Take a breath. Electricity has a reputation for being the chapter where numericals suddenly get serious, and plenty of students open it, see three formulas that all look alike, and quietly close it again. You are not going to do that here. We are going to build this chapter from absolute zero, one idea at a time, with a water-pipe picture in your head and a calculator in your hand. By the end you will be able to look at any circuit in the NCERT Class 10 Science textbook (Chapter 11) and say out loud what the current is doing, where the voltage is being spent, and how much the whole thing costs your family per month.
Here is the honest good news: Electricity is one of the most scoring chapters in the whole Class 10 Science paper. Unit IV, Effects of Current, carries 13 marks in the 80-mark theory paper, and Electricity supplies most of that. The questions repeat in shape year after year. Once you can do the standard moves, you get those marks almost mechanically. This page gives you the theory in plain English, sixteen fully worked examples with every step shown, circuit diagrams you can actually read, and a practice worksheet with answers you can check yourself.
What You’ll Learn
Jump straight to whichever piece you need. Every link lands on a full section with worked examples.
- Electric Charge, Electric Current And The Ampere
- Potential Difference, Voltage And The Volt
- Electric Circuits, Symbols And Circuit Diagrams
- Ohm’s Law Solved Examples And The V–I Graph
- Resistance, Resistivity And The Factors That Change Them
- Series Combination Of Resistors
- Parallel Combination Of Resistors
- Series Versus Parallel: Comparison And Daily Life Applications
- Mixed Networks: Electricity Class 10 Numericals With Solutions
- Heating Effect Of Electric Current And Joule’s Law
- Electric Power And The Interrelation Between P, V, I And R
- The Commercial Unit kWh And Reading Your Electricity Bill
- The Two Prescribed Practicals And How To Score In Them
- Electricity Class 10 Formula List And Unit Quick Table
- Electricity Class 10 Important Questions: Exam Strategy
- Practice Worksheet With Verified Answers
Your Game Plan
- Get the three basic quantities straight first — charge, current, potential difference. Do not touch numericals until these feel like ordinary words.
- Learn Ohm’s law and practise moving between V, I and R until you can do it without thinking.
- Master resistors in series, then in parallel, then mixed networks. Series first, always.
- Add the energy side — heating effect, power, and the kilowatt hour.
- Finish with the two practicals and the formula table, then attack the worksheet with a pen and no peeking.
Study Notes
Electric Charge, Electric Current And The Ampere
Picture a water pipe. Water flows through it. If you stand at one point and count how many litres go past you every second, you have measured the rate of flow. Electric current is exactly that idea, except the thing flowing is electric charge (विद्युत आवेश) and the pipe is a wire.
Charge is a basic property of matter, just like mass. It comes in two kinds, positive and negative, and it is measured in coulomb (C). One electron carries a tiny negative charge of 1.6 × 10−19 C. Flip that around and you get a fact worth memorising: it takes about 6.25 × 1018 electrons to make up one coulomb. So a coulomb is not a small amount of charge at all — it is an enormous crowd of electrons.
Electric current is the amount of charge passing through any cross-section of a conductor in one second.
An ampere is a fairly large current for everyday electronics, so you will often meet smaller units: 1 milliampere (mA) = 10−3 A and 1 microampere (µA) = 10−6 A. A pocket calculator draws microamperes; a room heater draws about ten amperes.
Why it works. In a metal, the outermost electrons of each atom are only loosely held. They drift about randomly all the time, in every direction, so no net charge goes anywhere. The moment you connect a cell, the whole crowd gets a gentle push in one direction and that lopsided drift is the current. The electrons themselves crawl — typically less than a millimetre per second — but the push travels through the wire almost instantly, which is why a bulb lights the moment you flip the switch.
Step 1 — fix the units. Time must be in seconds. t = 3 × 60 = 180 s.
Step 2 — apply I = Q/t. I = 90 C ÷ 180 s = 0.5 A.
Step 3 — count the electrons. n = Q ÷ e = 90 ÷ (1.6 × 10−19) = 5.625 × 1020 electrons.
Notice how a modest half-ampere still means hundreds of billions of billions of electrons. That is normal.
Potential Difference, Voltage And The Volt
Water does not flow through a level pipe on its own. You need a height difference, or a pump. In a circuit, the cell is the pump. What it creates is a potential difference (विभवांतर) between its two terminals, and that is what pushes charge around the loop.
Potential difference between two points is defined as the work done to move a unit positive charge from one point to the other.
Keep the water picture handy. Current is how much is flowing. Potential difference is how hard it is being pushed. They are different questions with different answers, and mixing them up is the single most expensive habit in this chapter.
Step 1 — write the formula. V = W / Q.
Step 2 — substitute. V = 60 J ÷ 5 C.
Step 3 — answer. V = 12 V.
Read that back in words: every single coulomb that crosses picks up 12 joules of energy. That is what “a 12 volt battery” is telling you.
Electric Circuits, Symbols And Circuit Diagrams
A circuit is simply a closed loop that charge can travel around without a break. Break the loop anywhere — open the switch, loosen a wire — and the current stops everywhere at once. That is why a single blown fuse can darken a whole room.
Nobody draws realistic pictures of cells and bulbs in physics. We use agreed symbols so that any student anywhere can read the diagram. These are the ones you must be able to draw from memory:
| Component | How the symbol looks | What it does |
|---|---|---|
| Cell | One long thin line (+) and one short thick line (−) | Maintains the potential difference |
| Battery | Two or more cells drawn side by side | A bigger potential difference |
| Switch / plug key | A small gap with a hinged line | Opens or closes the loop |
| Fixed resistor | A plain rectangle in the wire | Provides a known resistance |
| Variable resistor (rheostat) | A rectangle with a slanting arrow across it | Lets you change the current smoothly |
| Ammeter | Circle with the letter A | Measures current; goes in series |
| Voltmeter | Circle with the letter V | Measures potential difference; goes in parallel |
| Bulb | Circle with a cross inside | Converts electrical energy to light and heat |
If you have already worked through our Light — Reflection and Refraction notes for Class 10 Science, you will recognise the habit: physics rewards students who draw clean, labelled diagrams before they start calculating.
Ohm’s Law Solved Examples And The V–I Graph
Georg Simon Ohm asked a simple question: if I push harder, does more current flow, and by how much? He found that for a metal conductor at a steady temperature, the answer is beautifully tidy.
The phrase “at constant temperature” is not decoration. Heat a wire and its resistance changes, and then the neat straight line bends. That is why the law is stated with the condition attached, and why examiners give a mark for including it.
Reading the graph. Plot V on the vertical axis and I on the horizontal axis for an ohmic conductor and you get a straight line passing through the origin. The slope of that line is the resistance. Through the origin matters too — zero push gives zero current, which is exactly what common sense expects.
Step 1. I = V / R = 12 ÷ 8 = 1.5 A.
Step 2. With R = 16 Ω: I = 12 ÷ 16 = 0.75 A.
Read the pattern. Doubling the resistance halved the current, because at fixed V, current and resistance are inversely proportional. Spotting that relationship saves you time in objective questions.
I (A): 0.5, 1.0, 1.5, 2.0, 2.5 | V (V): 2, 4, 6, 8, 10
Step 1 — test each pair. V/I = 2/0.5 = 4; 4/1.0 = 4; 6/1.5 = 4; 8/2.0 = 4; 10/2.5 = 4.
Step 2 — conclude. The ratio V/I is the same every time, so the conductor obeys Ohm’s law and R = 4 Ω.
Step 3 — from the graph instead. Slope = (10 − 2) ÷ (2.5 − 0.5) = 8 ÷ 2 = 4 Ω. Same answer, as it must be.
Never find the slope by taking one point and dividing — always use the difference between two well-separated points, as done above.
Resistance, Resistivity And The Factors That Change Them
Back to the pipe. A short fat pipe lets water through easily. A long narrow one fights it. Resistance is the electrical version of that fight — the property of a conductor that opposes the flow of charge through it.
Four things decide how much resistance a wire has, and you should be able to list them with the direction of each effect:
- Length (l). Resistance is directly proportional to length. Double the length, double the resistance.
- Area of cross-section (A). Resistance is inversely proportional to area. A thicker wire has less resistance.
- Material. Copper and silver conduct beautifully; nichrome and constantan resist strongly. This is captured by the material’s resistivity.
- Temperature. For metals, resistance rises as temperature rises. This is why the law carries its “constant temperature” condition.
That last sentence is the whole point of having two words. Resistance belongs to a particular piece of wire. Resistivity belongs to the substance itself. Cut a copper wire in half and its resistance halves, but its resistivity does not budge.
Metals have resistivities of the order of 10−8 Ω m, alloys sit around 10−6 Ω m, and insulators such as rubber and glass run up to 1012 Ω m and beyond. That colossal range is why copper is used for wiring and rubber for the sleeve around it.
Step 1 — convert the radius to metres. r = 0.5 mm = 0.5 × 10−3 m = 5 × 10−4 m.
Step 2 — find the area of cross-section. A = πr² = 3.1416 × (5 × 10−4)² = 3.1416 × 25 × 10−8 = 7.854 × 10−7 m².
Step 3 — apply R = ρl/A. R = (1.68 × 10−8 × 2) ÷ (7.854 × 10−7).
Step 4. R = (3.36 × 10−8) ÷ (7.854 × 10−7) = 0.0428 Ω (about 4.3 × 10−2 Ω).
A two-metre copper wire barely resists at all. That is exactly why we treat connecting wires as having zero resistance in circuit problems.
Step 1 — what stays the same? Stretching does not add or remove metal, so the volume is unchanged: A₁l₁ = A₂l₂.
Step 2 — find the new area. If l₂ = 2l₁, then A₂ = A₁l₁ ÷ 2l₁ = A₁/2. The wire becomes half as thick in cross-section.
Step 3 — rebuild the resistance. R₂ = ρl₂/A₂ = ρ(2l₁) ÷ (A₁/2) = 4 × (ρl₁/A₁) = 4R₁.
Step 4. R₂ = 4 × 5 = 20 Ω.
Why it works. Stretching hurts you twice — longer path and narrower path. In general, stretching a wire n times its length multiplies its resistance by n².
Series Combination Of Resistors
Resistors are in series when they are joined end to end so that there is only one path for the current. Think of three narrow sections in a single pipe, one after another. Every drop of water must pass through all three.
Why it works. Since the same charge must cross all three resistors one after another, the energy it loses in each one adds together. Energy per unit charge is exactly what potential difference means, so the voltages add. Divide the total voltage by the shared current and you get the sum of the resistances.
(a) Equivalent resistance. Rs = 5 + 10 + 15 = 30 Ω.
(b) Current. I = V / Rs = 12 ÷ 30 = 0.4 A. This same 0.4 A flows through all three.
(c) Voltage across each.
V₁ = I × 5 = 0.4 × 5 = 2 V
V₂ = I × 10 = 0.4 × 10 = 4 V
V₃ = I × 15 = 0.4 × 15 = 6 V
Check your work. 2 + 4 + 6 = 12 V, which matches the battery exactly. Always run this check — it catches almost every arithmetic slip, and it takes four seconds.
Parallel Combination Of Resistors
Resistors are in parallel when both ends of each resistor are joined to the same two points, so the current has a choice of routes. Back at the water pipe: one wide channel splits into three side channels and then rejoins.
Why it works. Adding another branch is like opening another door in a crowded hall — more people get out per second, so the total current rises. More current for the same voltage means less resistance overall. That is why the combined value drops below every single branch value, which surprises students the first time they see it.
Step 1 — equivalent resistance. 1/Rp = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so Rp = 2 Ω.
Step 2 — total current. I = V / Rp = 6 ÷ 2 = 3 A.
Step 3 — branch currents. Each branch has the full 6 V across it.
I₁ = 6 ÷ 6 = 1 A through the 6 Ω resistor.
I₂ = 6 ÷ 3 = 2 A through the 3 Ω resistor.
Check. 1 + 2 = 3 A, matching the total. Also notice Rp = 2 Ω is smaller than 3 Ω, the smaller resistor — exactly as the rule predicts.
Step 1 — choose a common denominator. The LCM of 4, 6 and 12 is 12.
Step 2 — add the reciprocals. 1/Rp = 1/4 + 1/6 + 1/12 = 3/12 + 2/12 + 1/12 = 6/12 = 1/2.
Step 3 — invert. Rp = 2 Ω.
The step everyone forgets: you have calculated 1/Rp, not Rp. Turn it upside down before you write the final answer. Writing “Rp = 1/2 Ω” here is the single most common error in the entire chapter.
Series Versus Parallel: Comparison And Daily Life Applications
| Point of comparison | Series combination | Parallel combination |
|---|---|---|
| Number of paths | Exactly one | One per branch |
| Current | Same in every resistor | Divides; adds up to the total |
| Potential difference | Divides; adds up to the supply | Same across every branch |
| Equivalent resistance | Rs = R₁ + R₂ + R₃ | 1/Rp = 1/R₁ + 1/R₂ + 1/R₃ |
| Size of the result | More than the largest resistor | Less than the smallest resistor |
| If one device fails | Everything stops | The others keep working |
| Where it is used | Fuses, switches and ammeters in a line | Household wiring for lights, fans and sockets |
So why are homes wired in parallel? Three solid reasons, and the examiner wants all three. First, every appliance receives the full 220 V it was designed for. Second, each appliance can be switched on and off independently — you do not lose the fridge because a bulb burnt out. Third, the appliances draw very different currents, and parallel wiring lets each one take what it needs instead of forcing them all to share a single value.
Mixed Networks: Electricity Class 10 Numericals With Solutions
Real board questions rarely give you a pure series or a pure parallel circuit. They give you a mixture, and the marks go to whoever simplifies it in the right order. The method never changes:
- Find the group of resistors that are clearly in parallel and replace them with a single equivalent value.
- Redraw the circuit with that one value in place. Actually redraw it — do not try to hold it in your head.
- Now everything left is in series, so add.
- Use the total to find the main current from the supply.
- Work backwards through your redrawn diagrams to get individual voltages and branch currents.
Step 1 — simplify the parallel pair. 1/Rp = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3, so Rp = 3 Ω.
Step 2 — now it is a simple series circuit. Rtotal = 6 + 3 = 9 Ω.
Step 3 — main current. I = V / Rtotal = 18 ÷ 9 = 2 A.
Step 4 — voltage across the 6 Ω resistor. V = 2 × 6 = 12 V.
Step 5 — voltage across the parallel section. V = 2 × 3 = 6 V. (Check: 12 + 6 = 18 V. Correct.)
Step 6 — branch currents. Both branches share that 6 V.
Through the 4 Ω: 6 ÷ 4 = 1.5 A
Through the 12 Ω: 6 ÷ 12 = 0.5 A
Final check. 1.5 + 0.5 = 2 A, which is the main current. Everything is consistent.
Heating Effect Of Electric Current And Joule’s Law
Push electrons through a resistor and they keep colliding with the vibrating atoms of the metal. Every collision hands over a little energy, the atoms vibrate harder, and the wire warms up. That is the heating effect of electric current — unavoidable, sometimes a nuisance, and sometimes the entire point of the device.
Why the current is squared. Start from energy = charge × potential difference, so H = QV. Charge is Q = It, giving H = VIt. Now substitute V = IR and you get H = (IR)(I)(t) = I²Rt. Nothing mysterious — it is just Ohm’s law folded into the energy equation. Because of that square, doubling the current gives four times the heat.
Where we want the heat. Electric irons, room heaters, geysers, toasters and electric kettles all use a coil of nichrome, an alloy chosen because it has high resistivity, it does not oxidise easily even when red hot, and it has a high melting point. The filament of an incandescent bulb uses tungsten for much the same reason, at a much higher temperature.
Where the heat is a nuisance. In transmission cables and in the windings of a motor, heat is wasted energy. That is why household wiring uses thick copper, which has low resistivity, and why long-distance power lines carry electricity at very high voltage and therefore low current — small I means very small I²R losses.
The fuse. A fuse is a deliberately weak link: a short piece of wire of suitable material and thickness, placed in series with the appliance. If the current exceeds the safe value, the fuse wire heats up beyond its melting point, melts and breaks the circuit before anything expensive burns. Its rating must be slightly above the appliance’s normal working current.
Step 1 — convert time. t = 1 minute = 60 s.
Step 2 — substitute into H = I²Rt. H = (2)² × 25 × 60.
Step 3 — work through it. H = 4 × 25 × 60 = 100 × 60 = 6000 J, that is 6 kJ.
Note what would happen at 4 A instead of 2 A: the heat would be 24 000 J, four times as much, not twice. That squared term is the reason overloaded wiring becomes dangerous so quickly.
Step 1 — current from P = VI. I = P / V = 1100 ÷ 220 = 5 A.
Step 2 — resistance from Ohm’s law. R = V / I = 220 ÷ 5 = 44 Ω.
Step 3 — cross-check with R = V²/P. R = (220)² ÷ 1100 = 48 400 ÷ 1100 = 44 Ω. The two routes agree.
Follow-up thought. A 5 A current needs a fuse rated a little above 5 A, so a 6 A fuse would be the sensible choice for this heater.
Electric Power And The Interrelation Between P, V, I And R
Heat tells you how much energy was delivered. Power tells you how fast it is being delivered. It is the rate of doing work, or the rate at which electrical energy is consumed.
P = VI • P = I²R • P = V²/R • P = E/t
The unit is the watt (W), and 1 W = 1 V × 1 A = 1 J/s. Larger unit: 1 kilowatt (kW) = 1000 W.
All four forms describe the same thing. Which one you pick depends entirely on which two quantities the question hands you. Train yourself to scan the question first, note what is given, and then choose — that habit alone speeds up your numericals enormously.
| If the question gives you… | Use this form | Typical situation |
|---|---|---|
| Voltage and current | P = VI | Appliance rating plate |
| Current and resistance | P = I²R | Resistors in series |
| Voltage and resistance | P = V²/R | Resistors in parallel |
| Energy and time | P = E/t | Bill and consumption questions |
Step 1 — pick the right form. We are given P and V, so use P = V²/R, which rearranges to R = V²/P.
Step 2 — substitute. R = (220)² ÷ 60 = 48 400 ÷ 60.
Step 3. R = 806.67 Ω (to two decimal places).
Step 4 — current. I = P / V = 60 ÷ 220 = 0.27 A approximately.
What the marking means. “60 W, 220 V” means: when and only when this bulb is connected to exactly 220 V, it will consume 60 W. Connect it to a lower voltage and it consumes less than 60 W.
Step 1 — find each resistance.
R100 = (220)²/100 = 48 400/100 = 484 Ω
R60 = (220)²/60 = 48 400/60 = 806.67 Ω
Step 2 — total resistance. R = 484 + 806.67 = 1290.67 Ω.
Step 3 — the shared current. I = 220 ÷ 1290.67 = 0.1705 A.
Step 4 — actual power in each, using P = I²R.
P100 = (0.1705)² × 484 = 14.06 W
P60 = (0.1705)² × 806.67 = 23.44 W
Step 5 — answer. The 60 W bulb glows brighter.
Why it works. In series the current is common, so P = I²R makes power depend only on resistance — and the lower-wattage bulb has the higher resistance. Check: 14.06 + 23.44 = 37.5 W, and the supply delivers VI = 220 × 0.1705 = 37.5 W. It balances.
Step 1 — convert to watts. P = 2 kW = 2000 W.
Step 2 — find the working current. I = P / V = 2000 ÷ 220 = 9.09 A.
Step 3 — choose the next standard rating above it. A 10 A fuse is appropriate.
Why not a 5 A fuse? It would melt during normal use. Why not a 30 A fuse? It would happily let a dangerous fault current through without breaking. The rating must be just above the normal working current — not far above.
The Commercial Unit kWh And Reading Your Electricity Bill
The joule is far too small for household accounts. A single ceiling fan running for an evening consumes millions of joules. So the electricity board uses a bigger unit built out of power × time.
1 kWh = 1000 W × 3600 s = 3.6 × 106 J.
This is what your bill calls one unit. Energy in kWh = (power in kW) × (time in hours).
Step 1 — total power. P = 4 × 60 = 240 W = 0.24 kW.
Step 2 — total time. t = 5 × 30 = 150 hours.
Step 3 — energy. E = 0.24 kW × 150 h = 36 kWh, that is 36 units.
Step 4 — cost. 36 × 7.00 = ₹252.00.
Step 5 — the same energy in joules, if asked. 36 × 3.6 × 106 = 1.296 × 108 J.
Now redo this with LED bulbs of 9 W instead. The monthly energy drops to 4 × 0.009 × 150 = 5.4 kWh, costing ₹37.80. That is the whole argument for switching, in one line of arithmetic.
The Two Prescribed Practicals And How To Score In Them
The CBSE Class 10 Science curriculum for 2026-27 lists two practicals under this chapter, and both of them turn up as viva questions and as source-based written questions. Know them properly.
Practical 1 — Studying how V across a resistor depends on I, and finding its resistance. You connect the resistor in series with a cell, key, rheostat and ammeter, and put a voltmeter in parallel across the resistor. Move the rheostat to get five or six different currents, record the ammeter and voltmeter readings each time, and calculate V/I for every pair. You should find the ratio nearly constant. Then plot V against I; the graph is a straight line through the origin, and its slope is the resistance.
Practical 2 — Finding the equivalent resistance of two resistors in series and in parallel. Set up each combination in turn, measure the total current and the total voltage across the combination, and compute R = V/I. Then compare your measured value with the theoretical value from Rs = R₁ + R₂ and 1/Rp = 1/R₁ + 1/R₂.
Once you can handle these, test yourself under time pressure with the CBSE Class 10 Science Practice Paper 2026-27 Set 1, which mixes Electricity numericals with the rest of the syllabus exactly as the board does.
Electricity Class 10 Formula List And Unit Quick Table
Copy this table onto a single sheet by hand and stick it above your study table. Handwriting it once is worth reading it ten times.
| Quantity | Symbol | SI unit | Main formula |
|---|---|---|---|
| Charge | Q | coulomb (C) | Q = It ; Q = ne |
| Electric current | I | ampere (A) | I = Q/t |
| Potential difference | V | volt (V) | V = W/Q |
| Resistance | R | ohm (Ω) | R = V/I |
| Resistivity | ρ | ohm metre (Ω m) | R = ρl/A |
| Equivalent resistance, series | Rs | ohm (Ω) | Rs = R₁ + R₂ + R₃ |
| Equivalent resistance, parallel | Rp | ohm (Ω) | 1/Rp = 1/R₁ + 1/R₂ + 1/R₃ |
| Heat produced | H | joule (J) | H = I²Rt = VIt = V²t/R |
| Electric power | P | watt (W) | P = VI = I²R = V²/R |
| Commercial energy | E | kilowatt hour (kWh) | E = P(kW) × t(h); 1 kWh = 3.6 × 106 J |
Electricity Class 10 Important Questions: Exam Strategy
Electricity sits in Unit IV, Effects of Current, which carries 13 marks in the 80-mark theory paper alongside Magnetic Effects of Current. Nothing has been removed from the Electricity chapter itself in the 2026-27 curriculum — the whole of it is examinable. The formative-only note in the official document applies to Motor, Electromagnetic Induction and Electric Generator, which belong to the next chapter, so do not let a rumour convince you that any part of Electricity is optional. If you hear otherwise, check the curriculum PDF on cbseacademic.nic.in yourself before you skip anything.
These are the question shapes that repeat most often, and what each one is really testing:
- State Ohm’s law and draw the V–I graph. One mark for the statement with the temperature condition, one for the labelled straight line through the origin.
- Why is household wiring done in parallel? Give all three reasons — full voltage, independent switching, different current needs.
- Find the equivalent resistance of a mixed network. Simplify parallel groups first, then add. Show the redrawn circuit.
- A wire is stretched / cut / folded. Always begin from R = ρl/A and state that volume stays constant.
- Compare two appliances’ energy consumption and cost. Convert to kW and hours, then multiply.
- Explain the choice of nichrome or tungsten. High resistivity, high melting point, resists oxidation at high temperature.
- Assertion–reason questions on brightness in series versus parallel. Decide first which quantity is common, then choose P = I²R or P = V²/R accordingly.
Physics is not the only chapter that rewards this write-the-formula-first discipline. The same habit works in chemistry: see how it plays out in our Chemical Reactions and Equations notes and in Metals and Non-metals for Class 10. If organic chemistry is your weak spot, Carbon and its Compounds is built the same way as this page.
Practice Worksheet With Verified Answers
Do these with a pen, a rough sheet and a calculator, and write out every step the way the worked examples do. Only open an answer after you have committed to your own. Ten questions, roughly forty minutes.
Q1. A steady current of 0.75 A flows through a conductor for 10 minutes. Calculate the charge that passes through it.
Show Answer
Convert the time first: t = 10 × 60 = 600 s. Then Q = I × t = 0.75 × 600 = 450 C.
Q2. A wire of resistance 20 Ω is cut into four equal pieces, and all four pieces are then connected in parallel. Find the resistance of the combination.
Show Answer
Resistance is proportional to length, so each quarter-length piece has 20 ÷ 4 = 5 Ω. Four equal 5 Ω resistors in parallel: 1/R = 4/5, so R = 5/4 = 1.25 Ω. (Shortcut: n equal resistors of value r in parallel give r/n.)
Q3. You are given resistors of 2 Ω, 3 Ω and 6 Ω. What is the maximum resistance you can obtain by combining all three, and what is the minimum? Name the arrangement in each case.
Show Answer
Maximum comes from a series arrangement: 2 + 3 + 6 = 11 Ω. Minimum comes from a parallel arrangement: 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so R = 1 Ω.
Q4. A 10 Ω and a 15 Ω resistor are joined in parallel across a 6 V battery. Find the equivalent resistance, the total current drawn, and the current in each branch.
Show Answer
1/Rp = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6, so Rp = 6 Ω. Total current I = 6 ÷ 6 = 1 A. Branch currents: 6 ÷ 10 = 0.6 A and 6 ÷ 15 = 0.4 A. Check: 0.6 + 0.4 = 1 A.
Q5. An electric iron rated 750 W is used for 45 minutes every day for 30 days. Calculate the energy consumed in kWh and the cost at ₹6.50 per unit.
Show Answer
P = 750 W = 0.75 kW. Daily time = 45 min = 0.75 h, so total time = 0.75 × 30 = 22.5 h. Energy = 0.75 × 22.5 = 16.875 kWh. Cost = 16.875 × 6.50 = ₹109.69 (to the nearest paisa, ₹109.6875).
Q6. A current of 5 A passes through a 20 Ω resistor for 30 seconds. How much heat is produced?
Show Answer
H = I²Rt = (5)² × 20 × 30 = 25 × 20 × 30 = 15 000 J, that is 15 kJ.
Q7. A wire has a resistance of 4 Ω. A second wire of the same material has twice the length and twice the area of cross-section. What is its resistance? Explain briefly.
Show Answer
R = ρl/A. For the second wire, R′ = ρ(2l)/(2A) = ρl/A = R = 4 Ω. Doubling the length doubles the resistance, but doubling the area halves it, so the two effects cancel exactly.
Q8. A bulb is marked “100 W, 250 V”. Find its resistance and the current it draws when working at its rated voltage.
Show Answer
R = V²/P = (250)² ÷ 100 = 62 500 ÷ 100 = 625 Ω. Current I = P/V = 100 ÷ 250 = 0.4 A. Cross-check with Ohm’s law: 250 ÷ 625 = 0.4 A.
Q9. A 6 Ω and a 12 Ω resistor are connected in series across a 9 V battery. Find the current, and the power dissipated in each resistor.
Show Answer
Rs = 6 + 12 = 18 Ω. Current I = 9 ÷ 18 = 0.5 A, the same in both. Power in the 6 Ω resistor: P = I²R = 0.25 × 6 = 1.5 W. Power in the 12 Ω resistor: P = 0.25 × 12 = 3 W. Check the total: 1.5 + 3 = 4.5 W, and the supply gives VI = 9 × 0.5 = 4.5 W.
Q10. 24 J of work is done in moving 4 C of charge between two points in a circuit. Find the potential difference between them. If a 3 Ω resistor is then connected across those two points, what current flows through it?
Show Answer
V = W/Q = 24 ÷ 4 = 6 V. Then I = V/R = 6 ÷ 3 = 2 A.
Kaizen for today: solve just two numericals a day, every day, and by exam week you will have solved sixty — small steps, done daily, add up faster than any all-nighter ever will.