Current Electricity Class 10 ICSE — Selina Chapter 8
Current electricity class 10 icse is the chapter where most students first meet Ohm’s law and circuit calculations. In Selina’s Chapter 8, you will study how charge moves through a conductor, how current and potential difference are measured, and how resistance, e.m.f., electrical power and energy are defined. The chapter closes with the formulas used to calculate household electricity bills.
Download Current Electricity Class 10 ICSE Physics Chapter 8 PDF
Download the Current Electricity Class 10 ICSE Physics chapter PDF from the Selina Concise textbook to revise offline or print for practice. It is the full chapter as printed in the book — 37 pages, 21 sections, 113 questions.
What is inside Chapter 8
37 pages · 21 sections · 2 worked examples · 113 exercise questions
| Section | Title |
|---|---|
| 8.1 | CONCEPT OF CHARGE |
| 8.2 | CONCEPT OF CURRENT |
| 8.3 | CONCEPT OF POTENTIAL AND POTENTIAL DIFFERENCE (P.D.) |
| 8.4 | CONCEPT OF RESISTANCE |
| 8.5 | OHM'S LAW (V = IR) |
| 8.6 | EXPERIMENTAL VERIFICATION OF OHM'S LAW |
| 8.7 | OHMIC AND NON-OHMIC RESISTORS |
| 8.9 | SPECIFIC RESISTANCE (OR RESISTIVITY) |
| 8.10 | CHOICE OF MATERIAL OF A WIRE |
| 8.11 | SUPERCONDUCTORS |
| 8.12 | ELECTRO-MOTIVE FORCE (E.M.F.) OF A CELL |
| 8.13 | TERMINAL VOLTAGE OF A CELL |
| 8.14 | INTERNAL RESISTANCE OF A CELL |
| 8.15 | COMBINATION OF RESISTORS |
| 8.16 | ELECTRICAL ENERGY |
| 8.17 | MEASUREMENT OF ELECTRICAL ENERGY (EXPRESSION W = QV = VIt) |
| 8.18 | ELECTRICAL POWER AND ITS EXPRESSION P = W/t = VI |
| 8.19 | COMMERCIAL UNIT OF ELECTRICAL ENERGY |
| 8.20 | POWER RATING OF COMMON ELECTRICAL APPLIANCES |
| 8.21 | HOUSEHOLD CONSUMPTION OF ELECTRICAL ENERGY |
| 8.22 | HEATING EFFECT OF CURRENT |
- EXERCISE-8(A): 33 questions
- MULTIPLE CHOICE TYPE: 13 questions
- EXERCISE-8(B): 11 questions
- MULTIPLE CHOICE TYPE: 34 questions
- MULTIPLE CHOICE TYPE: 22 questions
8.1 CONCEPT OF CHARGE
Charge comes in two kinds. A body that gains electrons becomes negatively charged, while a body that loses electrons becomes positively charged. Like charges repel each other, and unlike charges attract each other.
The S.I. unit of charge is the coulomb (C). Smaller units are the millicoulomb, microcoulomb and nanocoulomb, where 1\ \text{mC} = 10^{-3}\ \text{C}, 1\ \mu\text{C} = 10^{-6}\ \text{C} and 1\ \text{nC} = 10^{-9}\ \text{C}.
The charge on a body is q = \pm ne, where n is the number of electrons in deficit or excess and e = 1.6 \times 10^{-19}\ \text{C}. One coulomb means a deficit of 6.25 \times 10^{18} electrons.
Worked pattern: a body with 2.5 \times 10^{19} electrons in excess carries q = 2.5 \times 10^{19} \times 1.6 \times 10^{-19} = 4\ \text{C} of negative charge.
8.2 CONCEPT OF CURRENT
Charge in motion constitutes electric current. Current is the rate of flow of charge, or the charge flowing per second through a conductor. If charge Q flows in time t, then I = \frac{Q}{t}.
Current is measured in amperes: 1\ \text{A} = 1\ \text{C}\,\text{s}^{-1}. Smaller units are 1\ \text{mA} = 10^{-3}\ \text{A} and 1\ \mu\text{A} = 10^{-6}\ \text{A}. Since Q = ne, we also write I = \frac{ne}{t}.
Current is a scalar quantity. The direction of current is opposite to the motion of electrons. An ammeter is connected in series, with its + terminal towards the positive terminal of the source.
Worked pattern: if 5 \times 10^{18} electrons pass in 4 s, then Q = 5 \times 10^{18} \times 1.6 \times 10^{-19} = 0.8\ \text{C}, so I = \frac{0.8}{4} = 0.2\ \text{A}.
8.3 CONCEPT OF POTENTIAL AND POTENTIAL DIFFERENCE (P.D.)
The potential at a point is the work done in bringing a unit positive charge from infinity to that point. It is denoted by V and is a scalar quantity. If work W is done on charge Q, then V = \frac{W}{Q}.
Potential is positive near a positive charge and negative near a negative charge. The work needed to move charge Q to a point at potential V is W = QV. The S.I. unit of potential is the volt: 1\ \text{volt} = 1\ \text{J}\,\text{C}^{-1}.
The potential difference between two points is the work done in moving a unit positive charge from one point to the other: V_A - V_B = \frac{W}{Q}. A voltmeter measures p.d. and is connected in parallel, with its + terminal to the higher potential point.
Worked pattern: moving 3 C of charge between two points takes 12 J of work, so the p.d. is \frac{12}{3} = 4\ \text{V}.
8.4 CONCEPT OF RESISTANCE
Resistance is the obstruction a conductor offers to the flow of current. The cause lies in the behaviour of free electrons inside the metal.
A metal contains free electrons and fixed positive ions. When a p.d. is applied, the electrons drift towards the positive end. During their movement they collide with the fixed positive ions and lose kinetic energy, then accelerate again. This repeats continuously.
The resistance of a conductor depends on the number of collisions the electrons suffer while moving from one end to the other. More collisions mean more resistance.
8.5 OHM’S LAW (V = IR)
Ohm’s law states that the current in a conductor is directly proportional to the potential difference across its ends, provided the physical conditions and temperature remain constant. If the p.d. is doubled, the current also doubles.
Mathematically, \frac{V}{I} = \text{constant} = R, so V = IR. The unit of resistance is the ohm (\Omega): 1\ \Omega = 1\ \text{V}/\text{A}, and 1\ \text{k}\Omega = 10^3\ \Omega. Conductance is the reciprocal of resistance, measured in siemens.
For a metallic conductor, the I-V graph is a straight line through the origin. Its slope is \frac{\Delta I}{\Delta V} = \frac{1}{R}. Ohm’s law is obeyed only when temperature stays constant.
Worked pattern: a conductor carries 0.5 A when 6 V is applied across it, so R = \frac{6}{0.5} = 12\ \Omega.
8.6 EXPERIMENTAL VERIFICATION OF OHM’S LAW
The verifying circuit has a battery, key, rheostat, ammeter and resistance wire connected in series, with a voltmeter in parallel across the wire. The rheostat controls the current, the ammeter reads I, and the voltmeter reads V.
Several sets of I and V readings are recorded, and the ratio \frac{V}{I} stays almost constant. The V-I graph plotted with V on the Y-axis and I on the X-axis is a straight line, which verifies the law.
To find the slope, take two points P and Q on the line, drop normals to both axes, and read \Delta V and \Delta I. The slope is R = \frac{\Delta V}{\Delta I}. Greater slope means greater resistance.
The two graph slopes differ, and students often confuse them:
| Graph | Slope | What it equals |
|---|---|---|
| I-V graph (I on Y, V on X) | \frac{\Delta I}{\Delta V} | \frac{1}{R}, the reciprocal of resistance |
| V-I graph (V on Y, I on X) | \frac{\Delta V}{\Delta I} | R, the resistance itself |
8.7 OHMIC AND NON-OHMIC RESISTORS
Ohmic resistors obey Ohm’s law. Their \frac{V}{I} ratio is constant for all values of V and I, and the V-I graph is a straight line through the origin. Examples are metallic conductors like copper and silver, nichrome, and electrolytes with suitable electrodes.
Non-ohmic resistors do not obey Ohm’s law. Their V-I graph is a curve, and resistance changes with V or I. Examples are LED, junction diode, transistor and the filament of a bulb. The resistance at any point is the slope of the tangent there, called the dynamic resistance.
Resistance depends on four factors: the material, length (R \propto l), thickness (R \propto \frac{1}{a}), and temperature. For metals, resistance rises with temperature, so the V-I line at a higher temperature is steeper.
8.9 SPECIFIC RESISTANCE (OR RESISTIVITY)
Combining R \propto l and R \propto \frac{1}{a} gives R = \rho \frac{l}{a}, where \rho is the specific resistance or resistivity. Its unit is \Omega\,\text{m}. Resistivity is the resistance of a wire of unit length and unit area of cross section.
Typical values are about 10^{-8}\ \Omega\,\text{m} for metals, 10^{-5}\ \Omega\,\text{m} for semiconductors, and 10^{13}\ \Omega\,\text{m} for insulators. Resistivity rises with temperature for metals, falls for semiconductors, and stays nearly constant for manganin and constantan. Conductivity is \sigma = \frac{1}{\rho}.
Stretched-wire pattern: a wire of resistance 5 Ω is stretched to double its length. Its length becomes 2l, and since volume is constant, its area halves to \frac{a}{2}. Then R' = \rho \frac{2l}{a/2} = 4\rho\frac{l}{a} = 4 \times 5 = 20\ \Omega. Resistance becomes four times the original value, while the specific resistance is unchanged.
8.10 CHOICE OF MATERIAL OF A WIRE
The material of a wire is chosen according to its purpose. Connection and power transmission wires use copper or aluminium because their resistivity is very small, and they are made thick to keep resistance low. Silver is avoided despite lower resistivity because of its high cost.
Standard resistors use alloys like manganin and constantan, whose resistivity is high and changes little with temperature. A fuse wire is a lead-tin alloy with a low melting point and resistivity higher than copper, so excess current melts it.
A bulb filament uses tungsten for its high melting point. Heating elements in heaters, toasters and ovens use nichrome, whose resistivity is high and rises strongly with temperature.
8.11 SUPERCONDUCTORS
A superconductor is a substance whose resistance becomes almost zero at a very low temperature near absolute zero. Zero resistance means infinite conductivity: once current starts flowing, it persists even without a potential difference.
Examples are mercury below 4.2 K, lead below 7.25 K, and niobium below 9.2 K. Superconductors are not in common use because such low temperatures are hard to achieve. If obtained at room temperature, they could shrink computers and allow power lines as thin as a single wire.
8.12 ELECTRO-MOTIVE FORCE (E.M.F.) OF A CELL
The e.m.f. of a cell, denoted \varepsilon, is the energy spent per unit charge in taking a positive test charge around the complete circuit, both outside and inside the cell. If work W is done on charge q, then \varepsilon = \frac{W}{q}. Its unit is the volt.
E.m.f. is measured in open circuit with a high-resistance voltmeter. It depends on the material of the electrodes and on the electrolyte, but is independent of the shape of the electrodes, the distance between them, and the amount of electrolyte.
Standard values are 1.08 V for a voltaic cell, 1.5 V for a Leclanche cell, and 1.08 V for a Daniel cell.
8.13 TERMINAL VOLTAGE OF A CELL
Terminal voltage is the p.d. between the electrodes when the cell is in a closed circuit. If work W' is done in carrying charge q only around the external circuit, then V = \frac{W'}{q}.
The voltage drop inside the cell is v = \frac{w}{q}, the energy spent per unit charge in the electrolyte. By conservation of energy, \varepsilon = V + v, so V = \varepsilon - v. Terminal voltage is therefore less than e.m.f. when current flows, and it falls further as the current drawn increases.
| E.m.f. of cell | Terminal voltage of cell |
|---|---|
| Work done per unit charge in the complete circuit, inside and outside the cell | Work done per unit charge in the external circuit only |
| Independent of the current drawn | Depends on the current drawn; more current means less terminal voltage |
| Equal to terminal voltage when no current flows; greater when current flows | Equal to e.m.f. when no current flows; less when current flows |
8.14 INTERNAL RESISTANCE OF A CELL
The internal resistance r is the resistance offered by the electrolyte inside the cell to the flow of current. Its unit is the ohm, and it causes the voltage drop v = Ir when current I is drawn.
For a cell of e.m.f. \varepsilon connected to external resistance R, the total resistance is R + r. The current is I = \frac{\varepsilon}{R + r}, and since V = IR, the internal resistance is r = \left(\frac{\varepsilon}{V} - 1\right)R.
Internal resistance is lower with larger electrode surface area and higher electrolyte temperature, and higher with greater electrode distance and higher electrolyte concentration.
Worked pattern: a cell of e.m.f. 6 V reads 4.5 V across a 9 Ω resistor, so r = \left(\frac{6}{4.5} - 1\right) \times 9 = 3\ \Omega.
8.15 COMBINATION OF RESISTORS
In series, the same current flows through each resistor and the p.d. divides. The equivalent resistance is R_s = R_1 + R_2 + R_3. For n equal resistors it is nR. In parallel, the p.d. across each resistor is the same and the current divides. Then \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}, and for two resistors R_p = \frac{R_1 R_2}{R_1 + R_2}.
The parallel equivalent is always less than the smallest resistor. In mixed circuits, solve each parallel branch first, then add the series parts. On a V-I graph, the series combination gives a steeper line than the parallel combination.
| Feature | Series | Parallel |
|---|---|---|
| Current | Same through every resistor | Divides among the branches |
| Potential difference | Divides across the resistors | Same across every resistor |
| Equivalent resistance | R_1 + R_2 + R_3 | \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}} |
| Result | Greater than any single resistor | Less than the smallest resistor |
Worked pattern: a 12 Ω and a 6 Ω resistor in parallel give R_p = \frac{12 \times 6}{12 + 6} = 4\ \Omega. Connect this in series with an 8 Ω resistor, and the total is 12 Ω. Across a 24 V battery, the current is I = \frac{24}{12} = 2\ \text{A}.
8.16 ELECTRICAL ENERGY
Electrical energy converts into other forms of energy in daily devices. In heaters and ovens, it becomes heat energy. In a bulb filament, it becomes heat and light energy.
In motors and fans, it changes mostly into mechanical energy with a small part as heat. Charging a lead accumulator turns electrical energy into chemical energy. A loudspeaker converts it to sound energy, and a coil around a soft iron bar converts it to magnetic energy.
8.17 MEASUREMENT OF ELECTRICAL ENERGY (EXPRESSION W = QV = VIt)
By the definition of potential difference, the work done in moving charge Q through a p.d. V is W = QV. Since Q = It, this becomes W = VIt.
Using Ohm’s law, two more forms follow: W = I^2Rt and W = \frac{V^2t}{R}. All four expressions are equivalent, and the S.I. unit of electrical energy is the joule.
Choose the convenient form by what you know. Use VIt when V and I are given, I^2Rt for series circuits, and \frac{V^2t}{R} for parallel circuits.
8.18 ELECTRICAL POWER AND ITS EXPRESSION P = W/t = VI
Electric power is the rate at which electrical energy is supplied. Since P = \frac{W}{t} and W = VIt, we get P = VI. By Ohm’s law, P = \frac{V^2}{R} and P = I^2R.
The S.I. unit of power is the watt: 1\ \text{W} = 1\ \text{V} \times 1\ \text{A} = 1\ \text{J}\,\text{s}^{-1}. Bigger units are 1\ \text{kW} = 10^3\ \text{W}, 1\ \text{MW} = 10^6\ \text{W}, and 1\ \text{GW} = 10^9\ \text{W}.
Worked pattern: a current of 2 A flows at 10 V, so P = 2 \times 10 = 20\ \text{W}.
8.19 COMMERCIAL UNIT OF ELECTRICAL ENERGY
Electrical energy consumed is W = P \times t. For practical billing, larger units are used. One watt-hour is 1\ \text{Wh} = 3600\ \text{J}. One kilowatt-hour is 1\ \text{kWh} = 3.6 \times 10^6\ \text{J}.
The kilowatt-hour is the unit in which household electricity consumption is measured and billed. This commercial unit is the basis of the cost calculations in this ICSE chapter.
Worked pattern: a 1000 W appliance run for 3 h consumes 1 \times 3 = 3\ \text{kWh} = 1.08 \times 10^7\ \text{J}.
8.20 POWER RATING OF COMMON ELECTRICAL APPLIANCES
Every electrical appliance carries a power rating, which states the power it draws when operated at its rated voltage. Common examples listed in the chapter include the electric bulb, fan, heater, toaster and television.
The energy consumed by an appliance in a given time is the rating multiplied by the time of use. The rating therefore lets you compare how much electricity different appliances consume.
8.21 HOUSEHOLD CONSUMPTION OF ELECTRICAL ENERGY
For each appliance, energy is power times time, expressed in kWh. The total household consumption is the sum over all appliances, and the cost is the total kWh multiplied by the rate per kWh.
Worked pattern: two 60 W bulbs run 5 h a day use 2 \times 0.06 \times 5 = 0.6\ \text{kWh}. A 1000 W heater run 2 h uses 1 \times 2 = 2\ \text{kWh}. The daily total is 2.6 kWh. Over 30 days that is 78 kWh. At ₹5 per kWh, the monthly cost is 78 \times 5 = ₹390.
8.22 HEATING EFFECT OF CURRENT
When current flows through a resistor, electrical energy converts into heat. The heat produced is given by Joule’s law of heating: H = I^2Rt, where I is the current, R the resistance, and t the time.
This effect has practical applications. Electric heaters and ovens use it directly. A fuse wire melts when excess current flows, and a bulb filament glows because of the heat produced.
Worked pattern: a 5 Ω resistor carrying 2 A for 60 s produces H = 2^2 \times 5 \times 60 = 1200\ \text{J}.
What to practise from Current Electricity Class 10 ICSE
This chapter carries a large question bank. Work through the blocks below in order, and check every numerical against the formulas in sections 8.5, 8.9, 8.14 and 8.15, since those sections feed most of the circuit problems.
- EXERCISE-8(A) — 33 questions
- MULTIPLE CHOICE TYPE — 13 questions
- EXERCISE-8(B) — 11 questions
- MULTIPLE CHOICE TYPE — 34 questions
- MULTIPLE CHOICE TYPE — 22 questions
For more practice across the syllabus, browse the Class 10 ICSE Physics book pages and the wider Class 10 book collection. The next chapter, Household Circuits, builds directly on the fuse and heating concepts from this one. The official CISCE syllabus is the reference for the exact scope of the board examination.
Current Electricity Class 10 ICSE frequently asked questions
Why is the terminal voltage of a cell less than its e.m.f.?
Some energy is spent in pushing charge through the electrolyte inside the cell. That energy appears as a voltage drop v = Ir. By conservation of energy, \varepsilon = V + v, so the terminal voltage V is less than the e.m.f. whenever current flows.
What is the difference between ohmic and non-ohmic resistors?
Ohmic resistors obey Ohm’s law, so \frac{V}{I} is constant and the V-I graph is a straight line through the origin. Non-ohmic resistors give a curved V-I graph, and their resistance changes with V or I. Metals are ohmic; LEDs and bulb filaments are non-ohmic.
How do I calculate the equivalent resistance of resistors in series and parallel?
In series, add the resistances directly: R_s = R_1 + R_2 + R_3. In parallel, add the reciprocals: \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}. For two parallel resistors, use R_p = \frac{R_1R_2}{R_1 + R_2}.
What does the slope of the V-I graph represent in Current Electricity Class 10 ICSE?
On a V-I graph, the slope \frac{\Delta V}{\Delta I} equals the resistance R of the conductor. A steeper line means a larger resistance. On an I-V graph, however, the slope \frac{\Delta I}{\Delta V} equals \frac{1}{R}, the reciprocal of resistance.
What is the commercial unit of electrical energy and how is it related to the joule?
The commercial unit is the kilowatt-hour. One kilowatt-hour equals 3.6 \times 10^6\ \text{J}, since 1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s}. Electricity bills are calculated by multiplying the total kWh consumed by the rate per kWh.
Reference: ICSE Class 10 Physics textbook material, chapter 8: Current Electricity — CISCE-aligned.