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CBSE · Class 10 · Science

Chapter 11: Electricity

5 concepts · 40 practice questions with answers · Free to read

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Electric current, circuit & potential difference

In daily life

A torch lights the moment you press its switch. The cell pushes charges around the wires, much as a pump pushes water around a pipe loop.

Simple definition

Electric current is the rate of flow of charge: the charge passing a point in one second. Unit: ampere (1 A = 1 C per s). Current flows only in a closed circuit. By convention it goes from the positive to the negative terminal outside the cell, while electrons actually move the opposite way. Potential difference is the work done to move one coulomb of charge between two points; unit volt (1 V = 1 J per C). A cell keeps it up by chemical action. An ammeter (very low resistance) is joined in series; a voltmeter (very high resistance) in parallel.

Equation / formula : I = Q/t · V = W/Q · 1 A = 1 C/s · 1 V = 1 J/C · charge of one electron = 1.6 × 10⁻¹⁹ C

Key ideas of a simple circuitKey ideas of a simple circuit⚡Current Iunit ampere (A)🔋Cell or batterykeeps the voltage🔌Closed circuitcurrent flows➡️Conventional current+ to −🔵Electrons move− to +📟Ammeterin series📏Voltmeterin parallel🎚️Rheostatchanges resistance

In the figure : Current needs a closed circuit; the ammeter sits in the line, the voltmeter sits across the two points.

Rule 1 : Current and charge — Current is the charge passing per second, I = Q/t. So the charge is Q = I × t, and the number of electrons is the charge divided by 1.6 × 10⁻¹⁹ C.

  • Easy : 150 C passes a point in 25 s: I = 150 ÷ 25 = 6 A.
  • Medium : A charger gives 1.2 A for 15 minutes: t = 900 s, so Q = 1.2 × 900 = 1080 C.
  • For exams : Electrons in 8 C: n = 8 ÷ (1.6 × 10⁻¹⁹) = 5 × 10¹⁹.

Rule 2 : Potential difference and work — Potential difference is the work done per coulomb, V = W/Q. So W = V × Q: a higher voltage gives more energy to every coulomb.

  • Easy : 48 J of work moves 6 C: V = 48 ÷ 6 = 8 V.
  • Medium : A 6 V battery moves 7 C: W = 6 × 7 = 42 J.
  • For exams : A bulb has 4.5 V across it and carries 0.3 A for 20 s. Charge Q = 0.3 × 20 = 6 C; work W = 4.5 × 6 = 27 J.
Remember

An ammeter has very low resistance and a voltmeter very high resistance, so the meters do not disturb the circuit they measure.

Common mistake

Joining an ammeter across the cell (in parallel). Its tiny resistance lets a huge current flow and can damage it. Ammeter in series only.

Shortcut

Ammeter: in the line (series). Voltmeter: across two points (parallel). Memory clue : "ASVP": Ammeter Series, Voltmeter Parallel.

Spot the difference

PointAmmeterVoltmeter
Measurescurrent (A)potential difference (V)
Connectedin series, very low resistancein parallel, very high resistance

Practice questions

Easy
  1. 1What is the SI unit of electric current?
  2. 2Which instrument measures current, and how is it connected?
  3. 3What is the conventional direction of current outside the cell?
Medium
  1. 4A charge of 120 C flows in 40 s. Find the current.
  2. 5How much work is done to move 5 C across a potential difference of 12 V?
  3. 6How many electrons make up a charge of 3.2 C?
Challenging
  1. 7A lamp has 12 V across it and carries 0.6 A for 5 minutes. Find the charge that flows and the work done.
  2. 8Why must a voltmeter have a very high resistance?
Show answers (try them yourself first)
  1. 1The ampere (1 A = 1 C per s).
  2. 2An ammeter, joined in series.
  3. 3From the positive terminal to the negative terminal (electrons go the other way).
  4. 4I = 120 ÷ 40 = 3 A.
  5. 5W = V × Q = 12 × 5 = 60 J.
  6. 63.2 ÷ (1.6 × 10⁻¹⁹) = 2 × 10¹⁹ electrons.
  7. 7t = 300 s, so Q = 0.6 × 300 = 180 C and W = 12 × 180 = 2160 J.
  8. 8So that it draws almost no current and does not disturb the circuit it measures.

Ohm’s law & resistance of a conductor

In daily life

A wide garden hose delivers more water than a thin one at the same tap pressure. In the same way a short, thick wire lets more current pass than a long, thin one.

Simple definition

Ohm’s law: at constant temperature the potential difference across a metallic wire is directly proportional to the current, so V ÷ I stays constant. This constant is the resistance R (unit ohm, 1 Ω = 1 V per A). A wire’s resistance grows with its length, falls with its area of cross-section and depends on its material; resistivity ρ is the property of the material (unit Ω m). Metals have very low resistivity, alloys such as nichrome higher, and insulators like glass and rubber enormously high. It also changes with temperature.

Equation / formula : V = IR · I = V/R · R = ρl/A · 1 Ω = 1 V/A · unit of ρ = Ω m

What decides resistanceWhat decides resistance📈V–I graphstraight line, slope = R📏Longer wiremore resistance🟤Thicker wireless resistance🔌Copper, silverlow ρ: power lines🔥Nichromehigh ρ: heater coils💡Tungstenhigh melting: filament🧤Glass, rubberinsulators, huge ρ🎚️Rheostatvariable resistance

In the figure : Resistance depends on length, area and material; resistivity depends only on the material (and temperature).

Live figure

Ohm’s law & resistance of a conductor

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Rule 1 : Ohm’s law, V = IR — For a resistor at constant temperature, V ÷ I is always the same number, R. Use V = IR, I = V ÷ R or R = V ÷ I.

  • Easy : 12 V drives 1.5 A: R = 12 ÷ 1.5 = 8 Ω.
  • Medium : An 18 Ω resistor has 27 V across it: I = 27 ÷ 18 = 1.5 A.
  • For exams : Readings 2 V, 4 V, 6 V give 0.2 A, 0.4 A, 0.6 A. V ÷ I = 10 Ω every time, so Ohm’s law holds; the V–I graph is a straight line through the origin with slope 10 Ω.

Rule 2 : Resistance of a wire, R = ρl/A — R = ρl ÷ A: R grows with length l, falls with area A, and depends on the material through ρ. Changing the length or area changes R, not ρ.

  • Easy : Length doubled, same wire: a 5 Ω piece becomes 10 Ω.
  • Medium : R = 4 Ω. Same material, length three times, area twice: R = 4 × 3 ÷ 2 = 6 Ω.
  • For exams : A wire has R = 8 Ω, l = 2 m, A = 1 × 10⁻⁷ m²: ρ = RA ÷ l = 8 × 10⁻⁷ ÷ 2 = 4 × 10⁻⁷ Ω m.
Remember

Resistance changes with temperature, so Ohm’s law is exact only when the temperature stays constant.

Common mistake

Saying that resistivity changes when a wire is cut or stretched. Only the resistance changes; ρ belongs to the material.

Shortcut

Long = more R. Thick = less R. Cover the unknown in the V-I-R triangle. Memory clue : The "VIR triangle": V on top, I and R below. Cover V to get I × R; cover I to get V ÷ R; cover R to get V ÷ I.

Spot the difference

PointResistance RResistivity ρ
Depends onlength, area, material, temperaturematerial and temperature only
Unitohm (Ω)ohm metre (Ω m)

Practice questions

Easy
  1. 1State Ohm’s law.
  2. 2Define 1 ohm.
  3. 3What happens to the resistance of a wire if its length is doubled?
Medium
  1. 4A resistor has 12 V across it and carries 0.4 A. Find its resistance.
  2. 5Why are heating coils made of an alloy like nichrome and not of a pure metal?
  3. 6A wire of resistance 9 Ω has its radius doubled (same length and material). Find the new resistance.
Challenging
  1. 7A wire 3 m long with area of cross-section 1.5 × 10⁻⁷ m² has a resistance of 12 Ω. Find the resistivity of its material.
  2. 8A 6 Ω resistor carries 2 A. The potential difference across it is tripled. Find the new current.
Show answers (try them yourself first)
  1. 1At constant temperature the potential difference across a metallic wire is proportional to the current through it (V = IR).
  2. 2The resistance of a conductor that carries 1 A when 1 V is applied across it.
  3. 3It also doubles.
  4. 4R = 12 ÷ 0.4 = 30 Ω.
  5. 5An alloy has higher resistivity and does not oxidise or burn easily when red hot.
  6. 6Area becomes 4 times, so R = 9 ÷ 4 = 2.25 Ω.
  7. 7ρ = RA ÷ l = 12 × 1.5 × 10⁻⁷ ÷ 3 = 6 × 10⁻⁷ Ω m.
  8. 8V = 2 × 6 = 12 V; tripled = 36 V; I = 36 ÷ 6 = 6 A.

Resistors in series & parallel

In daily life

An old festival string of lights goes completely dark when one bulb fails, but the lights in your home work independently. The first is a series circuit, the second a parallel circuit.

Simple definition

Series: resistors joined end to end give one path, so the current is the same in all, the voltages add up, and the total resistance is the sum (bigger than the largest). Parallel: resistors joined between the same two points give separate branches, so the voltage is the same across all, the branch currents add up, and the reciprocals of the resistances add (total smaller than the smallest). Houses are wired in parallel so each appliance gets full voltage and works on its own.

Equation / formula : Series: Rs = R1 + R2 + R3 · Parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 · two resistors: Rp = (R1 × R2) ÷ (R1 + R2)

Series and parallel at a glanceSeries and parallel at a glance➡️Seriesone path, same current➕Series totalRs = R1 + R2 + R3🔀Parallelbranches, same voltage➗Parallel total1/Rp = 1/R1 + 1/R2⬆️Series total islarger than the largest⬇️Parallel total issmaller than the smallest🏠Home wiringis parallel💡One bulb fusesseries off · parallel on

In the figure : Series adds resistance; parallel adds paths, so it lowers the total resistance.

Live figure

Resistors in series & parallel

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Rule 1 : Series combination — Same current through every resistor; Rs = R1 + R2 + R3. The battery voltage is shared: V = V1 + V2 + V3, and a larger resistor gets a larger share.

  • Easy : 2 Ω, 5 Ω and 8 Ω in series: Rs = 2 + 5 + 8 = 15 Ω.
  • Medium : The same set on 30 V: I = 30 ÷ 15 = 2 A; voltage across the 5 Ω = 2 × 5 = 10 V.
  • For exams : 4 Ω and R in series carry 2 A from an 18 V source. Total resistance = 18 ÷ 2 = 9 Ω, so R = 9 − 4 = 5 Ω; voltages 8 V and 10 V add to 18 V.

Rule 2 : Parallel combination — Same voltage across every branch; branch currents add up: I = I1 + I2. For resistors: 1/Rp = 1/R1 + 1/R2 + … The total is always below the smallest resistor.

  • Easy : Two 8 Ω in parallel: 1/Rp = 1/8 + 1/8 = 1/4, so Rp = 4 Ω.
  • Medium : 9 Ω and 18 Ω in parallel: Rp = (9 × 18) ÷ (9 + 18) = 162 ÷ 27 = 6 Ω.
  • For exams : A 24 V source across 9 Ω and 18 Ω in parallel: I1 = 24 ÷ 9 = 2.67 A and I2 = 24 ÷ 18 = 1.33 A; total 4 A, which matches 24 ÷ 6 = 4 A.
Remember

In parallel wiring each branch gets the full source voltage, and adding a branch lowers the total resistance, so the source supplies more current.

Common mistake

Adding resistances directly in a parallel circuit. Use reciprocals, and check that the answer is smaller than the smallest resistor.

Shortcut

Series: same current, add the resistances. Parallel: same voltage, add the currents. Memory clue : Series = "one road, same traffic" (same current). Parallel = "many roads, same start and end" (same voltage).

Spot the difference

PointSeriesParallel
Same in every partcurrentpotential difference
Total resistancesum, larger than eachless than the smallest

Practice questions

Easy
  1. 1Which quantity is the same in every resistor joined in series?
  2. 2Write the formula for the equivalent resistance in series.
  3. 3Why are the appliances in a house connected in parallel?
Medium
  1. 4Find the equivalent resistance of 3 Ω, 4 Ω and 11 Ω in series.
  2. 5Find the equivalent resistance of two 15 Ω resistors in parallel.
  3. 6Resistors of 6 Ω and 9 Ω are in series across 30 V. Find the current.
Challenging
  1. 7A 5 Ω resistor is in series with a parallel pair of 6 Ω and 12 Ω across an 18 V battery. Find the current from the battery.
  2. 8Resistors of 8 Ω and 24 Ω are in parallel across 24 V. Find the total current.
Show answers (try them yourself first)
  1. 1The current.
  2. 2Rs = R1 + R2 + R3.
  3. 3Each appliance gets the full voltage and works independently of the others.
  4. 4Rs = 3 + 4 + 11 = 18 Ω.
  5. 5Rp = 15 ÷ 2 = 7.5 Ω.
  6. 6I = 30 ÷ (6 + 9) = 2 A.
  7. 7Pair = 72 ÷ 18 = 4 Ω; total = 5 + 4 = 9 Ω; I = 18 ÷ 9 = 2 A.
  8. 824 ÷ 8 = 3 A and 24 ÷ 24 = 1 A, so the total is 4 A.

Heating effect of electric current

In daily life

A toaster wire glows red while the copper cord stays cool, although both carry the same current. The wire that resists more turns more electrical energy into heat.

Simple definition

When current flows, moving electrons collide with the atoms of the conductor and the electrical energy turns into heat (heating effect). Joule’s law: H = I²Rt, so heat grows with the square of the current, with the resistance and with the time. Heaters, irons, toasters and bulb filaments use it. A fuse is a thin wire of a low-melting alloy joined in series; if the current exceeds its rating it melts and breaks the circuit, protecting the appliance from overload and short circuit.

Equation / formula : H = I²Rt · H = VIt · H = V²t/R · SI unit of heat: joule (J)

How current makes heatHow current makes heatCurrentflows⟹Electrons hitatoms⟹Electrical energybecomes heat⟹H = I²RtSame current: more resistance, more heat.

In the figure : Heat produced depends on the current, the resistance and the time.

Rule 1 : Joule’s law of heating — H = I²Rt. Double the current and the heat becomes four times; double the resistance or the time and the heat doubles. With the voltage known, use H = V²t ÷ R.

  • Easy : 3 A through 6 Ω for 10 s: H = 3² × 6 × 10 = 540 J.
  • Medium : A coil gives 50 J. With the current doubled (R and t the same): H = 4 × 50 = 200 J.
  • For exams : A 25 Ω heater on 100 V for 2 minutes: I = 100 ÷ 25 = 4 A; H = 4² × 25 × 120 = 48 000 J = 48 kJ.

Rule 2 : The fuse — A fuse is joined in series so the whole current passes through it. It melts at a current just above the safe value. Choose the rating just above the normal current of the appliance.

  • Easy : A 5 A fuse melts if the current rises above 5 A.
  • Medium : A 660 W iron on 220 V draws I = 660 ÷ 220 = 3 A, so a 5 A fuse suits it; a 1 A fuse would blow at once.
  • For exams : Why must fuse wire have a low melting point and be joined in series? Low melting point: it must melt quickly when excess current flows. Series: all the circuit current must pass through it.
Remember

A heating element needs high resistivity and a high melting point (nichrome); a fuse wire needs a low melting point (lead–tin alloy).

Common mistake

Fitting a fuse of a very high rating "for safety". It will not melt when the current is dangerous, so the wires may overheat. Use the correct rating.

Shortcut

Heat ∝ I² × R × t. Double the current, four times the heat. Memory clue : "I-square-R-T": H = I² R t.

Spot the difference

PointHeater coilCopper cord
Resistancehighvery low
Same current givesred-hot glowstays cool

Practice questions

Easy
  1. 1What is the heating effect of electric current?
  2. 2State Joule’s law of heating.
  3. 3Why is the filament of an electric bulb made of tungsten?
Medium
  1. 4Find the heat produced when 2 A flows through 8 Ω for 15 s.
  2. 5Why is a fuse joined in series with an appliance?
  3. 6The resistance and time stay the same and the current becomes three times. What happens to the heat?
Challenging
  1. 7A 30 Ω coil is connected to 60 V for 1 minute. Find the heat produced.
  2. 8An appliance of 880 W works on 220 V. Which fuse suits it: 2 A, 5 A or 30 A? Why?
Show answers (try them yourself first)
  1. 1Electrical energy turns into heat when current flows through a conductor.
  2. 2Heat produced is proportional to I², to R and to t (H = I²Rt).
  3. 3Tungsten has a very high melting point, so it can glow white hot without melting.
  4. 4H = 2² × 8 × 15 = 480 J.
  5. 5So that the whole current passes through it and it can break the circuit when the current is too high.
  6. 6The heat becomes 9 times.
  7. 7I = 60 ÷ 30 = 2 A; H = 2² × 30 × 60 = 7200 J.
  8. 8I = 880 ÷ 220 = 4 A, so 5 A: a 2 A fuse would blow, and a 30 A fuse would not protect it.

Electric power & electrical energy

In daily life

Your electricity bill counts "units". A heater used for hours costs far more than a phone charger, because the bill depends on both the power of the appliance and the time it runs.

Simple definition

Electric power is the rate at which electrical energy is used: P = W ÷ t = VI = I²R = V² ÷ R. Unit: watt (1 W = 1 J per s = 1 V × 1 A). The energy used = power × time. The commercial unit is the kilowatt-hour (kWh), also called one "unit" of electricity: 1 kWh = 3.6 × 10⁶ J. A bulb marked "220 V, 60 W" uses 60 J of energy every second when joined to 220 V. Cost = units × rate per unit.

Equation / formula : P = VI = I²R = V²/R · E = P × t · 1 kWh = 3.6 × 10⁶ J

From power to the billFrom power to the billPower Pin kW⟹× time tin hours⟹Energy inkWh (units)⟹× rateRs per unitFirst divide watts by 1000 to get kW.

In the figure : Power in kW times hours gives units; units times the rate gives the bill.

Rule 1 : Power of a device — P = VI. With Ohm’s law the same power can be written P = I²R or P = V² ÷ R. A rating "120 V, 60 W" means the device uses 60 J every second at 120 V.

  • Easy : 12 V and 2.5 A: P = 12 × 2.5 = 30 W.
  • Medium : A heater of 50 Ω on 100 V: P = V² ÷ R = 10 000 ÷ 50 = 200 W.
  • For exams : Bulb "120 V, 60 W": I = P ÷ V = 60 ÷ 120 = 0.5 A; R = V² ÷ P = 14 400 ÷ 60 = 240 Ω.

Rule 2 : Energy and cost — Energy in kWh = power in kW × time in hours. Cost = units × rate. Change watts to kilowatts (÷ 1000) and minutes to hours (÷ 60) first.

  • Easy : An 800 W iron for 3 hours: 0.8 kW × 3 h = 2.4 kWh.
  • Medium : At Rs 7 per unit, 2.4 units cost 2.4 × 7 = Rs 16.8.
  • For exams : Three 20 W lamps for 6 hours daily for 30 days: 0.06 kW × 6 × 30 = 10.8 kWh; at Rs 5 per unit the cost is Rs 54.
Remember

1 kWh = 1 unit = 3.6 × 10⁶ J. The electricity meter at home records energy in kWh.

Common mistake

Forgetting to convert watts to kilowatts: 1500 W for 2 h is 1.5 × 2 = 3 kWh, not 3000 kWh.

Shortcut

Units = kW × hours. Cost = units × rate. Memory clue : "PIV": P = I × V. And "kilowatt × hour = one unit".

Spot the difference

PointWatt (W)Kilowatt-hour (kWh)
Measurespower (rate of using energy)energy (total used)
Equals1 joule per second3.6 × 10⁶ joule

Practice questions

Easy
  1. 1Define electric power.
  2. 2What is the SI unit of power, and what does 1 W mean?
  3. 3How many joules are there in 1 kWh?
Medium
  1. 4Find the power of a device that runs on 9 V and 2 A.
  2. 5Find the energy used by a 100 W fan in 5 hours.
  3. 6Find the cost of 4 kWh at Rs 7 per unit.
Challenging
  1. 7A 2 kW geyser runs for 45 minutes daily for 30 days. At Rs 6 per unit, find the monthly cost.
  2. 8Find the current and the resistance of a bulb marked "240 V, 60 W".
Show answers (try them yourself first)
  1. 1The rate at which electrical energy is used (P = W ÷ t).
  2. 2The watt; 1 W means 1 joule of energy used every second.
  3. 33.6 × 10⁶ J.
  4. 4P = VI = 9 × 2 = 18 W.
  5. 50.1 kW × 5 h = 0.5 kWh.
  6. 64 × 7 = Rs 28.
  7. 72 × 0.75 × 30 = 45 kWh; cost = 45 × 6 = Rs 270.
  8. 8I = 60 ÷ 240 = 0.25 A; R = 240² ÷ 60 = 57 600 ÷ 60 = 960 Ω.