Electricity

An electric circuit is a continuous and closed path along which electric charge can flow. Break the path anywhere, by opening a switch or a plug key, and the current stops. That is all the torch does: the switch closes the path between the cells and the bulb.

Electric current is the rate of flow of electric charge through a cross-section of a conductor. If a net charge Q flows through a cross-section in time t, then I = Q/t. The SI unit of charge is the coulomb (C). One coulomb is the charge carried by about 6 × 10¹⁸ electrons, because each electron carries 1.6 × 10⁻¹⁹ C. The SI unit of current is the ampere (A), named after André-Marie Ampère: one ampere is one coulomb of charge flowing per second, 1 A = 1 C/s. Smaller currents are written in milliampere (1 mA = 10⁻³ A) and microampere (1 μA = 10⁻⁶ A).

Two things about direction trip students up. In a metal the charges that actually move are electrons, which are negative. By convention, though, the direction of current is taken as the direction in which positive charge would flow, that is from the positive terminal of the cell, through the circuit, to the negative terminal. So the conventional current is opposite to the direction of electron flow. Say this clearly when asked.

Current is measured with an ammeter. It is always connected in series with the part of the circuit where the current is to be measured, so that the whole current passes through it, and it has a very low resistance so that it does not itself reduce the current. A typical one-mark numerical: a current of 0.5 A flows for 10 minutes; the charge that passes is Q = It = 0.5 × 600 = 300 C, and the number of electrons is 300 ÷ (1.6 × 10⁻¹⁹) = 1.875 × 10²¹. Convert minutes to seconds before you do anything else.

Charge does not flow through a wire on its own, just as water does not flow through a horizontal pipe unless there is a pressure difference between its ends. In a conductor the electrons need a difference of electric pressure, called the potential difference, to push them along. A cell or a battery maintains this potential difference by the chemical action inside it, and it keeps doing so as long as the chemicals last.

The potential difference between two points in an electric circuit carrying some current is defined as the work done to move a unit charge from one point to the other: V = W/Q. Its SI unit is the volt (V), after Alessandro Volta. One volt is the potential difference between two points when 1 joule of work is done to move a charge of 1 coulomb between them: 1 V = 1 J/C. So a 6 V battery gives 6 J of energy to every coulomb that passes through it, a favourite one-mark question.

The potential difference is measured with a voltmeter. It is always connected in parallel across the two points between which the potential difference is required, and it has a very high resistance so that hardly any current is diverted through it. Contrast this with the ammeter, which goes in series and has low resistance. Students regularly swap these two in circuit diagrams and in one-mark questions, so fix the pair now: ammeter, series, low resistance; voltmeter, parallel, high resistance.

A simple numerical: how much work is done in moving a charge of 2 C across two points having a potential difference of 12 V? W = VQ = 12 × 2 = 24 J. Keep the definition and the unit sentence ready; together they are a standard 2-mark answer.

A circuit diagram is a drawing of a circuit using conventional symbols instead of pictures of the real components. Examiners expect the NCERT symbols exactly, and a circuit drawn with the wrong symbols or wrong placement of meters can cost a full mark even when the calculation is right.

The symbols you must be able to draw: an electric cell is a long thin line next to a short thick line, the longer line being the positive terminal; a battery is two or more such cells in a row; a plug key or switch is drawn open (a gap with a plug beside it) or closed (the plug in the gap); a wire joint is a dot where the lines meet; wires crossing without joining are drawn as one line hopping over the other with no dot; an electric bulb is a circle with a loop inside it; a resistor of fixed resistance is a small rectangle (or a zigzag); a variable resistance or rheostat is a rectangle with an arrow through it; an ammeter is a circle with the letter A and a positive and negative terminal; a voltmeter is a circle with the letter V and its terminals.

Three placement rules decide the marks. The ammeter is in series so that the entire current passes through it. The voltmeter is in parallel across the component whose potential difference is wanted. The positive terminal of each meter must be connected towards the positive terminal of the cell, otherwise the pointer deflects the wrong way. The plug key is placed in series so that it can break the whole circuit.

When a question says 'draw a circuit diagram to show a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor and a 12 Ω resistor in series with a plug key and an ammeter', draw the loop neatly with straight lines and right-angle corners, label every component with its value, and put the ammeter anywhere in the loop, since in series the current is the same everywhere. Then, if asked, note that the ammeter reads 6 V ÷ 25 Ω = 0.24 A.

Ohm's law states that the potential difference V across the ends of a given metallic wire in an electric circuit is directly proportional to the current I flowing through it, provided its temperature remains the same. That is, V ∝ I, or V/I = constant = R, or V = IR. The constant R is the resistance of the conductor. It is a property of the conductor that opposes the flow of charge. Its SI unit is the ohm (Ω): if a potential difference of 1 V across the ends of a conductor makes a current of 1 A flow through it, its resistance is 1 Ω, so 1 Ω = 1 V/1 A. Both the statement of the law and the definition of the ohm carry marks, so write both when asked to state Ohm's law.

The verification is NCERT Activity 11.1 and the board asks for it as a 3- or 5-mark question. Set up a circuit with a nichrome wire XY, an ammeter, a voltmeter across XY, a plug key, and a cell. Take the reading of the ammeter and the voltmeter with one cell, then repeat with two, three and four cells in series. Tabulate V and I. You find that V/I is the same in every row. Plot V on the y-axis against I on the x-axis: the graph is a straight line passing through the origin, and its slope, V/I, gives the resistance of the wire. A steeper line (larger V for the same I) means a larger resistance. The board version often gives you the table or the graph and asks you to find R, or gives two graphs for wires A and B and asks which has more resistance: the steeper one when V is on the y-axis.

Rewriting the law as I = V/R shows why resistance matters: for the same potential difference, doubling the resistance halves the current. A component that gives a large resistance is a resistor; one with a small resistance is a good conductor. A variable resistance, or rheostat, is used in a circuit to change the current without changing the source.

Two conditions belong to the law and examiners look for them: it applies to metallic conductors, and the temperature and other physical conditions must stay constant. The filament of a bulb does not obey it once it heats up, because its resistance rises with temperature.

NCERT Activity 11.2 leads to three results. A longer wire of the same material and thickness gives a smaller current, so resistance is directly proportional to length: R ∝ l. A thicker wire of the same material and length gives a larger current, so resistance is inversely proportional to the area of cross-section: R ∝ 1/A. A wire of a different material with the same length and thickness gives a different current, so resistance depends on the nature of the material. Resistance also depends on temperature (it rises for metals as they get hotter). Combining the first two, R = ρl/A, where ρ (rho) is a constant of proportionality called the electrical resistivity of the material. Its SI unit is Ω m (from ρ = RA/l).

Resistivity depends only on the material and its temperature, not on the length or thickness of the sample; that is the answer to the one-mark question 'on what does the resistivity of a wire depend'. Metals and alloys have very low resistivity, of the order of 10⁻⁸ to 10⁻⁶ Ω m, and are good conductors. Insulators like rubber and glass have resistivity of the order of 10¹² to 10¹⁷ Ω m. From the NCERT table, copper is 1.62 × 10⁻⁸ Ω m, aluminium 2.63 × 10⁻⁸, tungsten 5.20 × 10⁻⁸, iron 10.0 × 10⁻⁸, mercury 94.0 × 10⁻⁸, and the alloy nichrome 100 × 10⁻⁸ Ω m, so iron conducts better than mercury.

Two application statements come up every year. The resistivity of an alloy is generally higher than that of its constituent metals, and alloys do not oxidise (burn) readily at high temperature, so alloys such as nichrome are used in electrical heating devices like toasters and electric irons. Copper and aluminium have among the lowest resistivities, so they are used for electrical transmission lines.

The favourite numerical trap is a wire that is stretched or cut. Stretching a wire to double its length halves its area (volume is constant), so R = ρl/A becomes ρ(2l)/(A/2) = 4R. Cutting a wire into n equal parts gives each part R/n; connecting those parts in parallel gives R/n². If a wire's length is doubled and you want the same resistance, the area must also be doubled.

Two or more resistors are in series when they are joined end to end so that the same current flows through each of them. NCERT Activity 11.4 shows this: moving the ammeter to different positions in a series circuit does not change its reading. The potential difference across the whole combination, however, is shared: Activity 11.5 shows that the voltmeter reading across the battery equals the sum of the voltmeter readings across the individual resistors.

The derivation the board rewards, step by step. Let three resistors R₁, R₂ and R₃ be connected in series across a battery of potential difference V, and let the current through the circuit be I. In series the current is the same through each resistor. By Ohm's law the potential differences across them are V₁ = IR₁, V₂ = IR₂ and V₃ = IR₃. The total potential difference is the sum: V = V₁ + V₂ + V₃. If the combination is replaced by a single equivalent resistance R_s that draws the same current I from the same battery, then V = IR_s. Equating, IR_s = IR₁ + IR₂ + IR₃, so R_s = R₁ + R₂ + R₃. The equivalent resistance of resistors in series equals the sum of the individual resistances. It is always greater than the largest individual resistance, and n identical resistors R in series give nR.

Worked example in the NCERT style: a 9 V battery is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. Total resistance = 13.4 Ω, so I = 9/13.4 = 0.67 A, and since the current is the same everywhere in series, 0.67 A flows through the 12 Ω resistor too. To find the potential difference across one resistor, multiply that resistor by the common current: across the 12 Ω resistor it is 0.67 × 12 ≈ 8.06 V.

The examiner is looking for four things in the derivation: the labelled circuit diagram, the statement that current is the same, the addition of potential differences, and the final boxed result. Skip any one and a mark goes.

Resistors are in parallel when they are connected between the same two points, so that the potential difference across each of them is the same, while the current from the battery divides among them (NCERT Activity 11.6).

The derivation. Let R₁, R₂ and R₃ be connected in parallel across a potential difference V. Let the currents through them be I₁, I₂ and I₃, and let the total current be I. The total current is the sum of the branch currents: I = I₁ + I₂ + I₃. By Ohm's law, I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃. If R_p is the equivalent resistance, then I = V/R_p. Substituting, V/R_p = V/R₁ + V/R₂ + V/R₃, and dividing by V, 1/R_p = 1/R₁ + 1/R₂ + 1/R₃. The reciprocal of the equivalent resistance in parallel is the sum of the reciprocals of the individual resistances. R_p is always less than the smallest resistance in the group. For two resistors, R_p = R₁R₂/(R₁ + R₂), and for n identical resistors R in parallel, R_p = R/n.

A typical numerical: 10 Ω, 15 Ω and 5 Ω in parallel. 1/R_p = 1/10 + 1/15 + 1/5 = (3 + 2 + 6)/30 = 11/30, so R_p = 30/11 = 2.73 Ω, smaller than 5 Ω as it must be. Combination questions ask you to get, say, 13.5 Ω and 6 Ω from three 9 Ω resistors: two in parallel (4.5 Ω) in series with the third gives 13.5 Ω; two in series (18 Ω) in parallel with the third gives 18 × 9/27 = 6 Ω.

Why domestic circuits use parallel and not series, a 2- or 3-mark question: in series the current is the same through every appliance, so a bulb and a heater that need very different currents cannot both work properly; if one appliance fails or is switched off, the circuit breaks and all the others stop; and the total resistance is high, so the current is small. In parallel each appliance gets the full supply voltage, can be switched on and off independently, and the total resistance is low.

A battery or cell is a source of electrical energy. Its chemical reaction maintains the potential difference that drives the current, and to keep the current going the source must keep spending energy. Part of that energy may do useful work (turn a motor, for instance); the rest is spent in heat that raises the temperature of the appliance. In a purely resistive circuit, one with only resistors connected to a battery, the whole of the source energy is dissipated as heat. This is the heating effect of electric current.

The derivation, which the board asks for directly. Consider a current I flowing through a resistor of resistance R, with potential difference V across it. Let a charge Q flow through it in time t. The work done in moving the charge Q through a potential difference V is W = VQ. Therefore the source supplies energy VQ in time t, and the power input is P = VQ/t = VI. The energy supplied in time t is Pt = VIt. This energy is dissipated in the resistor as heat, so H = VIt. Using Ohm's law V = IR, H = I²Rt. This is Joule's law of heating.

The law in words, three parts, each worth a mark: the heat produced in a resistor is directly proportional to the square of the current for a given resistance, directly proportional to the resistance for a given current, and directly proportional to the time for which the current flows.

Worked example, NCERT Example 11.9 style: an electric iron of resistance 20 Ω takes a current of 5 A. Heat developed in 30 s is H = I²Rt = 5² × 20 × 30 = 25 × 20 × 30 = 15000 J. Choose the form of the formula that uses the quantities given. If V is fixed and R changes, use H = V²t/R; halving R at the same voltage doubles the heat. If I is fixed, use I²Rt.

The heating of a conductor is often unwanted. It raises the temperature of components and changes their properties, and a lot of the energy in a bulb is lost as heat rather than light. But the same effect is used deliberately in the electric laundry iron, the immersion heater, the toaster, the oven, the kettle and the heater. Their heating elements are made of nichrome, an alloy, for three reasons that examiners like listed: nichrome has a high resistivity so that a short coil gives a lot of heat, a high melting point so that it does not melt at red heat, and it does not oxidise readily at high temperatures.

The electric bulb works on the same effect. The filament must retain as much of the heat as possible so that it gets very hot and emits light, so it must not melt at such high temperatures. Tungsten is used almost exclusively for the filament because it has a very high melting point (3380 °C) and a high resistivity. The bulbs are filled with chemically inactive gases like nitrogen and argon, which prolong the life of the filament by preventing it from oxidising.

The electric fuse is the safety application. It protects circuits and appliances by stopping the flow of any unduly high current. A fuse is a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron or lead. If a current larger than the specified value flows through the circuit, the fuse wire heats up and melts, which breaks the circuit. The fuse is placed in series with the device, on the live wire. Domestic fuses are rated 1 A, 2 A, 3 A, 5 A, 10 A and so on. To pick the rating, work out the normal current: an electric iron of 1 kW on a 220 V supply draws 1000/220 = 4.5 A, so a 5 A fuse is used.

Why does the heating element of an electric heater glow red while its connecting cord stays cool? Both carry the same current, but the element has a much higher resistance (nichrome, thin, long) than the copper cord, and heat produced is I²Rt, so the element produces far more heat and glows.

Electric power is the rate at which electrical energy is consumed or dissipated in a circuit: P = W/t. Since the energy delivered in time t is VIt, P = VI. Using Ohm's law this can be written as P = I²R = V²/R. The SI unit is the watt (W): the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V, so 1 W = 1 volt × 1 ampere. Practical amounts are given in kilowatt, 1 kW = 1000 W.

Which form to use is the whole art of these numericals. When an appliance is rated, say 100 W at 220 V, the two numbers give you its resistance from R = V²/P = 220²/100 = 484 Ω, and its normal current from I = P/V = 100/220 = 0.45 A. That resistance is a property of the bulb and stays the same if it is put on a different voltage; only the power changes. So a 220 V, 100 W bulb on 110 V gives P = V²/R = 110²/484 = 25 W, a quarter. A 100 W bulb has a lower resistance than a 40 W bulb of the same rating. In parallel on 220 V (the household case) the 100 W bulb draws more current and glows brighter. In series the current is the same through both, so heat produced is I²R and the 40 W bulb, with the higher resistance, glows brighter.

Electrical energy is power multiplied by time: W = Pt. The joule is too small for household use, so the commercial unit is the kilowatt hour: the energy consumed at the rate of 1 kW for 1 hour. 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J. This is the 'unit' on your electricity bill. Note carefully: the watt and kilowatt are units of power, the kilowatt hour is a unit of energy; confusing the two is a marked error.

Worked bill example: four 60 W bulbs used for 5 hours a day for 30 days consume 4 × 60 × 5 × 30 = 36000 Wh = 36 kWh. At ₹6 per unit the cost is ₹216.