Electricity

Start with a circuit mystery

A battery is connected to two resistors. Move the same components into a different arrangement and the current changes. Open one switch and sometimes everything stops; sometimes one branch keeps working. Nothing mysterious has happened to the resistors. The connections determine the paths available for charge and the way energy is transferred.

This guide follows Class 10 NCERT Science, Electricity, Chapter 11 in the 2026–27 reprint: current, voltage, circuit diagrams, Ohm’s law, resistance, combinations, heating and power. The complete NCERT chapter supplies the reference scope. The CBSE 2026–27 syllabus lists these concepts and related practicals, but its teacher note contains inconsistent assessment wording about electric effects. Use your school’s current assessment guidance; no chapter-specific marks or exclusion is promised here.

Follow three questions throughout: How much charge passes? How much energy is transferred per coulomb? What paths are connected? You can test the central comparison in Circuit Detective. All proposed investigations here use the screen, provided data or paper. They are not instructions for household wiring or for connecting real batteries directly with wire.

What is electric charge?

Electric charge is a property of matter. Electrons carry negative charge; protons carry positive charge. Like charges repel and unlike charges attract. A particle carries charge: charge is not itself a particle.

The SI unit is the coulomb, C. An electron has charge −e and a proton +e. The positive elementary-charge magnitude is exactly e = 1.602176634 × 10⁻¹⁹ C in SI; for school calculations we often use the rounded value 1.6 × 10⁻¹⁹ C. NIST records the exact value.

For a collection of n electrons, its charge is Q = −ne. To count electrons from a charge magnitude, use n = |Q|/e. With the rounded e, 1 C of charge magnitude corresponds to about 6.25 × 10¹⁸ electrons. Their combined charge is −1 C, not +1 C. The count is positive; the minus sign describes the type of charge.

A collection has charge −3.2 × 10⁻¹⁹ C. How many electrons does that represent in the rounded model?

n = |Q|/e = (3.2 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 2 electrons. Dividing two charges leaves a count with no coulomb unit. A negative charge does not mean a negative number of electrons.

What is electric current?

Imagine a checkpoint across a wire. Current tells you the rate at which charge passes that cross-section. For steady current, using charge magnitude in the chosen direction, I = Q/t. Equivalently, Q = It. Q is the charge that passes during the interval, not all the charge stored in the wire.

The SI unit of current is the ampere, A: 1 A = 1 C/s. A current of 0.20 A transfers 0.20 C past a checkpoint each second. In 3 minutes, t = 180 s, so Q = 0.20 × 180 = 36 C. Useful conversions are 1 mA = 10⁻³ A and 1 μA = 10⁻⁶ A.

In a metallic wire, moving electrons carry the current. Conventional current points in the direction positive charge would move, so it is opposite to electron drift in the metal. In the external circuit of a discharging battery, conventional current runs from its positive terminal towards its negative terminal. This arrow does not mean protons are travelling through the wire.

Other materials can have other carriers. In an ionic solution, positive and negative ions can both contribute. The general definition is charge flow, not “electrons only in every material”. OpenStax distinguishes these cases.

Charge checkpoint: predict the time for 1 C at 0.25 A, 0.50 A and 1 A.

Use t = Q/I. The times are 4 s, 2 s and 1 s. Doubling the rate halves the time needed for the same transferred charge. None of these values tells you how long one particular electron takes to complete a circuit.

Does the resistor use up the current?

In a steady single-path circuit, charge does not continuously pile up before a resistor or disappear inside it. The same current enters and leaves. A resistor transfers electrical energy to thermal energy; it does not consume electrons. This distinction between conserved charge and transferred energy will explain the series rule.

What is Potential Difference?

Potential difference, also called voltage, describes energy transferred per unit charge between two points. For the magnitudes used in these circuit examples, V = W/Q, where W is energy transferred, or work done. Thus 1 V = 1 J/C.

If 18 J is transferred when 3 C passes through a component, the potential difference is 18/3 = 6 V. If twice as much charge passes through that same 6 V drop, twice as much energy is transferred. Voltage and energy are connected, but they are different quantities.

A cell converts stored chemical energy and maintains a potential difference across its terminals. It can have a voltage when the external switch is open and there is no steady external current. A complete conducting path is needed for the simple circuit to carry current. The battery is the energy source; voltage is an energy-per-charge measure, not a substance flowing down the wire.

Two ideal batteries both maintain 6 V. Must they store the same energy?

No. The voltage specifies energy per coulomb, not how much charge a battery can move before its usable chemical energy is depleted. Their capacities and total available energies may differ. “Same volts” does not mean “same runtime”.

Read the connections and place the meters

A circuit diagram is a map of connections. The drawn angle, shape or length of a line does not by itself specify a real wire’s length or resistance. Identify the components and junctions before calculating. The companion reference redraws the chapter’s symbols with readable labels.

• A cell has unequal parallel lines; the longer line marks its positive terminal. Repeated cell pairs represent a battery.
• A closed switch joins its contacts. An open switch interrupts that path.
• A junction dot marks connected wires. A crossing drawn without a connection must remain electrically separate; use a clear bridge when sketching your own diagram.
• A resistor may use a rectangular or zigzag convention. A variable resistor has an adjustment arrow. A lamp symbol represents a lamp, not an assumption that its resistance is constant.

How can we measure electric current?

An ammeter measures current through a path and is connected in series with that path. Its resistance is designed to be small. In an ideal circuit calculation, we neglect it. Placing it directly across a voltage source would create a low-resistance path; the meter belongs in the path being measured.

How can we measure potential difference?

A voltmeter measures voltage across two points and is connected in parallel with the component or combination. Its resistance is large; an ideal voltmeter draws no current. For a DC meter with polarity markings, its positive terminal goes to the higher-potential side for a positive reading.

A voltmeter is across three series resistors. Does it read the voltage across just the middle one?

No. It reads the potential difference between the two points where its leads connect. If those points span all three resistors, it measures their total voltage drop. To investigate the middle resistor, the leads must span that resistor’s two ends. This is a diagram-reading question, not a wiring instruction.

OHM’S LAW

For an ohmic conductor under fixed physical conditions, including temperature, potential difference is directly proportional to current: V ∝ I. Their ratio is its resistance, R = V/I, so V = IR and I = V/R. One ohm is one volt per ampere: 1 Ω = 1 V/A.

At a fixed voltage, doubling resistance halves current. At a fixed resistance, doubling voltage doubles current. Each sentence needs its fixed condition; changing voltage and resistance together requires a fresh calculation.

Discover a pattern in data

An original virtual resistor gives these readings: at 1.5 V, current is 0.30 A; at 3.0 V, 0.60 A; at 4.5 V, 0.90 A. Each V/I ratio is 5 Ω. Predict the current at 6.0 V before continuing.

Check the prediction and what the graph means

I = 6.0/5 = 1.2 A if the same conditions continue. Plot current I horizontally and voltage V vertically: the points lie on a straight line through the origin, with slope ΔV/ΔI = 5 Ω. If the axes are exchanged, the slope is 1/R. Label axes and units before describing a “steeper” graph.

Ohm’s law is an observed relationship with limits. A filament lamp heats as current increases, so its resistance need not remain constant over a range of readings. A curved V–I graph is evidence to investigate, not automatically a failed meter. The formula R = V/I at one operating point does not prove that R stays fixed at every voltage. OpenStax explains ohmic and non-ohmic behaviour.

What is a resistor, conductor and insulator?

A resistor is a component chosen to provide resistance. Conductors allow current relatively readily; insulators offer much greater resistance under ordinary conditions. Compare pieces of similar geometry and at specified conditions when judging materials. These categories are useful, but an insulator is not a promise of protection at every voltage or temperature.

What is a Rheostat or Variable Resistance?

A rheostat changes the resistance included in a circuit. In a wire-wound model, its sliding contact changes the length of resistive wire in use. It can adjust current with the source voltage fixed. The component changes a circuit parameter; it does not destroy the unwanted charge.

Resistance and resistivity: change one thing

For a uniform conductor, R = ρl/A. Here l is length, A is cross-sectional area, and ρ is the material’s resistivity at the stated temperature. Resistance has unit Ω; resistivity has unit Ω m, because ρ = RA/l. They are not interchangeable names for the same quantity.

• Double length while holding material, temperature and area fixed: resistance doubles.
• Double area while holding material, temperature and length fixed: resistance halves.
• Change material: resistivity may change even if the dimensions do not.
• Change temperature: resistance and resistivity can change. For typical metals over ordinary ranges, resistance rises with temperature; this is not the rule R ∝ temperature.

For a round wire, A = πd²/4, where d is diameter. Doubling diameter makes the area four times as large, not twice as large. At unchanged length, material and temperature, resistance becomes one-quarter.

A uniform 8 Ω wire is compared with another of the same material and temperature, twice as long and twice the diameter. What is the new resistance?

Length contributes a factor 2; area contributes a factor 4. R′/R = 2/4 = 1/2, so R′ = 4 Ω. State both changes before applying the formula. Resistivity stays the same because the material and temperature were held fixed.

Unit check: a wire with ρ = 2.0 × 10⁻⁸ Ω m, l = 5.0 m and A = 1.0 mm² has A = 1.0 × 10⁻⁶ m². Its resistance is (2.0 × 10⁻⁸ × 5.0)/(1.0 × 10⁻⁶) = 0.10 Ω. The resistivity is a supplied value for this example, not a claim that one measurement uniquely identifies a metal.

Why are alloys used in electrical heating devices?

Connecting wires and heating elements do different jobs. Copper and aluminium have low resistivity, helping limit unwanted heating in conductors. Suitable resistance alloys combine relatively high resistivity with durability at their operating temperatures, allowing compact heating elements.

For scale, OpenStax lists copper at about 1.7 × 10⁻⁸ Ω m at 20°C. A manufacturer’s specific nickel-chromium heating alloy, Nikrothal 80, is about 1.1 × 10⁻⁶ Ω m at 20°C and is designed for oxidation resistance. These are examples with stated conditions, not universal numerical boundaries for all metals and alloys. Oxidation resistance can involve a protective oxide; it does not mean oxidation never occurs. Material comparison; manufacturer’s data.

Resistors in Series

Resistors are in series when current must pass through them one after another along a shared path, with no conducting branch between them. The current is the same through each. Their voltage drops add: V = V₁ + V₂ + V₃.

Using V₁ = IR₁ and the corresponding expressions for the other resistors gives IRₛ = IR₁ + IR₂ + IR₃. Thus Rₛ = R₁ + R₂ + R₃. Positive resistances in series produce a total greater than any one of them.

With an ideal 6 V source and 3 Ω plus 6 Ω in series, Rₛ = 9 Ω and I = 6/9 = 2/3 A. The drops are (2/3) × 3 = 2 V and (2/3) × 6 = 4 V. Their sum is 6 V. The larger resistor has the larger voltage drop because both carry the same current.

After the 3 Ω resistor, should an ammeter read less because some current has been used?

No. In the steady series path it reads the same 2/3 A. Energy has been transferred, represented by the 2 V drop, but charge has not vanished. If the only path is broken, current stops throughout that path.

Resistors in Parallel

Parallel branches connect the same two circuit nodes. Each has the same potential difference across it. Their currents add at the junction: I = I₁ + I₂ + I₃. They do not have to share current equally.

Substitute I = V/R into the current sum: V/Rₚ = V/R₁ + V/R₂ + V/R₃. For nonzero V, divide by V to obtain 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃. For two resistors, the useful rearrangement is Rₚ = R₁R₂/(R₁ + R₂). For multiple finite positive resistances, the equivalent parallel resistance is smaller than the smallest branch resistance.

Connect the same 3 Ω and 6 Ω resistors across 6 V. Their currents are 6/3 = 2 A and 6/6 = 1 A. The total is 3 A; Rₚ = 6/3 = 2 Ω. The smaller resistance takes more current because both branches have the same voltage.

Opening the switch beside 3 Ω leaves the 6 Ω path complete. In this ideal model it still carries 1 A. The open switch has 6 V across it; the disconnected 3 Ω resistor carries zero current and has zero voltage across the resistor itself. “Open circuit” is not zero resistance.

Parallel connections allow loads to use the same supply voltage while taking different currents, and branch switches can act independently. The principle appears in building circuits and vehicle electrical systems. Adding a positive-resistance parallel branch lowers equivalent resistance and increases current from an ideal fixed-voltage source. It does not increase the current in unchanged existing branches. Real sources and wiring have limits; the ideal model does not authorise adding appliances indiscriminately.

Combine the rules in stages

In an original mixed circuit, a 2 Ω resistor is in series with a parallel pair of 6 Ω and 3 Ω. The pair is 2 Ω, so the whole circuit is 4 Ω. Across an ideal 12 V source, total current is 3 A. The first resistor drops 6 V, leaving 6 V across the pair; branch currents are 1 A and 2 A. The totals provide two checks: 6 V + 6 V = 12 V and 1 A + 2 A = 3 A.

Design challenge: make 12 Ω and 3 Ω using two identical 6 Ω resistors.

Series gives 6 + 6 = 12 Ω. Parallel gives 6 × 6/(6 + 6) = 3 Ω. Sketch the two shared nodes in the parallel arrangement. A different drawing angle does not change those connections.

Heating Effect of Electric Current

A resistor converts electrical energy into thermal energy. For steady current and a fixed resistance over time t, H = VIt = I²Rt = V²t/R. H is energy in joules when voltage is in volts, current in amperes, resistance in ohms and time in seconds.

For I = 0.50 A, R = 8 Ω and t = 120 s, first find V = IR = 4 V. Then H = 4 × 0.50 × 120 = 240 J. The other forms agree: 0.50² × 8 × 120 = 240 J, and 4² × 120/8 = 240 J.

Joule’s heating rule contains fixed conditions. H ∝ I² when R and t stay fixed; H ∝ R when I and t stay fixed; H ∝ t when I and R stay fixed. These are not three invitations to change every variable at once.

Two people disagree: “More resistance means more heat” and “More resistance means less heat.” Can both be reasoning correctly?

Yes, under different controls. At fixed current and time, H = I²Rt increases with R. At fixed voltage and time, H = V²t/R decreases with R because the current falls. For 4 Ω versus 8 Ω at the same 2 A, power is 16 W versus 32 W. Across the same 8 V instead, power is 16 W versus 8 W. Ask what stayed fixed before choosing an equation.

Useful heat, unwanted heat and filament lamps

Heating is the intended output of an electric iron or resistance heater. In connecting wires and electronic components, unwanted heating wastes energy and can cause overheating. The element and its supply leads carry the same series current, but the element has much greater resistance and therefore releases much more heat per second.

An incandescent lamp makes a tungsten filament hot enough to emit visible light. Tungsten’s very high melting point makes this possible, but most input energy becomes heat or non-visible radiation. Hot tungsten can oxidise in air, so the lamp excludes oxygen; designs use a vacuum or a suitable gas fill. This explains an incandescent filament, not every modern lighting technology. Royal Society of Chemistry: tungsten.

Electric Fuse

A fuse is an overcurrent protection device placed in series with the circuit it protects. Excessive current heats its metal or alloy element until it melts and opens the path. A short circuit provides an unintended low-resistance path; an overload can demand more current than a circuit is designed to carry.

A fuse’s ampere rating describes the current it can carry under specified test conditions. Real selection also involves operating voltage, starting current, temperature, fault current and opening time. The rule is not simply “choose any rating a little larger than the appliance current”. Eaton explains fuse ratings and their conditions. Read this as the science of protection, not a procedure for selecting or replacing a household fuse.

Electric Power

Power is the rate of energy transfer: P = E/t. For the steady circuit quantities here, P = VI. For a resistor, substituting V = IR gives P = I²R = V²/R. The SI unit is the watt, W: 1 W = 1 J/s = 1 V × 1 A. One kilowatt is 1000 W.

The 8 Ω resistor carrying 0.50 A uses 2 W: 4 × 0.50 = 0.50² × 8 = 4²/8 = 2. In 120 s it transfers 2 × 120 = 240 J. Power says how fast; energy says how much over the interval.

A stated appliance rating belongs to stated operating conditions. A hypothetical resistive load rated 24 W at 12 V draws 2 A there and has resistance 6 Ω. If that resistance stays fixed at 6 V, its power is 6²/6 = 6 W, one-quarter of 24 W. Real hot filaments change resistance with temperature, so the constant-resistance prediction needs its qualification.

Energy, bills and a fair comparison

For constant power, E = Pt. Watts multiplied by seconds give joules; kilowatts multiplied by hours give kilowatt-hours. 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J. A billing “unit” in the NCERT context means 1 kWh. The h is multiplication by time, not division by time.

This measurement travels across borders. NCERT’s Indian billing examples and the US Energy Information Administration’s consumer explanation both use kWh for energy over time. Currency, tariffs and billing rules can differ while the physical unit stays the same. A 0.10 kWh transfer means the same amount of energy in either account.

Which uses more energy: a 120 W load for 45 minutes or a 900 W load for 6 minutes?

First: 0.120 kW × 0.75 h = 0.090 kWh. Second: 0.900 kW × 0.10 h = 0.090 kWh. They use equal energy, 324,000 J, despite different powers and durations. These are constant-power hypothetical loads, not claims about the actual cycling of named appliances.

To estimate a simple energy charge, multiply kWh by the stated price per kWh. At an invented classroom rate of ₹6 per kWh, 2.5 kWh would cost ₹15 for energy alone. This is not a current tariff or a complete bill: fixed charges, taxes and other rules may apply.

Surprise: if electrons are not used up, what are we paying for?

Energy transfer. Charges already exist throughout a conducting circuit, and current describes their motion. The source maintains conditions that let electrical energy be transferred to devices. A bill is not a count of electrons delivered and destroyed. NCERT makes this distinction in its enrichment box on printed page 191.

Make a prediction worth testing

Use Circuit Detective with its fixed 6 V source and 3 Ω/6 Ω resistors. First predict which arrangement draws more current. Then open the 3 Ω path and compare what survives. Say your prediction aloud or sketch it before revealing a result.

For a paper challenge, invent a claim with a missing condition: “A thicker wire always carries more current,” “A larger resistance always heats more,” or “The most powerful device always uses the most energy.” Give a friend enough extra information to decide whether it holds. Revise the statement so that its conditions are explicit.

• Current is charge per time; voltage is energy per charge; power is energy per time.
• In a metal, conventional current and electron drift have opposite directions. Charges are not consumed by a resistor.
• Ohm’s law requires suitable fixed conditions. A single V/I calculation is not proof of an unchanging resistance.
• Resistance depends on geometry and material conditions; resistivity characterises the material at stated conditions.
• Series shares current and adds voltage drops. Parallel shares voltage and adds branch currents.
• Choose heating and power formulas by what remains fixed. Convert minutes, milliamperes and square millimetres carefully.
• Check answers against conservation, units and sensible limits before trusting a calculation.

Sources and what comes next

The complete NCERT 2026–27 Electricity chapter and CBSE Science curriculum were checked. The explanations, fictional numerical examples and discovery prompts are original. This is a Class 10 reference, not a claim that every board prescribes identical content.

Additional primary-publisher checks cover potential difference, resistor combinations and energy and power. Material, unit and protection sources are linked where used. Explore the next connection in Magnetic Effects of Electric Current: current can create a magnetic field as well as transfer energy.