Chapter 3 · Electricity

3.3 Resistors in Series and in Parallel

Resistors in series share the same current and add up, $R_s=R_1+R_2+R_3$; resistors in parallel share the same potential difference and their reciprocals add, $\frac1{R_p}=\frac1{R_1}+\frac1{R_2}+\frac1{R_3}$. Any network built from the two can be replaced by one equivalent resistor.

Real circuits almost never contain just one resistor. A string of festival lights may hold fifty little lamps, a phone charger's circuit board holds dozens of resistors, and the wiring of a single flat feeds lamps, fans, a fridge and a geyser from the same two mains wires. To predict the current drawn from the supply — and the current and potential difference for each part — you need rules for combining resistors. There are only two basic ways to join two components: one after the other, so that the same charge must pass through both (series), or side by side between the same two points, so that the charge has a choice of paths (parallel).

In this section we find how resistances add in each case, using nothing more than Ohm's law, $V=IR$, and two facts about circuits: charge does not pile up anywhere, and the potential difference between two points is the same whichever path you follow between them. With the two rules you can reduce any series–parallel network, step by step, to a single equivalent resistance — and you will see why every house is wired in parallel while a cheap string of fairy lights is often wired in series.

📘 Definition 3.6 — Equivalent resistance

The equivalent resistance $R_{\text{eq}}$ of a combination of resistors between two points is the resistance of the single resistor that, connected between the same two points across the same potential difference $V$, would draw the same current $I$ from the supply: $R_{\text{eq}}=\dfrac VI$. Replacing the combination by $R_{\text{eq}}$ changes nothing for the rest of the circuit.

Resistors in Series

Resistors are in series when they are joined end to end, so that there is only one path for the charge through them. Connect resistors $R_1$, $R_2$ and $R_3$ in series to a battery, and put an ammeter anywhere in the loop — before $R_1$, between $R_1$ and $R_2$, after $R_3$. It reads the same everywhere: charge cannot be stored or lost along the way, so whatever flows into $R_1$ each second must flow out of it and on into $R_2$. Now use a voltmeter. The potential differences $V_1$, $V_2$, $V_3$ across the three resistors are different (the largest across the largest resistor), and they add up to the potential difference $V$ across the whole combination. Each coulomb of charge gets $V$ joules from the battery and gives up part of it in each resistor.

📐 Law 3.3 — Resistors in series

For resistors $R_1$, $R_2$, $R_3$ connected in series, (i) the same current $I$ flows through each, and (ii) the potential difference across the combination is the sum of the potential differences across the resistors, $V=V_1+V_2+V_3$. Applying Ohm's law to each resistor, $V_1=IR_1$, $V_2=IR_2$, $V_3=IR_3$, and to the equivalent resistor, $V=IR_s$:

$$IR_s=IR_1+IR_2+IR_3,$$ $$R_s=R_1+R_2+R_3 .$$

The equivalent resistance of resistors in series is the sum of their resistances, so it is larger than the largest of them. The battery's potential difference is shared in proportion to the resistances: $V_1=\dfrac{R_1}{R_s}V$, and so on.

Why it makes sense. Joining resistors in series is like joining wires end to end: the charge has a longer obstacle course to get through. From Section 3.2, a longer wire has a proportionally larger resistance — so lengths, and resistances, simply add. Series connection has one big drawback: if any one component breaks (a lamp filament burns out, a wire comes loose), the only path is broken and the current stops everywhere.

✏️ Example 1 — Three resistors in series
Resistors of $2.0\ \Omega$, $3.0\ \Omega$ and $5.0\ \Omega$ are connected in series to a $12\ \text{V}$ battery. Find (a) the equivalent resistance, (b) the current, and (c) the potential difference across each resistor.
(a) $R_s=2.0+3.0+5.0=10\ \Omega$.
(b) $I=\dfrac V{R_s}=\dfrac{12}{10}=1.2\ \text{A}$ — the same through every resistor.
(c) $V_1=1.2\times2.0=2.4\ \text{V}$, $V_2=1.2\times3.0=3.6\ \text{V}$, $V_3=1.2\times5.0=6.0\ \text{V}$.
Check $2.4+3.6+6.0=12\ \text{V}$, the battery's potential difference. ✓ The biggest resistor takes the biggest share.
✏️ Example 2 — A series resistor for a lamp
A small lamp is designed to work at $6.0\ \text{V}$, when it carries $0.50\ \text{A}$. You only have a $9.0\ \text{V}$ battery. What resistor must be connected in series with the lamp so that it works normally?
Idea In series the lamp and the resistor carry the same current, $0.50\ \text{A}$, and their potential differences add up to $9.0\ \text{V}$.
Share The resistor must take $9.0-6.0=3.0\ \text{V}$, so $R=\dfrac{3.0}{0.50}=6.0\ \Omega$.
Check Lamp resistance $\dfrac{6.0}{0.50}=12\ \Omega$; total $12+6.0=18\ \Omega$; $I=\dfrac{9.0}{18}=0.50\ \text{A}$. ✓

Resistors in Parallel

Resistors are in parallel when each one is connected directly between the same two points — the same two junctions. Now a voltmeter connected across $R_1$, across $R_2$ or across $R_3$ gives the same reading, $V$: each resistor is connected straight across the same two points. An ammeter in the main wire reads a current $I$; ammeters in the three branches read $I_1$, $I_2$, $I_3$, and the main current is their sum. Charge arriving at the junction splits between the branches — most of it through the branch with the smallest resistance — and joins up again at the far junction.

📐 Law 3.4 — Resistors in parallel

For resistors $R_1$, $R_2$, $R_3$ connected in parallel, (i) the same potential difference $V$ acts across each, and (ii) the total current is the sum of the branch currents, $I=I_1+I_2+I_3$. By Ohm's law $I_1=\dfrac V{R_1}$, $I_2=\dfrac V{R_2}$, $I_3=\dfrac V{R_3}$ and $I=\dfrac V{R_p}$, so

$$\frac V{R_p}=\frac V{R_1}+\frac V{R_2}+\frac V{R_3},$$ $$\frac1{R_p}=\frac1{R_1}+\frac1{R_2}+\frac1{R_3} .$$

The reciprocal of the equivalent resistance is the sum of the reciprocals, and $R_p$ is smaller than the smallest of the resistances. For two resistors, $R_p=\dfrac{R_1R_2}{R_1+R_2}$ ("product over sum"); for $n$ equal resistors $R$, $R_p=\dfrac Rn$.

Why it makes sense. Adding a branch in parallel is like making a wire thicker: it opens an extra lane for the charge. From Section 3.2, more cross-sectional area means less resistance — so every resistor added in parallel, however large, lowers the equivalent resistance and increases the current drawn from the supply. Note too that the branch currents are in the inverse ratio of the resistances: for two branches, $\dfrac{I_1}{I_2}=\dfrac{R_2}{R_1}$.

✏️ Example 3 — Two resistors in parallel
A $6.0\ \Omega$ and a $3.0\ \Omega$ resistor are connected in parallel across a $6.0\ \text{V}$ battery. Find the equivalent resistance, the current in each resistor and the current drawn from the battery.
$R_p$ $\dfrac1{R_p}=\dfrac16+\dfrac13=\dfrac{1+2}6=\dfrac36$, so $R_p=2.0\ \Omega$. (Product over sum: $\dfrac{6\times3}{6+3}=2.0\ \Omega$.)
Branches Each has $6.0\ \text{V}$ across it: $I_1=\dfrac{6.0}{6.0}=1.0\ \text{A}$, $I_2=\dfrac{6.0}{3.0}=2.0\ \text{A}$.
Total $I=1.0+2.0=3.0\ \text{A}$.
Check $\dfrac V{R_p}=\dfrac{6.0}{2.0}=3.0\ \text{A}$. ✓ The smaller resistor carries the larger current, in the ratio $\dfrac{I_2}{I_1}=\dfrac{6.0}{3.0}=2$.
✏️ Example 4 — Three branches
Resistors of $10\ \Omega$, $20\ \Omega$ and $20\ \Omega$ are connected in parallel to a $10\ \text{V}$ supply. Find the equivalent resistance and the total current.
$R_p$ $\dfrac1{R_p}=\dfrac1{10}+\dfrac1{20}+\dfrac1{20}=\dfrac{2+1+1}{20}=\dfrac15$, so $R_p=5.0\ \Omega$ — less than the smallest, $10\ \Omega$.
Current $I=\dfrac{10}{5.0}=2.0\ \text{A}$.
Check Branch currents $1.0$, $0.50$, $0.50\ \text{A}$ add up to $2.0\ \text{A}$. ✓
Interactive 1 — The Resistor Network Builder

Choose how three resistors are wired and set their values. The blue dots are free electrons: they move faster where the current is larger, split at every junction and join again. Each resistor glows according to the energy it converts every second, and is labelled with its own current and potential difference. Play cycles through the four ways of wiring them.

4612
$R_{\text{eq}}$—
battery current—
$R_1$—
$R_2$—
$R_3$—

Mixed Networks: Reducing Step by Step

Most circuits mix the two kinds of connection. The trick is never to try everything at once. Look for a group that is purely series (one path, nothing branching off in between) or purely parallel (every member joined directly between the same two junctions), replace that group by its single equivalent resistor, redraw, and repeat. Each step leaves a simpler circuit that behaves exactly like the original as far as the battery is concerned.

🧭 Problem-solving strategy — series–parallel networks
  1. Label the junctions (points where three or more wires meet). Two resistors are in parallel only if both their ends are joined to the same two junctions.
  2. Reduce from the inside out: combine the innermost pure series or pure parallel group, redraw, and repeat until one resistor $R_{\text{eq}}$ is left.
  3. Total current: $I=\dfrac V{R_{\text{eq}}}$ from the battery.
  4. Work back out: the current is the same through series parts; find the potential difference across each part with $V=IR$; in a parallel group every branch has that same potential difference, so each branch current is $\dfrac V{R_{\text{branch}}}$.
  5. Check: currents into a junction equal currents out; potential differences round any loop add up to the battery's.
✏️ Example 5 — A mixed circuit
A $4.0\ \Omega$ resistor $R_1$ is connected in series with a parallel pair, $R_2=6.0\ \Omega$ and $R_3=12\ \Omega$, across a $12\ \text{V}$ battery. Find the equivalent resistance, the battery current, and the current and potential difference for each resistor.
Parallel pair $R_{23}=\dfrac{6.0\times12}{6.0+12}=\dfrac{72}{18}=4.0\ \Omega$.
Series $R_{\text{eq}}=R_1+R_{23}=4.0+4.0=8.0\ \Omega$, so $I=\dfrac{12}{8.0}=1.5\ \text{A}$.
Work back $V_1=1.5\times4.0=6.0\ \text{V}$; the pair has $1.5\times4.0=6.0\ \text{V}$ across it, so $I_2=\dfrac{6.0}{6.0}=1.0\ \text{A}$ and $I_3=\dfrac{6.0}{12}=0.50\ \text{A}$.
Check $I_2+I_3=1.5\ \text{A}=I$ ✓ and $V_1+V_{23}=6.0+6.0=12\ \text{V}$ ✓. Set Interactive 1 to "R₁ + (R₂ ∥ R₃)" with $4$, $6$, $12\ \Omega$ to watch this circuit.
Interactive 2 — The Network Reducer

Pick a network. Each step, the reducer outlines one pure series or pure parallel group in purple, then replaces it by its equivalent resistor (shown in green) — exactly the strategy above. Use Step › to go at your own pace, or Play to watch. Try to say each value before it appears.

step—
this step—
resistors left—
$R_{\text{eq}}$ and current from 12 V—

Series or Parallel at Home?

Every lamp, fan and appliance in a house is designed to work at the mains potential difference, $220\ \text{V}$ in India. If they were connected in series, the $220\ \text{V}$ would be shared among them, so none would get its proper voltage; switching one off (or one failing) would break the only path and switch off everything; and since the resistances add, the current would be far too small for a heater or iron to warm up. In parallel, each appliance is connected directly across the two mains wires, so

Series connection still has its uses: a resistor in series to limit the current through a lamp or an LED (Example 2), a rheostat in series to control the current, and the cheapest decorative light strings, where many low-voltage lamps share the mains voltage between them.

✏️ Example 6 — Appliances on the mains
A fan of resistance $440\ \Omega$, a lamp of resistance $880\ \Omega$ and a television of resistance $220\ \Omega$ are connected in parallel to the $220\ \text{V}$ mains. (a) Find the current in each and the total current. (b) Find the equivalent resistance. (c) If, by mistake, the three were connected in series across $220\ \text{V}$, what would be the potential difference across the lamp?
(a) Each has $220\ \text{V}$ across it: fan $\dfrac{220}{440}=0.50\ \text{A}$, lamp $\dfrac{220}{880}=0.25\ \text{A}$, TV $\dfrac{220}{220}=1.0\ \text{A}$; total $I=1.75\ \text{A}\approx1.8\ \text{A}$.
(b) $R_{\text{eq}}=\dfrac VI=\dfrac{220}{1.75}=126\ \Omega$. (Check: $\dfrac1{440}+\dfrac1{880}+\dfrac1{220}=\dfrac{2+1+4}{880}=\dfrac7{880}$, and $\dfrac{880}{7}=126\ \Omega$ ✓ — less than the smallest, $220\ \Omega$.)
(c) In series $R_s=440+880+220=1540\ \Omega$, $I=\dfrac{220}{1540}=0.143\ \text{A}$, and the lamp gets $0.143\times880=126\ \text{V}$ — far below its $220\ \text{V}$, so it glows dimly, while the TV gets only $31\ \text{V}$ and does not work at all.
Meaning Parallel wiring gives every appliance the full mains voltage; series wiring would starve them all.
⚠️ Common mistakes

1. Forgetting to flip. $\dfrac1{R_p}=\dfrac16+\dfrac13=\dfrac12$ gives $R_p=2\ \Omega$, not $\dfrac12\ \Omega$. Always take the reciprocal at the end — and check that $R_p$ is smaller than the smallest resistor.

2. "Product over sum" for three. $\dfrac{R_1R_2}{R_1+R_2}$ works only for two resistors. For three, add the reciprocals, or combine two first and then the third with the result.

3. Mixing up what is shared. Series: same current, potential differences add. Parallel: same potential difference, currents add.

4. Seeing parallel where there is none. Two resistors drawn side by side are in parallel only if both ends meet at the same two junctions. Two resistors in the same branch, one after another, are in series even if the branch is part of a parallel group.

5. Using the battery voltage for every resistor. In a mixed circuit only the parallel group's own potential difference acts across its branches — find it from $V=IR_{\text{group}}$ first.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — optional: the resistor cube

In JEE Main, NEET and JEE Advanced 2026; not in the CBSE Class 10 syllabus.

Twelve equal resistors $R$ form the edges of a cube. What is the resistance between two opposite corners A and B? The network is not series–parallel, but symmetry cracks it. A current $I$ entering at A meets three identical edges, so it splits equally: $\tfrac I3$ in each. Each of those three corners passes its current on through two identical edges: $\tfrac I6$ in each of the six middle edges. Three edges then carry $\tfrac I3$ each into B. Follow any path from A to B (one edge of each kind) and add the potential differences:

$$V=\frac I3R+\frac I6R+\frac I3R=\frac56IR,\qquad R_{AB}=\frac VI=\frac56R .$$

With $R=6.0\ \Omega$, $R_{AB}=5.0\ \Omega$. Symmetry arguments like this — points at the same potential carry no current between them — are a favourite of entrance examinations.

Exercises

I
Resistors in series.
  • 1 Find the equivalent resistance of $4.0\ \Omega$, $6.0\ \Omega$ and $10\ \Omega$ connected in series.
  • 2 The three resistors of part 1 are connected to a $10\ \text{V}$ battery. Find the current and the potential difference across each resistor.
  • 3 Three identical $15\ \Omega$ resistors are joined in series. Find the equivalent resistance.
  • 4 A $5.0\ \Omega$ resistor and an unknown resistor $R$ in series draw $0.80\ \text{A}$ from a $12\ \text{V}$ battery. Find $R$.
  • 5 A $3.0\ \Omega$ and a $9.0\ \Omega$ resistor are connected in series to a $6.0\ \text{V}$ battery. What does a voltmeter across the $9.0\ \Omega$ resistor read?
  • 6 A string of $20$ identical lamps, each of resistance $22\ \Omega$, is connected in series across $220\ \text{V}$. Find the current and the potential difference across each lamp.
  • 7 Why does an ammeter give the same reading wherever it is placed in a series circuit?
  • 8 A lamp rated $3.0\ \text{V}$, $0.25\ \text{A}$ is to be run from a $4.5\ \text{V}$ battery. What resistor should be connected in series with it?
1 $R_s=4.0+6.0+10=20\ \Omega$.
2 $I=\tfrac{10}{20}=0.50\ \text{A}$; $V=0.50\times4.0=2.0\ \text{V}$, $0.50\times6.0=3.0\ \text{V}$, $0.50\times10=5.0\ \text{V}$ (total $10\ \text{V}$ ✓).
3 $3\times15=45\ \Omega$.
4 $R_s=\tfrac{12}{0.80}=15\ \Omega$, so $R=15-5.0=10\ \Omega$.
5 $I=\tfrac{6.0}{12}=0.50\ \text{A}$; $V=0.50\times9.0=4.5\ \text{V}$.
6 $R_s=20\times22=440\ \Omega$, $I=\tfrac{220}{440}=0.50\ \text{A}$; each lamp gets $0.50\times22=11\ \text{V}$ ($=\tfrac{220}{20}$ ✓).
7 There is only one path, and charge is neither stored nor lost in the resistors, so the same charge passes every point each second: the current is the same everywhere.
8 The resistor takes $4.5-3.0=1.5\ \text{V}$ at $0.25\ \text{A}$: $R=\tfrac{1.5}{0.25}=6.0\ \Omega$.
II
Resistors in parallel.
  • 1 Find the equivalent resistance of $4.0\ \Omega$ and $12\ \Omega$ in parallel.
  • 2 Three $9.0\ \Omega$ resistors are connected in parallel. Find the equivalent resistance.
  • 3 Find the equivalent resistance of $2.0\ \Omega$, $3.0\ \Omega$ and $6.0\ \Omega$ in parallel.
  • 4 A $10\ \Omega$ and a $15\ \Omega$ resistor are connected in parallel across $6.0\ \text{V}$. Find the current in each, the total current and the equivalent resistance.
  • 5 A total current of $3.0\ \text{A}$ divides between a $4.0\ \Omega$ and an $8.0\ \Omega$ resistor in parallel. Find the current in each.
  • 6 What resistance must be connected in parallel with $12\ \Omega$ to give an equivalent resistance of $4.0\ \Omega$?
  • 7 How many $100\ \Omega$ resistors must be connected in parallel to give $20\ \Omega$?
  • 8 Explain why the equivalent resistance of a parallel combination is smaller than the smallest resistance in it.
Check: enter up to four resistances (in Ω) and see their series and parallel combinations
series $R_s$—
parallel $R_p$—
1 $\tfrac{4.0\times12}{4.0+12}=\tfrac{48}{16}=3.0\ \Omega$.
2 $\tfrac{9.0}{3}=3.0\ \Omega$.
3 $\tfrac12+\tfrac13+\tfrac16=\tfrac{3+2+1}{6}=1$, so $R_p=1.0\ \Omega$.
4 $I_1=\tfrac{6.0}{10}=0.60\ \text{A}$, $I_2=\tfrac{6.0}{15}=0.40\ \text{A}$, $I=1.0\ \text{A}$; $R_p=\tfrac{6.0}{1.0}=6.0\ \Omega$ (product over sum: $\tfrac{150}{25}=6.0\ \Omega$ ✓).
5 $R_p=\tfrac{4.0\times8.0}{12}=\tfrac83\ \Omega$, so $V=3.0\times\tfrac83=8.0\ \text{V}$; $I_1=\tfrac{8.0}{4.0}=2.0\ \text{A}$, $I_2=\tfrac{8.0}{8.0}=1.0\ \text{A}$ (inverse ratio $2:1$ ✓).
6 $\tfrac1R=\tfrac1{4.0}-\tfrac1{12}=\tfrac{3-1}{12}=\tfrac16$, so $R=6.0\ \Omega$.
7 $\tfrac{100}{n}=20$ gives $n=5$.
8 Each branch adds an extra path, so for the same potential difference the total current is larger than the current through any single branch. Since $R_p=\tfrac VI$, a larger $I$ means a smaller $R_p$ — smaller even than the branch that carries the most current, the smallest resistor. (Like making a wire thicker.)
III
Mixed networks.
  • 1 A $2.0\ \Omega$ resistor is in series with a parallel pair of $3.0\ \Omega$ and $6.0\ \Omega$. Find the equivalent resistance.
  • 2 A $4.0\ \Omega$ and an $8.0\ \Omega$ resistor in series are connected in parallel with a $12\ \Omega$ resistor. Find the equivalent resistance.
  • 3 The network of part 1 is connected to an $8.0\ \text{V}$ battery. Find the battery current, the potential difference across the $2.0\ \Omega$ resistor and the current in the $3.0\ \Omega$ and $6.0\ \Omega$ resistors.
  • 4 You have four $4.0\ \Omega$ resistors. What are the largest and the smallest equivalent resistances you can make using all four?
  • 5 List all the equivalent resistances that can be made using all three of three $6.0\ \Omega$ resistors.
  • 6 How would you connect three $5.0\ \Omega$ resistors to get (a) $7.5\ \Omega$, (b) $3.3\ \Omega$?
  • 7 A $3.0\ \Omega$ resistor is in series with a parallel pair, $6.0\ \Omega$ and $R$. With a $6.0\ \text{V}$ battery the current drawn is $1.2\ \text{A}$. Find $R$.
  • 8 Two resistors give $9.0\ \Omega$ when connected in series and $2.0\ \Omega$ when connected in parallel. Find them.
1 $3.0\parallel6.0=\tfrac{18}{9}=2.0\ \Omega$; $R_{\text{eq}}=2.0+2.0=4.0\ \Omega$.
2 $4.0+8.0=12\ \Omega$ in parallel with $12\ \Omega$: $\tfrac{12}{2}=6.0\ \Omega$.
3 $I=\tfrac{8.0}{4.0}=2.0\ \text{A}$; $V_{2\Omega}=2.0\times2.0=4.0\ \text{V}$; the pair has $2.0\times2.0=4.0\ \text{V}$, so $I_3=\tfrac{4.0}{3.0}=1.3\ \text{A}$ and $I_6=\tfrac{4.0}{6.0}=0.67\ \text{A}$ (sum $2.0\ \text{A}$ ✓).
4 Largest: all in series, $16\ \Omega$. Smallest: all in parallel, $\tfrac{4.0}{4}=1.0\ \Omega$.
5 All series $18\ \Omega$; all parallel $2.0\ \Omega$; one in series with a parallel pair $6.0+3.0=9.0\ \Omega$; two in series in parallel with the third $\tfrac{12\times6.0}{18}=4.0\ \Omega$.
6 (a) One in series with the other two in parallel: $5.0+2.5=7.5\ \Omega$. (b) Two in series, in parallel with the third: $\tfrac{10\times5.0}{15}=3.3\ \Omega$.
7 $R_{\text{eq}}=\tfrac{6.0}{1.2}=5.0\ \Omega$, so the pair is $5.0-3.0=2.0\ \Omega$; $\tfrac1R=\tfrac1{2.0}-\tfrac1{6.0}=\tfrac13$, $R=3.0\ \Omega$.
8 $R_1+R_2=9.0$ and $\tfrac{R_1R_2}{R_1+R_2}=2.0$, so $R_1R_2=18$. Two numbers with sum $9$ and product $18$: $3.0\ \Omega$ and $6.0\ \Omega$.
IV
Circuits at home and harder problems.
  • 1 Give two reasons why household appliances are connected in parallel rather than in series.
  • 2 A lamp ($1100\ \Omega$), a fan ($440\ \Omega$) and a heater ($55\ \Omega$) are connected in parallel to $220\ \text{V}$. Find the current in each, the total current and the equivalent resistance.
  • 3 The three appliances of part 2 are wrongly connected in series across $220\ \text{V}$. Find the current and the potential difference across the heater. Comment.
  • 4 One of several lamps connected in parallel to the mains burns out. What happens to the other lamps, and to the current drawn from the mains?
  • 5 A decorative string has $40$ identical lamps in series on $220\ \text{V}$. What potential difference does each lamp get? What happens if one lamp burns out?
  • 6 Challenge. A uniform wire of resistance $12\ \Omega$ is bent into a circle. Find the resistance between two diametrically opposite points.
  • 7 Challenge. A $10\ \Omega$ wire is cut into five equal pieces, which are then connected in parallel. Find the equivalent resistance.
1 In parallel each appliance gets the full mains voltage and so draws its designed current; each can be switched on or off independently, and a fault in one does not cut off the others.
2 $I=\tfrac{220}{1100}=0.20\ \text{A}$, $\tfrac{220}{440}=0.50\ \text{A}$, $\tfrac{220}{55}=4.0\ \text{A}$; total $4.7\ \text{A}$; $R_{\text{eq}}=\tfrac{220}{4.7}=47\ \Omega$.
3 $R_s=1100+440+55=1595\ \Omega$, $I=\tfrac{220}{1595}=0.14\ \text{A}$; heater $0.138\times55=7.6\ \text{V}$ instead of $220\ \text{V}$ — it would hardly warm up. Series wiring is useless for appliances.
4 The others stay lit at full brightness (each still has $220\ \text{V}$ across it); the mains current falls by the current that lamp used to take.
5 $\tfrac{220}{40}=5.5\ \text{V}$ each. If one burns out the circuit is broken and all $40$ go out.
6 The two halves, $6.0\ \Omega$ each, are in parallel between the two points: $\tfrac{6.0}{2}=3.0\ \Omega$.
7 Each piece is $2.0\ \Omega$; five in parallel: $\tfrac{2.0}{5}=0.40\ \Omega$ — $25$ times smaller than the original.
🎯 Exam corner — board and entrance-exam practice BoardNEETJEE Main

E1. Board · 3 marks Three resistors $R_1$, $R_2$, $R_3$ are connected in parallel to a battery of potential difference $V$. Derive an expression for their equivalent resistance.

Facts In parallel each resistor has the same potential difference $V$ across it, and the total current divides: $I=I_1+I_2+I_3$.
Ohm's law $I_1=\dfrac V{R_1}$, $I_2=\dfrac V{R_2}$, $I_3=\dfrac V{R_3}$, and for the equivalent resistor $I=\dfrac V{R_p}$.
Result $\dfrac V{R_p}=\dfrac V{R_1}+\dfrac V{R_2}+\dfrac V{R_3}$; dividing by $V$, $\dfrac1{R_p}=\dfrac1{R_1}+\dfrac1{R_2}+\dfrac1{R_3}$. Exam tip: draw the circuit diagram — it usually carries a mark.

E2. Board · 3 marks A $5.0\ \Omega$ resistor is connected in series with two $10\ \Omega$ resistors that are in parallel with each other, across a $6.0\ \text{V}$ battery. Find (a) the total resistance, (b) the current through each $10\ \Omega$ resistor, (c) the potential difference across the $5.0\ \Omega$ resistor.

(a) $10\parallel10=5.0\ \Omega$; total $5.0+5.0=10\ \Omega$.
(b) $I=\dfrac{6.0}{10}=0.60\ \text{A}$, which splits equally between the equal branches: $0.30\ \text{A}$ each.
(c) $V=0.60\times5.0=3.0\ \text{V}$. Exam tip: equal resistors in parallel share the current equally — no calculation needed.

E3. Board · Assertion–Reason Assertion (A): In a house, the lamps and appliances are connected in parallel. Reason (R): In a parallel circuit every appliance gets the same potential difference and can be switched on or off independently. (A) Both A and R are true and R explains A. (B) Both are true but R does not explain A. (C) A is true but R is false. (D) A is false but R is true.

Idea Both statements are true, and R gives exactly the reasons for parallel wiring.
Answer (A). Exam tip: a reason that is true but about something else gives (B) — check that R answers "why A?".

E4. Board · Case study A student has three identical $12\ \Omega$ resistors and a $9.0\ \text{V}$ battery. (i) She connects two of them in parallel and the third in series with the pair. Find the equivalent resistance and the battery current. [2 marks] (ii) What are the largest and the smallest resistances she can make with all three? [1 mark] (iii) Find the battery current when all three are in parallel. [1 mark]

(i) Pair: $\dfrac{12}{2}=6.0\ \Omega$; total $6.0+12=18\ \Omega$; $I=\dfrac{9.0}{18}=0.50\ \text{A}$.
(ii) Largest $36\ \Omega$ (series); smallest $\dfrac{12}{3}=4.0\ \Omega$ (parallel).
(iii) $I=\dfrac{9.0}{4.0}=2.25\ \text{A}\approx2.2\ \text{A}$. Exam tip: $n$ equal resistors: $nR$ in series, $\dfrac Rn$ in parallel.

E5. NEET $n$ identical resistors have an equivalent resistance $x$ when connected in series and $y$ when connected in parallel. The ratio $\dfrac xy$ is (A) $n$ (B) $n^2$ (C) $\dfrac1{n^2}$ (D) $1$

Working $x=nR$, $y=\dfrac Rn$, so $\dfrac xy=\dfrac{nR}{R/n}=n^2$.
Answer (B). Exam tip: test with $n=2$: $\dfrac{2R}{R/2}=4=2^2$. ✓

E6. JEE Main A uniform wire of resistance $16\ \Omega$ is bent into a square. The resistance between two adjacent corners is (A) $3\ \Omega$ (B) $4\ \Omega$ (C) $8\ \Omega$ (D) $12\ \Omega$

Idea Each side is $4\ \Omega$. Between adjacent corners there are two paths: one side ($4\ \Omega$) and the other three sides in series ($12\ \Omega$), in parallel.
Working $R=\dfrac{4\times12}{4+12}=3\ \Omega$.
Answer (A). Exam tip: between opposite corners it would be $8\parallel8=4\ \Omega$ — read which corners are meant.
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. How are LED light strips wired so that they run from a $12\ \text{V}$ supply, can be cut to any length, and do not go completely dark when one LED fails?

Why it matters. LED strips light shop windows, stadium screens, aircraft cabins and homes worldwide. Designers combine series and parallel wiring: a few LEDs and a current-limiting resistor in series form one "segment", and hundreds of segments are connected in parallel along two supply rails. The choice decides the brightness, the current the power supply must deliver, and what happens when a part fails.

What to do.

  1. With a multimeter, measure the resistance of three resistors (for example $100\ \Omega$, $220\ \Omega$, $470\ \Omega$). Connect them on a breadboard in series, in parallel and in two mixed ways, and measure each combination. Compare with $R_s$ and $R_p$ calculated from your measured values.
  2. Model one strip segment: three LEDs that together take about $9.0\ \text{V}$ and a series resistor, on a $12\ \text{V}$ supply, designed to carry $20\ \text{mA}$. Calculate the resistor needed (the resistor takes the remaining voltage).
  3. A $5\ \text{m}$ strip has $100$ such segments in parallel. Calculate the total current the supply must give.
  4. Predict what happens to the rest of the strip if (a) one LED in a segment fails open, (b) a whole segment is cut off. Explain with series and parallel rules.
  5. Find the specification sheet of a real LED strip online and compare its "current per metre" with your model.

Tools: resistors, breadboard, digital multimeter, battery pack, a spreadsheet; optionally a short LED strip.

Hand in: A table of measured and calculated combinations with percentage differences, your segment design, and a half-page explanation of why the strip is wired as it is.

Going further. Long strips get dimmer towards the far end. Using what you know about the resistance of copper tracks (Section 3.2), suggest why, and how installers fix it.

Instructor note: the resistor in step 2 is $\frac{12-9.0}{0.020}=150\ \Omega$; $100$ segments draw $100\times0.020=2.0\ \text{A}$. Measured combinations usually agree with calculation within the resistors' tolerance (often $\pm5\%$, or $\pm1\%$ for metal-film resistors). An open LED darkens only its own segment (series within it, parallel to the rest). Voltage drop along the thin copper rails explains the dimming of long runs; installers feed the strip from both ends.

🧪 Self-check quiz

0/5 correct

Q1. Resistors of $2\ \Omega$, $3\ \Omega$ and $5\ \Omega$ are connected in series. The equivalent resistance is

Q2. A $6\ \Omega$ and a $3\ \Omega$ resistor are connected in parallel. The equivalent resistance is

Q3. In a series circuit, the quantity that is the same for every resistor is

Q4. One lamp in a set of lamps connected in parallel to the mains burns out. Then

Q5. Using all three of three $6\ \Omega$ resistors, which equivalent resistance can you not obtain?

← 3.2 Ohm's Law and the Factors Affecting Resistance3.4 The Heating Effect of Current and Electric Power →