CLASS 10 SCIENCE · CHAPTER 11 · STUDY NOTES

Electricity

A comprehensive study of electric current, potential difference, Ohm's law, resistance, and the heating effects of electricity in modern circuits.

In this chapterFollow the closed pathCompare energy for the same chargeChange voltage and read the slopeSeparate geometry from material

EXPLORE

Follow the closed path

Distinguish charge flow from energy transfer.

MAKE A COMPLETE PATH

Current needs a closed circuit.

The gap stops a sustained current. Closing the switch completes the conducting loop.

Charge already exists throughout the wire; the battery supplies energy. The fixed dots mark carriers already present in the wires, not travelling packets or real speed. The lamp is in the conducting path; no wire bypasses it.

Electric current: Net charge passing a cross-section per unit time; I = Q/t.

A cell supplies energy to charges already present in a conductor. Closing the switch makes a continuous path through the lamp; opening any part stops sustained current in this simple circuit.

Current measures a rate: I = Q/t, with 1 A = 1 C/s. Conventional current goes from the positive terminal through the external circuit to the negative terminal. Electrons in metal drift in the opposite direction.

Go deeper: Read a circuit without consuming its charges

The long line of a cell symbol marks its positive terminal. A circle containing a cross represents a lamp; a zigzag represents a resistor. A filled dot marks a wire junction. Crossing wires without a junction symbol need not connect.

An ammeter goes in series so the measured branch current passes through it. Milliamperes and microamperes are 10⁻³ A and 10⁻⁶ A. An original example: 0.20 A for 30 s transfers Q = It = 6 C; the lamp transfers energy without using up that charge.

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Question

An arrow points downward along the left wire in the pictured circuit. With the positive terminal at the bottom-left, does this show conventional current or electron drift in metal? Explain.

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NCERT reference: chapter PDF pages 1, 2, 4, 5.

EXPLORE

Compare energy for the same charge

Explain volts and place a voltmeter across two points.

ENERGY FOR EACH COULOMB

Voltage is energy transferred per charge.

Potential difference6 V

V = W/Q · 6 V = 6 J ÷ 1 C

A 1-coulomb amount of charge transfers 6 joules between these two points. Voltage measures energy per charge; it is not the amount of charge.

The bulb represents energy transfer. The single 1 C symbol is a quantity marker, not a single electron. Meter view shows ideal connection topology, without meter resistance or wiring instructions.

Potential difference: Work transferred per unit charge between two points; V = W/Q.

A potential difference of 4 V means 4 J transferred for each coulomb. Voltage is energy per charge, not the quantity of charge and not the speed of an electron.

A cell maintains a potential difference using its chemical energy. A voltmeter connects in parallel across the two points being compared; an ammeter belongs in the current path.

W = VQ = 4 V × 3 C = 12 J.

Go deeper: A voltage can exist without a complete current path

An open switch can leave the cell’s terminal voltage present while the simple loop carries no sustained current. The complete path and the potential difference have different jobs.

In an original example, 3 C transferred across 4 V corresponds to W = VQ = 12 J. The 1 C marker in the picture is an amount of charge, not one enormous electron or a parcel that disappears in a bulb.

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Question

An open switch gives zero current. A learner concludes that the cell must also have zero terminal voltage. Why does that not follow?

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NCERT reference: chapter PDF pages 3, 4, 5.

EXPLORE

Change voltage and read the slope

Use an ohmic relationship with stated conditions.

CHANGE V · READ I

A straight-line relationship at constant temperature.

Voltage6 VResistance3 Ω

I = V/R = 6/3 = 2.00 A

Doubling voltage doubles current when resistance and temperature remain constant. Raising resistance makes the I–V line less steep.

A calculated graph for an ideal ohmic resistor. The vertical axis is current, so the slope is 1/R. Real devices need not obey this line.

Ohm’s law: For an ohmic conductor at constant temperature, V = IR with R constant.

Graph axes	Slope	Condition
V vertically; I horizontally	R in Ω	Ohmic component, constant temperature
I vertically; V horizontally	1/R in A/V	Same physical resistor

An ideal ohmic resistor gives proportional current and voltage when temperature stays fixed. Doubling V doubles I; at fixed V, doubling R halves I.

This graph puts current vertically and voltage horizontally, so its slope is 1/R. The source’s opposite axis arrangement has slope R. Read the axes before interpreting steepness.

Go deeper: Resistance is not a universal straight-line rule

The resistance unit is the ohm: 1 Ω = 1 V/A. A variable resistor or rheostat changes resistance to regulate current. Its symbol includes a moving-contact arrow; this is not a claim that the supply voltage changes.

Real components may heat or behave non-ohmically, so a straight line is conditional. For an original fixed 8 Ω resistor, 4 V gives 0.50 A and 8 V gives 1.00 A. The ratio V/I remains 8 Ω for nonzero readings.

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Question

Two plots show the same resistor, but one uses V vertically and the other I vertically. Must their numerical slopes match?

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NCERT reference: chapter PDF pages 5, 6, 7.

EXPLORE

Separate geometry from material

Predict resistance while holding material and temperature fixed.

CHANGE THE WIRE

Longer resists more. Thicker resists less.

Length2 mCross-section2 mm²

R = ρL/A = 2.0 Ω

Keep material and temperature fixed. More length means a longer conducting path; more area means more parallel paths for charge.

Illustrative material: ρ = 2 × 10⁻⁶ Ω m. Cross-sectional areas are converted from mm² to m². The radius scales as √A, so doubling diameter quadruples area. Length and thickness use different illustration scales. The model material is hypothetical, not a named textbook value.

Resistivity: A material property represented by ρ, with SI unit Ω m; R = ρL/A.

For a uniform wire at a fixed temperature, R = ρL/A. A longer wire has more resistance; a larger cross-sectional area has less. Resistivity describes the material rather than the wire’s chosen length.

Area is not diameter. For a round wire, A = πd²/4, so twice the diameter gives four times the area and one-quarter the resistance at unchanged length.

Go deeper: Check units and choose materials for their role

Convert square millimetres to square metres: 1 mm² = 10⁻⁶ m². The model uses a hypothetical ρ = 2 × 10⁻⁶ Ω m. A 2 m wire with area 2 mm² therefore has R = 2 Ω.

Temperature affects resistance and resistivity. Copper and aluminium are useful conductors; heating alloys tolerate high temperatures and resist oxidation. Tungsten’s high melting point suits an incandescent filament. Do not treat one hypothetical value as a property of every alloy.

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Question

Two same-material wires have equal length. One has twice the diameter. A learner predicts half the resistance. Repair the reasoning.

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NCERT reference: chapter PDF pages 7, 8, 9, 10.

Electricity Class 10 Notes

Chapter notes

Electricity

Follow the closed path

Current needs a closed circuit.

Compare energy for the same charge

Voltage is energy transferred per charge.

Change voltage and read the slope

A straight-line relationship at constant temperature.

Separate geometry from material

Longer resists more. Thicker resists less.

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