12 + 9 + 3 = 24 W = V I = 12 \times 2$ .
Q2 · 3 marks · Short answerHarder board level
Four coils of $3 Ω$ , $6 Ω$ , $9 Ω$ and $18 Ω$ are available. (a) Using all four each time, find the highest and the lowest total resistance you can get. (b) If each arrangement is connected to a 9 V battery, find the current drawn in each case.
Hint
Highest: all in series. Lowest: all in parallel; use LCM 18.
Answer
Highest $= 3 + 6 + 9 + 18 = 36 Ω$ , current $9/36 = 0.25 A$ . Lowest: $\frac{1}{R _{p}} = \frac{6 + 3 + 2 + 1}{18} = \frac{12}{18}$ , so $R_{p} = 1.5 Ω$ , current $9/1.5 = 6 A$ .
Q3 · 3 marks · Short answerHarder board level
A heater has three coils A, B and C of $22 Ω$ each. Coil A is connected in parallel with the series pair B and C, and the arrangement is joined to a 220 V supply. Find the total current, the current in and power used by each coil, and say which coil heats the most. Check that the total power equals VI.
Hint
Coil A gets the full 220 V; B and C share 220 V equally.
Answer
A: $I_{A} = \frac{220}{22} = 10$ A, $P_{A} = 220 \times 10 = 2200$ W. B and C in series: $44 Ω$ , $I = \frac{220}{44} = 5$ A, so each has 110 V and uses $110 \times 5 = 550$ W. Total current $= 10 + 5 = 15$ A; total power $= 2200 + 550 + 550 = 3300$ W $= 220 \times 15$ . Coil A heats the most (four times as much as B or C).
Q4 · 4 marks · Case-basedHOTS / competency
Read the passage and answer the questions.
Imagine a lab board on which a $4 Ω$ resistor is joined in series with a parallel pair made of a $6 Ω$ resistor and an unknown resistor R. The whole arrangement is connected to an 18 V battery through a plug key. A student finds that the $6 Ω$ resistor uses a power of 6 W.
(a) Find the current through, and the p.d. across, the $6 Ω$ resistor.
(b) Find the p.d. across the $4 Ω$ resistor and the current from the battery.
(c) Find the current through R and the value of R.
(d) Check your answer by finding the equivalent resistance of the whole circuit.
Hint
Use $P = I^{2} R$ for the $6 Ω$ resistor; the parallel pair shares one p.d.
Answer
(a) $I^{2} \times 6 = 6$ , so $I = 1 A$ ; $V = 1 \times 6 = 6 V$ . (b) $18 - 6 = 12 V$ across $4 Ω$ ; battery current $= 12/4 = 3 A$ . (c) Current in R $= 3 - 1 = 2 A$ ; $R = 6/2 = 3 Ω$ . (d) $6$ and $3 Ω$ in parallel give $2 Ω$ , total $6 Ω$ , $I = 18/6 = 3 A$ , which matches.
Q5 · 3 marks · Short answerHarder board level
Two resistors give $25 Ω$ in series and $4 Ω$ in parallel. Find each resistance. A third identical resistor to the larger one is then added in parallel to the pair; find the new equivalent resistance.
Hint
Product = R_p × R_s; find two numbers with that sum and product.
Answer
$R_{1} R_{2} = 4 \times 25 = 100$ , $R_{1} + R_{2} = 25$ , so $5 Ω$ and $20 Ω$ . Adding another $20 Ω$ : $\frac{1}{R} = \frac{4 + 1 + 1}{20} = \frac{6}{20}$ , $R = \frac{10}{3} \approx 3.33 Ω$ .
Q6 · 1 mark · MCQBoard level
Resistors of $6 Ω$ and $3 Ω$ are joined in series across a 9 V battery. If the $3 Ω$ resistor is replaced by a $6 Ω$ resistor, the current from the battery
(a) increases to 1.5 A
(b) decreases to 0.75 A
(c) stays at 1 A
(d) decreases to 0.5 A
Hint
In series, a larger resistor makes the total larger.
Answer
(b). At first $I = \frac{9}{9} = 1$ A. Afterwards the total is $12 Ω$ , so $I = \frac{9}{12} = 0.75$ A. (Option (a) is the trap of thinking as if the resistors were in parallel.)
Q7 · 1 mark · MCQBoard level
n identical resistors, each of resistance R, give a total $R_{s}$ when joined in series and $R_{p}$ when joined in parallel. The ratio $\frac{R _{s}}{R _{p}}$ is
(a) $n$
(b) $n^{2}$
(c) $\frac{1}{n ^{2}}$
(d) $1$
Hint
$R_{s} = n R$ and $R_{p} = \frac{R}{n}$ .
Answer
(b) $n^{2}$ , since $\frac{n R}{R / n} = n^{2}$ .
Q8 · 1 mark · Assertion-reasonBoard level
Assertion (A): The bulbs in a house are connected in parallel.
Reason (R): In parallel, each bulb gets the full supply voltage and can be switched on or off without affecting the others.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Think about what happens to the other bulbs when one is switched off.
Answer
(a). Both are true, and R is the reason houses are wired in parallel: every bulb works at 220 V and draws its own current, and each works independently.
Q9 · 2 marks · Very short answerBoard level
Give two reasons why a series arrangement is not used for the appliances in a house.
Hint
Think about a broken bulb and about the voltage each appliance gets.
Answer
(i) In series, if one appliance fails or is switched off, the circuit is broken and all the others stop working (1 mark). (ii) The supply voltage is shared, so no appliance gets the full 220 V, and the same current flows through all, though a bulb and a heater need very different currents (1 mark).

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.What stays the same through every resistor in series?Show answer
The current.
2.What stays the same across every resistor in parallel?Show answer
The potential difference.
3.Formula for series equivalent resistance?Show answer
$R_{s} = R_{1} + R_{2} + R_{3}$
4.Formula for parallel equivalent resistance?Show answer
$\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$
5.Equivalent of n identical resistors R in parallel?Show answer
$R / n$
6.Why do fairy lights in series all go off when one bulb fuses?Show answer
The single path breaks, so current stops everywhere.

Resistors in Series and Parallel: frequently asked questions

Is the derivation of series and parallel resistance in the Class 10 syllabus?

The NCERT chapter derives both formulas from Ohm's law, so learn the derivations as well as the formulas.

Why is total resistance in parallel less than the smallest resistor?

Each branch gives charge an extra path, so more current flows for the same voltage. More current at the same voltage means less resistance overall.

Why are homes wired in parallel and not in series?

In parallel, every appliance gets the full supply voltage and draws the current it needs. Each can be switched on or off on its own, and if one fails the rest keep working.

What do I do in the series and parallel practical?

You connect two resistors in series, then in parallel, with a battery, plug key, ammeter in series and voltmeter across the combination. You find resistance as V divided by I and compare it with the value from the formulas.

Is the flow of charges inside a wire box examined?

No. That box item has been dropped and is not in the 2026-27 syllabus.