Chapter 11
Electricity
ELECTRIC CURRENT is the rate of flow of electric charge: I = Q / t. Unit: the AMPERE (A). 1 A = 1 coulomb per second. Current is measured with an AMMETER, which is always connected in SERIES.
CHARGE IS QUANTISED: Q = n × e, where n is the number of electrons and e = 1.6 × 10⁻¹⁹ C is the magnitude of the charge on ONE electron. So 1 coulomb is the charge carried by about 6 × 10¹⁸ electrons (1 ÷ 1.6 × 10⁻¹⁹ = 6.25 × 10¹⁸). Combined with I = Q/t this lets you find how many electrons flow in a given time — a standard board numerical.
CONVENTION (a favourite exam trap): inside a metal the ELECTRONS move from the negative terminal to the positive. But the direction of CONVENTIONAL CURRENT is taken as the direction positive charge would flow — from the POSITIVE terminal to the NEGATIVE. So conventional current is OPPOSITE to the direction of electron flow.
POTENTIAL DIFFERENCE between two points = the WORK DONE to move a UNIT charge from one point to the other: V = W / Q. Unit: the VOLT (V). 1 V = 1 joule per coulomb. Measured with a VOLTMETER, always connected in PARALLEL (across the component).
OHM'S LAW: the potential difference across a conductor is DIRECTLY PROPORTIONAL to the current through it, PROVIDED THE TEMPERATURE REMAINS CONSTANT. So V = IR, where R is the resistance. Unit of resistance: the OHM (Ω). Dropping the temperature condition is the commonest way to lose the mark.
The V–I GRAPH of an ohmic conductor is a STRAIGHT LINE THROUGH THE ORIGIN, and its SLOPE gives the resistance R (since R = V / I). A steeper line means a larger resistance.
RESISTANCE is the property of a conductor that OPPOSES the flow of charge through it. R = V / I. DEFINITION OF ONE OHM (an examable definition): if a potential difference of 1 VOLT across a conductor drives a current of 1 AMPERE through it, its resistance is 1 OHM. So 1 Ω = 1 V / 1 A.
ELECTRICAL ENERGY = POWER × TIME. This is the relation you actually use for bill numericals: work out the power in kilowatts, multiply by the hours of use, and the answer is already in kilowatt hours (units).
FOUR FACTORS AFFECTING RESISTANCE: (1) LENGTH — R is directly proportional to L, so a longer wire has MORE resistance; (2) AREA OF CROSS-SECTION — R is inversely proportional to A, so a thicker wire has LESS resistance; (3) the MATERIAL of the conductor; (4) its TEMPERATURE (the resistance of a metal increases as temperature rises).
R = ρL/A, where ρ (rho) is the RESISTIVITY of the material. Unit: the OHM METRE (Ω m). Resistivity depends only on the MATERIAL and its TEMPERATURE — it does NOT depend on the length or thickness of the wire, whereas resistance does. (Heating a metal RAISES ρ, and that is exactly how a rise in temperature raises R.)
Metals such as COPPER and ALUMINIUM have very low resistivity, so they are used for electrical wires. Rubber, glass and ebonite have enormous resistivity and are INSULATORS.
WHY ALLOYS ARE USED IN HEATING ELEMENTS (toasters, irons, electric heaters): alloys such as NICHROME have HIGHER resistivity than pure metals AND do not oxidise (burn) readily at high temperature. That is why heating elements are made of nichrome and not of copper.
RESISTORS IN SERIES: joined end to end, so there is only ONE path. (a) The SAME current flows through every resistor. (b) The potential differences ADD: V = V₁ + V₂ + V₃. (c) The equivalent resistance is the SUM: Rs = R₁ + R₂ + R₃. Rs is always LARGER than the largest individual resistance.
RESISTORS IN PARALLEL: all joined between the SAME TWO POINTS. (a) The SAME potential difference is across each resistor. (b) The CURRENTS add: I = I₁ + I₂ + I₃. (c) The reciprocals add: 1/Rp = 1/R₁ + 1/R₂ + 1/R₃. Rp is always SMALLER than the smallest individual resistance.
WHY DOMESTIC CIRCUITS USE PARALLEL CONNECTION: each appliance gets the FULL supply voltage; each can be switched on and off INDEPENDENTLY; if one appliance fails the others keep working; and each appliance can draw the different current it needs.
HEATING EFFECT — JOULE'S LAW OF HEATING: H = I²Rt. The heat produced is directly proportional to (a) the SQUARE of the current, (b) the resistance, and (c) the time. Note the square: DOUBLING the current gives FOUR times the heat.
THE ELECTRIC BULB: the filament is made of TUNGSTEN because it has a very high melting point (about 3380 °C) and high resistivity. The bulb is filled with a chemically INACTIVE gas such as NITROGEN or ARGON, which makes the filament last longer.
THE ELECTRIC FUSE: a short piece of wire of HIGH resistance and LOW melting point, connected in SERIES in the circuit. If the current exceeds a safe value, the fuse wire MELTS, the circuit breaks, and the appliances are protected from damage.
ELECTRIC POWER is the rate at which electrical energy is consumed: P = VI. Using Ohm's law this gives two more forms that are essential in numericals: P = I²R and P = V²/R. Unit: the WATT (W); 1 W = 1 joule per second. Choose the form that matches the quantities the question gives you.
COMMERCIAL UNIT OF ELECTRICAL ENERGY: the KILOWATT HOUR (kW h), which is what is called '1 UNIT' on an electricity bill. 1 kW h = 3.6 × 10⁶ J (that is, 1000 W × 3600 s). A bill for 200 units means 200 kW h of energy was consumed.
Electric current
I = Q / t
Unit: ampere (A). 1 A = 1 C/s. Ammeter connected in SERIES.
Quantisation of charge
Q = n e, e = 1.6 × 10⁻¹⁹ C
1 C = the charge on about 6 × 10¹⁸ electrons. Chain with I = Q/t to count electrons.
Potential difference
V = W / Q
Unit: volt (V). 1 V = 1 J/C. Voltmeter connected in PARALLEL.
Ohm's law
V = I R
Valid only at CONSTANT TEMPERATURE. The V–I graph is a straight line whose slope is R.
Resistance of a wire
R = ρ L / A
ρ = resistivity (Ω m) — depends on the MATERIAL and its TEMPERATURE, not on the wire dimensions. Longer → more R; thicker → less R.
Resistors in series
Rs = R₁ + R₂ + R₃
Same current through each; potential differences add. Rs > the largest resistance.
Resistors in parallel
1/Rp = 1/R₁ + 1/R₂ + 1/R₃
Same p.d. across each; currents add. Rp < the smallest resistance.
Two resistors in parallel (shortcut)
Rp = R₁R₂ / (R₁ + R₂)
Handy for exactly TWO resistors — product over sum. For three or more, use the reciprocal form.
Joule's law of heating
H = I² R t
Heat depends on the SQUARE of the current — double I gives 4× the heat.
Electric power
P = V I = I² R = V² / R
Unit: watt (W). Use whichever form matches the given quantities.
Commercial unit of energy
1 kW h = 3.6 × 10⁶ J
1 kW h = '1 unit' on the electricity bill = 1000 W × 3600 s.
The direction of conventional current in a circuit is:
easyTry answering on paper first — then reveal the model answer. 📄
Solve in your notebook. Stuck? Take the hint before the solution. ✏️
1. A 4 Ω, a 6 Ω and a 12 Ω resistor are all connected in PARALLEL across a 12 V battery. Find (a) the equivalent resistance, (b) the current through each resistor, and (c) the total current drawn. Verify your answer for the total current in two different ways.
medium2. A wire of resistance 10 Ω is cut into five equal pieces, and these five pieces are then joined in parallel. Find the resistance of the combination.
hard3. An electric heater of 1500 W is used for 4 hours a day. If electricity costs ₹6 per unit, calculate the cost of running it for 30 days.
medium4. Two lamps rated 100 W, 220 V and 60 W, 220 V are connected in parallel to a 220 V supply. Find the current drawn by each and the total current from the supply.
medium5. You are given three resistors of 2 Ω, 3 Ω and 6 Ω. Show how you would connect them to obtain a total resistance of (a) 11 Ω and (b) 1 Ω.
medium6. A copper wire has a resistance of 5 Ω. A second copper wire has twice the length and twice the diameter of the first. What is the resistance of the second wire? Explain each step.
hard7. A fuse rated 5 A is fitted in a household circuit operating at 220 V. Determine the maximum total power of appliances that can be safely run, and state whether a 1500 W heater together with a 750 W iron can be used at the same time.
hard8. A current of 0.5 A flows through a conductor for 4 minutes. Calculate (a) the charge that passes through it and (b) the number of electrons that constitute that charge.
medium