CBSE Class 10 Science · 2026–27
Why does a torch stop when one connection opens, while switching off one classroom fan leaves the lights working? Follow charge, energy and circuit paths to answer both questions. Work through the short topics, check your ideas as you go, then practise complete numerical solutions.
- Distinguish current, potential difference, resistance, power and energy using definitions and units.
- Read circuit nodes and connect meters correctly in a teacher-approved low-voltage model.
- Apply Ohm’s law and interpret a labelled graph under constant physical conditions.
- Use resistance/resistivity and geometry relationships with correct area conversions.
- Reduce series, parallel and simple mixed networks and check current/voltage consistency.
- Calculate Joule heating, device power, kWh and clearly hypothetical energy charges.
01A complete path lets charge move
A torch lights when its switch completes a conducting path from one battery terminal, through the lamp, to the other terminal. Open the path anywhere and the steady current stops. The battery transfers energy; it does not manufacture the electrons already present in the metal wires.
Electric current is the rate of flow of charge: I = Q/t. Here I is in amperes (A), Q in coulombs (C), and t in seconds (s). One ampere means one coulomb passes a chosen cross-section each second. A current of 0.30 A for 20 s transfers Q = It = 6.0 C.
In a metal, mobile electrons drift through the material. Conventional current in the external circuit is drawn from the positive terminal towards the negative terminal. Electron drift is the opposite way. An electron need not travel all the way from the battery to a lamp before the lamp responds.
Charge is conserved: a lamp transfers electrical energy into other forms; it does not use up current. In one unbranched steady circuit, the same charge per second passes every point. Convert minutes to seconds before using I = Q/t, and remember 1 mA = 0.001 A.
Pause. Think. Try.
02Voltage tells us energy transferred per coulomb
Two cells may drive different currents through the same lamp because they provide different potential differences. Potential difference is work done, or energy transferred, per unit charge between two points: V = W/Q. W is measured in joules (J), Q in coulombs, and V in volts. Thus 1 V = 1 J/C.
A 6 V source transfers 6 J for each coulomb passing through it in the simple model. If 4 C passes through a component with 6 V across it, the transferred energy is W = VQ = 24 J. Voltage is not the amount of charge and is not a rate of flow.
A useful distinction is current through a component and potential difference across it. A cell can have a potential difference across its terminals even when the switch is open and no steady external current flows.
A water-pressure comparison can help us imagine a driving difference, but it is only a model: electric charge is not water, and the battery is an energy-transfer device. Use the definitions and units to check the model. If a question gives energy per coulomb, it gives voltage; if it gives coulombs per second, it gives current.
Pause. Think. Try.
03Read a circuit before choosing a formula
Start a circuit diagram by finding the source, the complete path and any junctions. A cell symbol has a longer positive line and a shorter negative line; a battery contains more than one cell. Wires drawn as connected at a marked junction belong to the same conducting node. A crossing with no connection must be drawn clearly.
An ammeter goes in series with the component whose current is measured. It has very low resistance so that it changes the current as little as possible. An ideal ammeter has zero resistance. A voltmeter goes in parallel across the chosen component; its high resistance means it takes very little current. An ideal voltmeter has infinite resistance.
Connecting an ammeter directly across a cell can make a low-resistance path and cause overheating. For school practicals use a teacher-approved low-voltage circuit with a suitable resistor, switch and meter range. Keep the switch open while connections are checked, close it briefly for readings, and open it between readings. Do not use a wall socket or household appliance for this activity.
Before calculating: label the measured current I, the two points defining V, and the component R. Do not assume two resistors are in series because they appear beside each other; follow the actual connections.
Pause. Think. Try.
04Discover Ohm’s law from a table and graph
For an ohmic conductor at constant temperature and unchanged physical conditions, current is proportional to the potential difference across it: V = IR. The ratio V/I is its resistance R, measured in ohms (Ω). One ohm means a p.d. of 1 V produces a current of 1 A.
| I (A) | V (V) | V/I (Ω) |
|---|---|---|
| 0.10 | 0.50 | 5.0 |
| 0.20 | 1.00 | 5.0 |
| 0.30 | 1.50 | 5.0 |
| 0.40 | 2.00 | 5.0 |
Plot I horizontally and V vertically. The straight line passes through the origin, and its gradient ΔV/ΔI is 5 Ω. Check the axes: if I is vertical and V horizontal, the gradient is 1/R, not R.
For a teacher-led investigation, take several low-voltage readings, changing the applied p.d. while keeping the same resistor. Record units and plot the points; draw a best-fit line rather than forcing every measured point onto it. Brief readings help limit temperature rise.
Ohm’s law is a conditional material relationship, not a rule that every device obeys at every temperature. A filament heating strongly can change its resistance, so a curved graph need not mean the student divided incorrectly.
Pause. Think. Try.
05Change a wire’s length or thickness
Two copper wires need not have equal resistance. Resistance depends on length, cross-sectional area, material and temperature. To compare one factor fairly, keep the others fixed.
For a uniform wire of the same material at the same temperature, R is proportional to length L. Twice the length gives twice the resistance: charge must pass through a longer resistive path. Resistance is inversely proportional to cross-sectional area A. Twice the area gives half the resistance: a wider wire provides more conducting pathways.
Area is not diameter. For a circular wire, A = πr² = πd²/4. Doubling the diameter makes the area four times as large, so the resistance becomes one quarter if length, material and temperature stay unchanged.
For example, a 4 Ω wire is replaced by one made from the same material, at the same temperature, with twice the length and twice the area. The two effects cancel: R′/R = 2/2 = 1, so R′ = 4 Ω.
Stretching is a different problem because both length and area may change. Only if a question states constant volume can we use LA = constant. Doubling L then halves A and makes resistance four times as large, assuming material and temperature are unchanged. Do not apply that result to simply joining another piece of wire.
Pause. Think. Try.
06Separate a material property from a wire’s resistance
The geometry relationships combine into R = ρL/A. The symbol ρ, pronounced “rho”, is the material’s resistivity at a stated temperature. Rearranging gives ρ = RA/L, with SI unit Ω m. Resistance has unit Ω; resistivity has unit Ω m.
A long thin sample and a short thick sample of the same material can have different resistance but the same resistivity at the same temperature. Resistivity helps us compare materials without confusing their shapes. It is not absolutely temperature-independent.
Copper and aluminium have low resistivity and are useful conductors. Heating elements often use a suitable high-resistivity alloy such as nichrome, which can also withstand high temperatures and resist oxidation. Material choice involves more than resistance alone.
Unit care: convert L to metres and A to square metres before using SI resistivity. Since 1 mm = 10⁻³ m, 1 mm² = 10⁻⁶ m². Do not convert an area as though it were a length.
Suppose ρ = 2.0 × 10⁻⁸ Ω m, L = 3.0 m and A = 0.50 mm² = 5.0 × 10⁻⁷ m². Then R = (2.0 × 10⁻⁸ × 3.0)/(5.0 × 10⁻⁷) = 0.12 Ω. This is an original model calculation; a real sample’s temperature and dimensions need measuring.
Pause. Think. Try.
07One path: resistors in series
Resistors are in series when they form one unbranched path. The same steady current passes through every resistor. The potential differences add because the total energy transferred per coulomb is the sum of the transfers in the individual components.
For two resistors, V = V₁ + V₂ = IR₁ + IR₂ = I(R₁ + R₂). Therefore Rs= R₁ + R₂ + …. The equivalent resistance is larger than any one positive resistance in that series chain.
For 2 Ω and 4 Ω in series across an ideal 12 V source, Rs= 6 Ω and I = 12/6 = 2 A. The p.d.s are V₁ = 2 × 2 = 4 V and V₂ = 2 × 4 = 8 V. Check: 4 + 8 = 12 V. Current is the same; voltage need not be. Equal series resistors, however, share equal voltage.
Opening one point stops current in the whole series path. This explains why independently controlled household loads are not arranged as one long series chain.
A reliable drawing habit: start at one resistor terminal and follow the wire. If current can branch between the two components, do not call them series merely because they sit next to each other in the picture.
Pause. Think. Try.
08Several paths: resistors in parallel
Resistors are in parallel when each is connected between the same two nodes. Every branch therefore has the same potential difference. Charge flow divides at a junction and recombines later: I = I₁ + I₂ + … . It is conserved, not lost.
For two resistors, I = V/R₁ + V/R₂. Dividing by V gives 1/Rp= 1/R₁ + 1/R₂. For exactly two, Rp= R₁R₂/(R₁ + R₂). Do not use that two-resistor shortcut unchanged for three resistors.
For 3 Ω and 6 Ω across 6 V, the branch currents are 2 A and 1 A. Total current is 3 A, so Rp= 6/3 = 2 Ω. The equivalent is less than the smallest branch resistance. Adding a conducting branch gives an extra path and lowers the equivalent resistance.
If one branch opens, other intact branches can continue to work. Conceptually this is why domestic appliances have parallel supply connections and individual switches. Real household wiring and protection are designed by qualified people; this is a circuit explanation, not a wiring activity.
Check the common mix-up: parallel means equal voltage, not equal current. Equal branch currents occur only for equal resistances under the same p.d.
Pause. Think. Try.
09Reduce a mixed circuit in sensible steps
A mixed circuit combines series and parallel groups. Redraw it around its nodes before doing arithmetic. A straight-looking line on a page does not prove a series connection, and a rectangular shape does not prove parallel.
Consider three 6 Ω resistors. One branch has a single 6 Ω resistor. The other branch has two 6 Ω resistors in series. Both branches join the same two battery terminals.
- Reduce the series branch: 6 + 6 = 12 Ω.
- Combine the two branches in parallel: R = (6 × 12)/(6 + 12) = 4 Ω.
- For an ideal 12 V source, the battery current is 12/4 = 3 A.
- The single-resistor branch carries 12/6 = 2 A; the series branch carries 12/12 =1 A.
- Each resistor in the series branch has 1 × 6 =6 V across it.
Check at both levels: the branch currents add to 3 A; the two series voltage drops add to 12 V; and 4 Ω is less than 6 Ω. These checks catch many reciprocal and topology errors.
Ignore wire and battery resistance only when that is part of the question’s ideal model. If a component is bypassed by an ideal wire joining the same nodes, the p.d. across it is zero; it should not be added as another ordinary series resistance.
Pause. Think. Try.
10Why current warms a resistor
In a resistor, electrical energy is transferred into internal energy through interactions between moving charge carriers and the material. If the electrical work is entirely dissipated as heat, H = VQ. Using Q = It and V = IR gives H = VIt = I²Rt = V²t/R. With SI units the answer is in joules.
The conditions determine the comparison. At fixed current and time, doubling R doubles H. At fixed voltage and time, doubling R halves H because the current falls. Saying “more resistance always means more heat” misses this difference.
For example, a 5 Ω resistor carrying 2 A for 30 s dissipates H =2² ×5 ×30 =600 J. The same 5 Ω resistor across a constant 10 V source also has I =2 A, so VIt gives 10 ×2 ×30 =600 J. Agreement checks the substitution.
A heater’s element is designed to tolerate high temperature. A fuse uses heating to interrupt excessive current; replacing it with an arbitrary thicker wire defeats protection. Learners should explain the principle on diagrams, not heat wires or repair protective devices.
For a motor or other energy-converting device, VIt is the electrical energy transferred into the device; it is not necessarily all heat. Use the heat formula only when the model or wording justifies it.
Pause. Think. Try.
11Power rates the speed of energy transfer
A 60 W device transfers 60 J of energy each second under its specified operating conditions. Power is the rate of energy transfer: P = W/t. Since W = VIt, P = VI. For an ohmic resistive load, combine this with V = IR to obtain P = I²R = V²/R.
Power is measured in watts (W), not joules. 1 kW =1000 W. A high-power device can use less total energy than a low-power one if it runs for much less time.
A resistive lamp marked “12 V, 24 W” has a rated current 24/12 =2 A and an operating resistance 12²/24 =6 Ω at that rating. The rating specifies the intended voltage as well as the power. It does not mean the device transfers 24 J regardless of its supply voltage.
For two resistive lamps at the same rated voltage, the one with higher rated power has lower operating resistance because R=V²/P. That conclusion does not mean every brighter modern lamp consumes more electrical power: efficiency and light output also matter. Use the stated electrical model, not brightness alone.
If a fixed resistor is connected to half its original voltage, its power becomes one quarter because V is squared. A real hot filament may change resistance as its temperature changes, so that simple comparison needs the fixed-resistance assumption.
Pause. Think. Try.
12Turn a device rating into an energy estimate
Energy depends on both power and running time: E = Pt. Use watts and seconds for joules, or kilowatts and hours for kilowatt-hours. 1 kWh =1000 W ×3600 s =3.6 ×10⁶ J. A “unit” of electrical energy on a bill commonly means 1 kWh; it is energy, not power and not kW per hour.
A model 60 W fan used for 5 h transfers (60/1000) ×5 =0.30 kWh. If an invented classroom tariff is ₹8 per kWh, the energy charge for that use is 0.30 ×8 =₹2.40. This is not a quoted local tariff; real bills may include fixed charges, slabs and other items.
To compare several devices, calculate each P×t separately and add the energies. For variable-power devices, use measured energy or a suitable average rather than assuming the maximum rating applies continuously.
Try a desk audit without touching wiring: use teacher-provided photographs of labels and an agreed table of operating hours. Estimate energy, identify one practical saving, and explain which assumption most affects it. Keep necessary ventilation, visibility and learning needs in the decision.
Before finishing the chapter, connect the chain: Q/t gives current; W/Q gives voltage; V/I gives resistance; VI gives power; and Pt gives energy. Units help you decide which relationship answers the question.
Pause. Think. Try.
Your formula shelf
| Idea | Relation | Remember |
|---|---|---|
| Current | I = Q/t | I in A, Q in C, t in s; steady/average rate over the stated interval. |
| Potential difference | V = W/Q | V in V, W in J, Q in C. |
| Ohm’s law | V = IR | Ohmic conductor at unchanged temperature and physical conditions. |
| Uniform-wire resistance | R = ρL/A | SI units: ρ in Ω m, L in m, A in m²; temperature/material specified. |
| Series | Rs = R₁ + R₂ + … | One unbranched path; current equal, voltage drops add. |
| Parallel | 1/Rp = 1/R₁ + 1/R₂ + … | Branches between the same two nodes; voltage equal, branch currents add. |
| Power | P = VI = I²R = V²/R | P in W; resistance forms assume the applicable V=IR relationship. |
| Electrical energy | E = Pt = VIt = VQ | Watts×seconds gives J; kilowatts×hours gives kWh. |
| Joule heating | H = I²Rt | All calculated electrical energy is dissipated as heat in the stated resistor model. |
| Energy conversion | 1 kWh = 3.6 × 10⁶ J | kWh measures energy, not power or power per hour. |
Work through the numbers
A 36 C charge passes through a component in 90 s while its p.d. is 9 V. Find current and transferred energy.
- I=Q/t=36/90=0.40 A.
- E=VQ=9×36=324 J.
- Check: P=VI=3.6 W and Pt=3.6×90=324 J.
Find the resistance of a uniform wire with ρ=1.6×10⁻⁸ Ω m, L=5 m and area 0.40 mm².
- A=0.40×10⁻⁶=4.0×10⁻⁷ m².
- R=ρL/A=(1.6×10⁻⁸×5)/(4.0×10⁻⁷)=0.20 Ω.
- The answer is resistance in Ω; ρ retains its Ω m unit.
Three 6 Ω resistors form one single-resistor branch parallel to a branch of two in series across 12 V. Find equivalent resistance, total current and each branch current.
- Series branch: 6+6=12 Ω.
- Parallel equivalent: (6×12)/(6+12)=4 Ω.
- Battery current: 12/4=3 A.
- Single 6 Ω branch: 12/6=2 A; two-resistor branch: 12/12=1 A.
- Each series resistor has 1×6=6 V. Branch currents 2+1=3 A.
An ideal 4 Ω resistor carries 3 A for 2 min. Calculate heat, assuming all electrical energy becomes heat.
- t=2×60=120 s.
- H=I²Rt=3²×4×120=4320 J.
- P=I²R=36 W; Pt=36×120=4320 J.
A model 1.2 kW heater runs 25 min each day for 12 days. Find energy and the energy charge at an invented ₹7.50/kWh, excluding other charges.
- Total time=(25/60)×12=5 h.
- E=Pt=1.2×5=6.0 kWh.
- Energy charge=6.0×7.50=₹45.00.
- In joules: E=6×3.6×10⁶=2.16×10⁷ J.
Bring it all together
Mix the ideas. Try each question before opening its explanation.
Write it. Explain it.
Use paper for a calculation, a labelled diagram or a short explanation. Then compare the reasoning. Suggested marks help you check how complete your answer is.
A learner writes, “Current falls after a lamp because the lamp uses electricity.” Correct the statement for one steady, unbranched circuit, distinguishing charge from energy. · 2 marks
The same current flows before and after the lamp: charge is conserved. The lamp transfers electrical energy into light and heat, so energy per charge changes across it; charge is not consumed.
- 1 mark: same current/charge conservation.
- 1 mark: distinguishes energy transfer from charge consumption.
A resistor gives (V, I) readings (1.2 V, 0.20 A), (2.4 V, 0.40 A), (3.6 V, 0.60 A). Find R, describe a V-against-I graph, and state one condition for applying Ohm’s law. · 3 marks
Each V/I=6 Ω. With I horizontally and V vertically, the ideal data form a straight line through the origin of gradient 6 Ω. Temperature and other physical conditions remain constant.
- 1 mark: 6 Ω with V/I.
- 1 mark: correctly named axes and line/gradient.
- 1 mark: constant temperature/physical conditions.
Two ideal resistors 3 Ω and 6 Ω are across the same 6 V battery. Find both branch currents, total current and equivalent resistance. Find electrical energy transferred in 10 min; ignore battery/wire resistance. · 5 marks
- Both branch p.d.s are 6 V.
- I₁=6/3=2 A; I₂=6/6=1 A.
- I=2+1=3 A; R=6/3=2 Ω.
- P=VI=6×3=18 W.
- t=600 s; E=18×600=10800 J=0.003 kWh.
- 1 mark: equal branch p.d./parallel identification.
- 1 mark: 2 A and 1 A.
- 1 mark: 3 A and 2 Ω.
- 1 mark: 18 W and time conversion.
- 1 mark: 10800 J or 0.003 kWh with reasoning.
Two identical resistors can be connected in series or parallel. Compare total heating power first at the same applied voltage and then at the same total current. Explain why the answers reverse. · 3 marks
If each resistance is R, series gives 2R and parallel R/2. At fixed V, P=V²/Req: parallel power is 4 times series power. At fixed total I, P=I²Req: series power is 4 times parallel power. Different controlled quantities produce different comparisons.
- 1 mark: 2R and R/2.
- 1 mark: parallel/series power ratio 4 at fixed V.
- 1 mark: series/parallel ratio 4 at fixed total I with correct relation.
A school has four hypothetical 50 W fans used 3 h/day for 20 days. Estimate energy and the energy charge at an invented ₹8/kWh. Give one reason why a real bill may differ. · 3 marks
Total power=4×50 W=200 W=0.20 kW. Time=3×20=60 h. Energy=0.20×60=12 kWh. Energy charge=12×8=₹96. Actual hours/power may differ, or a real bill may include tariffs, slabs or fixed charges.
- 1 mark: 0.20 kW and 60 h.
- 1 mark: 12 kWh and ₹96.
- 1 mark: relevant real-world assumption or additional bill item.