Class 10 Science – Chapter 11: Electricity

1. Introduction to Electricity

When charges flow through a conductor we get an electric current. In a metal wire, the free (mobile) electrons carry the charge. Their drift from the negative terminal to the positive terminal of a cell is what we call current inside the wire.

Electric circuit: a continuous, closed path made of conducting wires and devices such as cells, switches and bulbs, along which electric current flows. If the path is broken anywhere (switch open, or a loose wire), the current stops — this is an open circuit. When the path is complete, current flows — a closed circuit.

Conventional direction of current: by convention, current is taken to flow from the positive terminal of the cell, through the external circuit, to the negative terminal. This is opposite to the direction of electron flow (electrons actually move from negative to positive). The convention was fixed before electrons were discovered, and it is still used while drawing circuit diagrams.

Examples:

• If 2 C of charge flows through a wire in 4 s, then \(I = 2/4 = 0.5\ \text{A}\).
• A 0.2 A current for 10 s carries \(Q = I t = 0.2 \times 10 = 2\ \text{C}\) of charge.

Basic Circuit Components

Every simple circuit needs a source of energy (cell or battery), conducting wires, a control device (switch or key) and a load (such as a bulb or resistor).

Circuit diagram: a picture of a circuit drawn with standard symbols is called a circuit diagram. Using symbols keeps the drawing simple and universal. The standard symbols used for the cell, battery, switch, bulb, resistor, ammeter, voltmeter, rheostat and fuse are collected in section 17.

2. Electric Potential and Potential Difference

Potential difference: the difference in electric potential between two points in a circuit. It is the work done in moving a unit positive charge from one point to the other. A cell (or battery) creates a potential difference across a circuit, and charges flow because of this difference — from higher to lower potential outside the cell. The potential difference across the two terminals of a cell is also called the voltage of the cell.

Examples:

• If 5 J of work is done in moving 2 C of charge between two points, the potential difference is \(V = 5/2 = 2.5\ \text{V}\).
• A 12 V battery supplies 12 J of energy to every coulomb of charge that passes through it.

A battery is a combination of one or more cells. The potential difference of a cell is fixed by the chemical reactions inside it (for example, a common dry cell gives about 1.5 V). The higher the potential difference, the more work is available to push charge around the circuit, so a bulb connected to a higher-voltage source gets more energy per coulomb.

3. Electric Circuit and Its Components

A simple electric circuit consists of a cell (or battery), a switch, connecting wires, a load such as a resistor (or bulb), and two meters commonly used in experiments — an ammeter and a voltmeter. The next three points tell you exactly how each meter must be connected.

In Fig. 2 and Fig. 3 notice the two simple rules again: the ammeter is in series (current enters and leaves the ammeter on the same line) and the voltmeter is in parallel (it forms a separate branch across the component). Reversing these connections gives wrong readings and can even damage the meters.

4. Electric Current

Charge and current relationship: if a net charge \(Q\) flows in time \(t\), the current is \(I = Q/t\), and a steady current \(I\) in time \(t\) carries charge \(Q = It\). If each electron carries \(e = 1.6 \times 10^{-19}\ \text{C}\), the number of electrons that flow when charge \(Q\) passes is

\[n = \frac{Q}{e}\]

• Example: charge \(Q = 3\ \text{C}\) flowing in \(1.5\ \text{s}\) gives \(I = 3/1.5 = 2\ \text{A}\).
• Example: the number of electrons crossing a point each second for a 1 A current is \(n = 1/(1.6 \times 10^{-19}) \approx 6.25 \times 10^{18}\).

Current in series circuits: in a series circuit the same current flows through every component, because there is only one path. If the current entering the circuit is \(I\), the same \(I\) flows through each resistor. This is why an ammeter placed anywhere in series reads the same value.

Measuring Current with an Ammeter

An ammeter is a device that measures current. It is connected in series with the circuit component whose current we want; the meter’s reading is the current flowing through it (and hence through the series circuit). An ideal ammeter has zero resistance so that it does not disturb the circuit. A milliammeter measures small currents (in mA).

5. Ohm’s Law

Meaning of resistance: resistance is the property of a conductor that opposes the flow of current. The larger the resistance, the smaller the current for a given potential difference. By rearranging \(V = IR\),

\[R = \frac{V}{I}\]

SI unit: ohm (Ω). \(1\ \text{Ω} = 1\ \text{V/A}\). A conductor has a resistance of 1 ohm if a potential difference of 1 V across it produces a current of 1 A.

Varying the rheostat changes the current. For each setting, the voltmeter records \(V\) and the ammeter records \(I\). For an ohmic (metallic) conductor at constant temperature, the ratio \(V/I\) stays the same, and the V–I graph is a straight line passing through the origin, as shown below. The slope of this line gives the resistance.

Some conductors (for example a filament lamp as its temperature rises, or a semiconductor diode) do not follow Ohm’s law; their V–I graph is not a straight line. Such devices are called non-ohmic conductors.

• Example: a current of 2 A flows through a 5 Ω resistor, so \(V = I R = 2 \times 5 = 10\ \text{V}\).
• Example: if 12 V across a conductor gives 3 A, then \(R = 12/3 = 4\ \text{Ω}\).

6. Resistance

Factors on which the Resistance of a Conductor Depends

For a wire of length \(l\) and uniform cross-sectional area \(A\), the resistance depends on four factors:

• Length \(l\): at constant temperature, resistance is directly proportional to length — \(R \propto l\). A longer wire offers more resistance because the charges to collide with atoms along a longer path.
• Cross-sectional area \(A\): resistance is inversely proportional to the area of cross-section — \(R \propto 1/A\). A thicker wire offers less resistance because there are more free electrons available.
• Nature of material: for the same length and thickness, different materials have different resistances; this is described by their resistivity \(\rho\).
• Temperature: for most metallic conductors the resistance increases as temperature rises (due to more vigorous vibration of atoms). For alloys like nichrome and manganin, the change of resistance with temperature is very small.

Worked reference: a copper wire of length \(l = 10\ \text{m}\) and area \(A = 1\ \text{mm}^2 = 1 \times 10^{-6}\ \text{m}^2\) has \(R = (1.62 \times 10^{-8})(10)/(1 \times 10^{-6}) = 0.162\ \text{Ω}\). The same dimensions in nichrome give about \(1.0 \times 10^{-6}(10)/(1 \times 10^{-6}) = 10\ \text{Ω}\) — over sixty times larger, which is why nichrome is used for heating coils.

• Example: doubling the length of a wire doubles its resistance (if the area is unchanged).
• Example: doubling the diameter doubles the radius, so the area becomes 4 times and the resistance becomes \(\frac{1}{4}\) of the original.

7. Resistivity

Notice the difference: the resistance of a wire changes with its length and thickness, but resistivity of the material stays the same for the same material at the same temperature.

Silver has the lowest resistivity and is the best conductor, but copper is cheaper, so electric wiring is mostly copper or aluminium. Alloys like nichrome have high resistivity and do not oxidise (burn) easily even at high temperature, so they are used to make heating elements.

Examples: a wire of length 50 m and area \(0.5\ \text{mm}^2\) (\(5 \times 10^{-7}\ \text{m}^2\)) has resistance \(R = 5\ \text{Ω}\). Its resistivity is \(\rho = RA/l = (5 \times 5 \times 10^{-7})/50 = 5 \times 10^{-8}\ \text{Ω m}\).

8. Resistors in Series

When resistors are connected one after the other so that the same current passes through each, they are said to be in series.

In a series circuit:

• The same current flows through every resistor (one path only).
• The potential difference divides among the resistors — \(V = V_1 + V_2 + V_3\), where each \(V_i = I R_i\).
• The total potential difference of the source equals the sum of the drops across the resistors.

Examples:

• Resistors of 5 Ω, 10 Ω and 15 Ω in series give \(R = 5 + 10 + 15 = 30\ \text{Ω}\).
• A battery of 12 V joined to 4 Ω and 8 Ω in series gives \(R = 12\ \text{Ω}\) and circuit current \(I = 12/12 = 1\ \text{A}\). The drops are \(1 \times 4 = 4\ \text{V}\) and \(1 \times 8 = 8\ \text{V}\), adding to 12 V.

Disadvantage in household wiring: if any appliance in a series circuit fails or is switched off, the circuit breaks and every other appliance stops working. That is why household circuits use parallel wiring instead (section 10).

9. Resistors in Parallel

When two or more resistors are connected between the same two points, each providing an independent path for the current, they are said to be in parallel.

In a parallel circuit:

• The same potential difference exists across every branch (each branch is joined to the same two points).
• The current divides among the branches — \(I = I_1 + I_2 + I_3\), with a larger current in the branch of smaller resistance.
• Adding a resistor in parallel gives the current an extra path, so the total resistance of the combination decreases.

Examples:

• Two resistors of 2 Ω and 3 Ω in parallel give \(R = (2 \times 3)/(2 + 3) = 6/5 = 1.2\ \text{Ω}\), which is smaller than both 2 Ω and 3 Ω.
• Two 12 Ω resistors in parallel give \(R = 144/24 = 6\ \text{Ω}\). For \(n\) equal resistors \(R\) each, the parallel combination gives \(R/n\).

Advantage in household wiring: every appliance receives the full mains voltage and works independently; switching one appliance on or off does not affect the others.

10. Series and Parallel Combinations — Comparison

Choosing between series and parallel depends on what we need. Decorative lights in a string are sometimes in series; household wiring is always in parallel.

11. Heating Effect of Electric Current

Notice the heat is proportional to the square of the current — doubling the current makes the heat four times larger. The formula \(I^2 R t\) works for any circuit; the forms \(VIt\) and \(V^2 t/R\) follow by substituting Ohm’s law.

• Example: a current of 2 A through a 5 Ω resistor for 60 s produces \(H = 4 \times 5 \times 60 = 1200\ \text{J}\).
• Example: an electric iron draws 5 A at 220 V for 2 minutes (120 s), giving \(H = 220 \times 5 \times 120 = 132{,}000\ \text{J}\).
• Example: across a 6 Ω coil with 12 V applied for 5 minutes (300 s), \(H = (144/6) \times 300 = 7200\ \text{J}\).

12. Applications of the Heating Effect

The heating effect of current is put to use in many household devices:

The rating of a fuse (for example, 5 A, 10 A) is the maximum safe current it can pass. The fuse must have a rating slightly above the normal current drawn by the appliance, so that it does not melt during normal use but melts during a fault.

13. Electric Power

How to choose the right power formula:

• Use \(P = VI\) when \(V\) and \(I\) are known (or easy to find).
• Use \(P = I^2 R\) when \(I\) and \(R\) are known (for example, a resistor carrying a known current).
• Use \(P = V^2/R\) when \(V\) and \(R\) are known (for example, an electric bulb rated in volts and watts).

• Example: a bulb draws 0.5 A at 220 V, so \(P = 220 \times 0.5 = 110\ \text{W}\).
• Example: a 2 kW heater on a 220 V supply draws \(I = P/V = 2000/220 \approx 9.1\ \text{A}\).
• Example: a device rated 100 W operating at 220 V has resistance \(R = V^2/P = 220^2/100 = 484\ \text{Ω}\).

14. Electrical Energy and the Kilowatt-Hour

Electrical energy is the total work (or heat) done by the current over a period of time. It is power multiplied by time in which the power is used:

Commercial unit: The joule is a very small unit for household consumption, so electricity boards measure energy in the kilowatt-hour (kWh), also written as one "unit".

\[1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^6\ \text{J}\]

• Example: a 100 W bulb used for 10 h consumes \(100\ \text{W} \times 10\ \text{h} = 1000\ \text{Wh} = 1\ \text{kWh}\).
• Example: 1 kWh in joules is \(1\ \text{kWh} = 3.6 \times 10^6\ \text{J}\).
• Example: a 2 kW air-conditioner run for 5 h consumes \(2 \times 5 = 10\ \text{kWh}\).

15. Electricity Bills and Energy Consumption

To find the energy consumed by appliances, we convert the power of each appliance to kilowatts, multiply by the time used (in hours), and add the kilowatt-hours of all appliances. The reading of an electricity meter (difference between two readings) gives the units consumed for billing.

Worked example: A household uses a 1000 W iron for 2 h a day, four 60 W bulbs for 5 h a day, and a 40 W fan for 8 h a day. Energy per day:

• Iron: \(1000\ \text{W} = 1\ \text{kW}\), energy = \(1 \times 2 = 2\ \text{kWh}\).
• Bulbs: \(4 \times 60\ \text{W} = 240\ \text{W} = 0.24\ \text{kW}\), energy = \(0.24 \times 5 = 1.2\ \text{kWh}\).
• Fan: \(40\ \text{W} = 0.04\ \text{kW}\), energy = \(0.04 \times 8 = 0.32\ \text{kWh}\).
• Total per day = \(2 + 1.2 + 0.32 = 3.52\ \text{kWh}\).

For 30 days the total is \(3.52 \times 30 = 105.6\ \text{kWh}\). At a hypothetical tariff of ₹6 per unit, the bill is \(105.6 \times 6 =\) ₹633.6. (Real tariffs vary by state and slab; any tariff used in an exam or worksheet is always given in the question.)

16. Important Formulas

These are the formulas of the whole chapter in one place. Remember the SI unit written beside each one.

1. Electric current

\[I = \frac{Q}{t}\]   \(\Rightarrow I\) in ampere (A), \(Q\) in coulomb (C), \(t\) in second (s). \(1\ \text{A} = 1\ \text{C/s}\).

2. Potential difference

\[V = \frac{W}{Q}\]   \(\Rightarrow V\) in volt (V), \(W\) in joule (J), \(Q\) in coulomb (C). \(1\ \text{V} = 1\ \text{J/C}\).

3. Ohm’s law

\[V = I R\]   \(\Rightarrow R = V/I\) in ohm (Ω). \(1\ \text{Ω} = 1\ \text{V/A}\).

4. Resistance of a wire

\[R = \rho \frac{l}{A}\]   \(\Rightarrow \rho\) in ohm-metre (Ω m), \(l\) in metre (m), \(A\) in m².

5. Resistors in series

\[R = R_1 + R_2 + R_3\]

6. Resistors in parallel

\[\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]

7. Two resistors in parallel

\[R = \frac{R_1 R_2}{R_1 + R_2}\]

8. Heating effect (Joule’s law of heating)

\[H = I^2 R t = V I t = \frac{V^2 t}{R}\]   \(\Rightarrow H\) in joule (J).

9. Electric power

\[P = \frac{W}{t} = V I = I^2 R = \frac{V^2}{R}\]   \(\Rightarrow P\) in watt (W). \(1\ \text{kW} = 1000\ \text{W}\).

10. Electrical energy

\[E = P t\]   \(\Rightarrow\) in kWh when \(P\) is in kW and \(t\) in hours.

11. kWh conversion

\[1\ \text{kWh} = 3.6 \times 10^6\ \text{J}\]

17. Circuit Symbols and Important Diagrams

Drawing a circuit with standard symbols keeps the picture simple and correct. Learn these symbols:

Every important circuit diagram of this chapter appears (numbered) at the place where the concept is explained. Use this index to practise drawing them:

18. Solved Numerical Examples

Attempt each problem on your own first, then check the working. Always write the Given, the Formula, then substitute with units, and end with a boxed final answer with its unit.

Electric Current and Charge

Electric Potential and Potential Difference

Ohm’s Law and Resistance

Resistance of a Wire and Combinations

Heating Effect

Electric Power and Electrical Energy

19. Application-Based Examples

These questions connect the formulas to real appliances and house wiring. Explain the physics, then compute.

20. Competency and Reasoning Questions

21. Case-Based Learning

Read each case, answer the questions step by step, and check the working. All tariffs are hypothetical.

22. Important Comparison Tables

23. Common Mistakes to Avoid

24. Exam Tips

• Solve numericals systematically: write Given (with units), choose the formula, substitute, calculate, and state the final answer with its unit. This layout earns marks even if the last step errs.
• Identify the formula from the data: look at which quantities you are given. V and I → \(P = VI\); I and R → \(P = I^2R\); V and R → \(P = V^2/R\). The same logic works for \(H\) and \(R\).
• Draw circuit diagrams neatly: use straight lines for wires, standard symbols (learn Fig. 11), label every component, and indicate the ammeter in series and voltmeter in parallel.
• Equivalent resistance: check whether the resistors are chained (series → add) or between the same two points (parallel → reciprocals). Use \(R_1R_2/(R_1+R_2)\) for two parallel resistors.
• Handle units carefully: convert all lengths to metres, areas to m², power to kilowatts (in billing sums), and seconds/hours according to the unit asked.
• Electricity-bill questions: find kWh by kW × hours, add for all appliances, multiply by the (given) tariff per unit. The tariff is always provided in the question.
• Present answers clearly: box the final value with its unit, and add a one-line interpretation ("so the 5 A fuse is unsuitable") to show understanding.
• Interpreting V–I graphs: remember slope = resistance; steeper line = larger resistance; a straight line through the origin = ohmic conductor at constant temperature.

25. Quick Revision (5–10 Minutes)

One-line pointers

• Conventional current: positive to negative; electrons flow the opposite way.
• Ammeter in series; voltmeter in parallel.
• V–I graph is a straight line through the origin for an ohmic conductor.
• Equivalent resistance is larger than the largest resistor in series, smaller than the smallest in parallel.
• Heat depends on \(I^2\), so doubling current quadruples heating.
• Household wiring is parallel so appliances work independently at full voltage.
• Fuse (series, live wire) melts when current exceeds its rating.
• Energy (kWh) = power (kW) × time (h); multiply by the tariff for the bill.
• Alloys like nichrome (high resistivity, high melting point) are used for heating elements.

26. Self-Practice Questions

27. Answer Key and Hints