Electricity

This chapter builds electric circuits from first principles, one linked idea at a time. It opens by defining electric current as the rate of flow of charge and introduces the ammeter, then asks what drives that flow, introducing potential difference and the voltmeter. It states Ohm's law, the proportionality between potential difference and current that defines resistance, then asks what physical properties fix that resistance, arriving at resistivity and $R = \rho l/A$, and explains why copper, nichrome and tungsten each suit different jobs. It then works out what happens when resistors combine, first in series and then in parallel, deriving the equivalent-resistance formula for each and explaining why household wiring uses parallel combinations. Finally it asks where the electrical energy goes, deriving Joule's law of heating ($H = I^2Rt$) and defining electric power ($P = VI$), ending with the kilowatt hour and applications such as fuses and bulb filaments.

Electric current and electric charge

Just as water current is the flow of water through a river, electric current is the flow of electric charge through a conductor. In a torch, the cell provides a continuous, closed path — an electric circuit — along which charge can flow; break that path anywhere and the current stops immediately, not gradually. Current was historically taken to flow in the direction positive charge would move, and this convention has stuck even though it is negatively charged electrons that actually drift through a wire — so conventional current is always opposite to actual electron flow. If a net charge $Q$ flows across any cross-section of a conductor in time $t$, the current $I$ through that cross-section is the charge flowing per unit time.

where: $I$ is current in amperes (A), $Q$ is charge in coulombs (C), and $t$ is time in seconds (s). One coulomb is the charge of roughly $6\times10^{18}$ electrons, each carrying $1.6\times10^{-19}\ \text{C}$. One ampere is one coulomb flowing per second, $1\ \text{A} = 1\ \text{C/s}$; smaller currents use milliampere ($10^{-3}\ \text{A}$) or microampere ($10^{-6}\ \text{A}$). Current is measured with an ammeter, always in series so the full current is forced through it.

Potential difference and the volt

Charge does not flow through a wire on its own, just as water in a perfectly horizontal, closed tube does not flow by itself; it flows only when one end sits at a higher level, creating a pressure difference. Similarly, electrons move only when there is a difference of "electrical pressure" between two points, called the potential difference, which a cell maintains using its stored chemical energy. We define potential difference between two points as the work done in moving a unit positive charge from one point to the other.

where: $V$ is potential difference in volts (V), $W$ is work done in joules (J), and $Q$ is charge in coulombs (C). One volt is the potential difference when one joule of work moves one coulomb of charge, $1\ \text{V} = 1\ \text{J/C}$. Potential difference is measured with a voltmeter, always connected in parallel — the opposite rule to the ammeter, and a common circuit-drawing mistake.

Ohm's law and resistance

Is there a fixed relationship between the potential difference across a conductor and the current through it? Varying the number of cells across a nichrome wire and plotting potential difference against current shows the ratio $V/I$ stays constant, and the graph is a straight line through the origin. This is Ohm's law: the potential difference across a given metallic conductor is directly proportional to the current through it, provided its temperature stays the same.

where: $R$ is a constant for the conductor at a given temperature, called its resistance — the property of opposing charge flow. Its SI unit is the ohm, symbol $\Omega$; resistance is one ohm if one volt across a conductor drives one ampere through it, $1\ \Omega = 1\ \text{V/A}$. Rearranged, $R = V/I$ gives resistance from measured values, while $I = V/R$ shows current is inversely proportional to resistance at fixed voltage. A low-resistance component is a good conductor; one with appreciable resistance is a resistor; a still higher one is an insulator. A rheostat is a variable resistor used to change resistance, and hence current, without changing the source voltage.

Resistivity: what fixes a conductor's resistance

Why do different components allow different currents on the same source? Swapping a nichrome wire for one of twice the length, then a thicker wire, then a copper wire of identical dimensions shows resistance depends directly on length, inversely on area of cross-section, and on the material itself.

where: $l$ is the length of the conductor (m), $A$ is its area of cross-section ($\text{m}^2$), and $\rho$ (rho) is the resistivity of the material, a property independent of shape, SI unit ohm metre ($\Omega\,\text{m}$). Metals and alloys have low resistivity, $10^{-8}$ to $10^{-6}\ \Omega\,\text{m}$; insulators like rubber and glass reach $10^{12}$ to $10^{17}\ \Omega\,\text{m}$. Both resistance and resistivity change with temperature. This explains material choices: nichrome (nickel, chromium, manganese and iron) resists oxidising at high temperature, so heating devices use it rather than pure metal; tungsten's high melting point, about $3380\ ^{\circ}\text{C}$, suits bulb filaments; and low-resistivity copper and aluminium are used for wiring and transmission lines, where wasted heat matters.

Combining resistors: series and parallel

Real circuits rarely use just one resistor. Resistors joined end to end, so the same current passes through each in turn, are connected in series; resistors connected between the same pair of points, giving current several paths, are connected in parallel. The two arrangements behave very differently, worked out fully in the derivations below.

where: $R_s$ is the series equivalent resistance, always greater than even the largest individual resistance; $R_p$ is the parallel equivalent, always smaller than even the smallest. Series circuits are impractical for household use — a bulb and a heater (needing very different currents) cannot share a series loop, and one failed component breaks the whole circuit. Parallel circuits avoid both problems: every appliance gets the full potential difference, switches independently, and total resistance decreases as more appliances are added.

Heating effect of current and electric power

Whenever charge flows through a resistor, the source that drives it must keep supplying energy, because the resistor continuously opposes the flow. If a charge $Q$ moves through a potential difference $V$ in time $t$, the work done by the source is $W = VQ$, and if the circuit is purely resistive, this entire input energy is dissipated as heat — the heating effect of electric current.

where: $H$ is the heat produced (J), $I$ is the current (A), $R$ is resistance ($\Omega$), and $t$ is time (s). This is Joule's law of heating: heat is directly proportional to the square of the current, to the resistance, and to the time current flows. The rate at which this energy is delivered is electric power.

where: $P$ is power in watts (W), consumed by a device carrying $1\ \text{A}$ at $1\ \text{V}$, so $1\ \text{W} = 1\ \text{V}\times 1\ \text{A}$. Since the watt is small for household use, electrical energy is billed in kilowatt hours (kWh), or "units": one kWh is the energy used by a $1000\ \text{W}$ appliance running for one hour, $1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$. Two everyday devices exploit this physics. A fuse, in series with a circuit, is a thin wire with a specific melting point; if current exceeds its rated value, Joule heating melts it and breaks the circuit — a $1\ \text{kW}$, $220\ \text{V}$ iron draws about $4.5\ \text{A}$, so needs a $5\ \text{A}$ fuse, not a $2\ \text{A}$ one. A bulb filament exploits heating deliberately: tungsten's very high melting point lets it glow brightly without melting, enclosed in inert gas to slow its decay.

Board level - charge, resistance and heat from a torch bulb (extending Example 11.1)

A torch bulb draws a steady current of 0.5 A for 10 minutes from a 3 V battery. Find (a) the charge that flows, (b) the filament's resistance, and (c) the heat produced.

Step 1. Convert time to seconds and find the charge using Eq. (1): $t = 10\ \text{min} = 600\ \text{s}$, so $Q = It = 0.5\ \text{A} \times 600\ \text{s} = 300\ \text{C}$, matching NCERT's own Example 11.1.

Step 2. Find the resistance using Ohm's law, Eq. (3), rearranged as $R = V/I$: $R = 3\ \text{V}/0.5\ \text{A} = 6\ \Omega$.

Step 3. Find the heat produced using Joule's law, Eq. (7): $H = I^2Rt = (0.5)^2 \times 6 \times 600 = 900\ \text{J}$. Check via power: $P = VI = 1.5\ \text{W}$, $H = Pt = 900\ \text{J}$ — same answer. Answer: Q = 300 C, R = 6 Ω, and H = 900 J.

Board level - a lamp and a resistor in series (extending Example 11.7)

A lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected in series to a 6 V battery. Find (a) the total resistance, (b) the current, (c) the potential difference across each, and (d) the power dissipated in each.

Step 1. Total resistance in series, Eq. (5): $R_s = R_1 + R_2 = 20\ \Omega + 4\ \Omega = 24\ \Omega$.

Step 2. Current from Ohm's law applied to the whole loop: $I = V/R_s = 6\ \text{V}/24\ \Omega = 0.25\ \text{A}$, the same current through both components since they are in series.

Step 3. Potential difference across each: $V_1 = IR_1 = 0.25 \times 20 = 5\ \text{V}$ across the lamp, $V_2 = IR_2 = 0.25 \times 4 = 1\ \text{V}$ across the conductor. Check: $V_1 + V_2 = 6\ \text{V}$, matching the battery, confirming Eq. (5).

Step 4. Power dissipated, using $P = I^2R$: lamp $P_1 = (0.25)^2 \times 20 = 1.25\ \text{W}$; conductor $P_2 = (0.25)^2 \times 4 = 0.25\ \text{W}$. Sum $1.5\ \text{W}$ equals $P = VI = 1.5\ \text{W}$. Answer: Rs = 24 Ω, I = 0.25 A, the lamp gets 5 V and 1.25 W, the conductor gets 1 V and 0.25 W.

Board HOTS / foundation JEE-NEET - a mixed series-parallel network (extending Example 11.9)

$R_1 = 10\ \Omega$ and $R_2 = 40\ \Omega$ are in parallel, and this combination is in series with a second parallel group of $R_3 = 30\ \Omega$, $R_4 = 20\ \Omega$, $R_5 = 60\ \Omega$, across a 12 V battery. Find (a) total resistance and current, (b) current through each resistor, and (c) power in R1.

Step 1. Reduce each parallel group using Eq. (6). For $R_1, R_2$: $1/R' = 1/10 + 1/40 = 5/40$, so $R' = 8\ \Omega$. For $R_3, R_4, R_5$: $1/R'' = 1/30 + 1/20 + 1/60 = 6/60$, so $R'' = 10\ \Omega$.

Step 2. $R'$ and $R''$ are in series, so total resistance $R = R' + R'' = 18\ \Omega$, and total current $I = V/R = 12/18 = 0.67\ \text{A}$.

Step 3. Find the voltage across each parallel group, since in series the total voltage splits between them: $V' = IR' = 5.33\ \text{V}$ across the $R_1$-$R_2$ group, $V'' = IR'' = 6.67\ \text{V}$ across the $R_3$-$R_4$-$R_5$ group. Check: $5.33 + 6.67 = 12.0\ \text{V}$.

Step 4. Within each group every resistor sees its group's voltage, so apply Ohm's law branch by branch: $I_1 = 5.33/10 = 0.53\ \text{A}$, $I_2 = 5.33/40 = 0.13\ \text{A}$ (sum $0.67\ \text{A}$, matches); $I_3 = 6.67/30 = 0.22\ \text{A}$, $I_4 = 6.67/20 = 0.33\ \text{A}$, $I_5 = 6.67/60 = 0.11\ \text{A}$ (sum $0.67\ \text{A}$, matches). Power in $R_1$: $P_1 = I_1^2R_1 = (0.53)^2 \times 10 = 2.8\ \text{W}$. Answer: R = 18 Ω, total I ≈ 0.67 A; branch currents are I1 ≈ 0.53 A, I2 ≈ 0.13 A, I3 ≈ 0.22 A, I4 ≈ 0.33 A, I5 ≈ 0.11 A; and P1 ≈ 2.8 W. Timing note: budget about four minutes. Always collapse every parallel group into a single equivalent resistance first, add the series pieces for one total resistance and current, then work backwards — voltage across each group, then current through each branch. Finding a branch current before its group's voltage is the most common error on this problem type.

• Swapping ammeter and voltmeter connections. Ammeter in series, voltmeter in parallel — the wrong way gives a meaningless or dangerous reading.
• Mixing up conventional current with electron flow. Conventional current flows positive to negative through the external circuit — opposite to actual electron drift.
• Treating parallel resistance like series resistance. $R_p$ comes from $1/R_p = 1/R_1 + 1/R_2 + \dots$, not from simply adding; forgetting the final reciprocal is a common slip.
• Losing track of which quantity stays constant. Series: same current, voltages add. Parallel: same voltage, currents add. Swapping these causes most network mistakes.
• Forgetting to square the current in Joule's law. $H = I^2Rt$, not $H = IRt$ — doubling current quadruples heat, not just doubles it.
• Confusing power and energy. A watt is a rate; a kilowatt hour is energy — a bill charges for kilowatt hours used, never a device's kilowatt rating alone.
• Getting area wrong from a given diameter. $A = \pi r^2$ needs the radius, not the diameter.

• Define electric current and potential difference in terms of charge and work, and state their SI units and measuring instruments.
• State Ohm's law, including its constant-temperature condition, and use $V = IR$ to find any one of $V$, $I$ or $R$.
• Explain how a conductor's resistance depends on length, area and material, and use $R = \rho l/A$ in problems.
• Derive the series and parallel equivalent-resistance formulas from Ohm's law, and explain why one increases and the other decreases total resistance.
• Explain why household circuits use parallel wiring, using both the independent-operation and failure-tolerance arguments.
• State and apply Joule's law, $H = I^2Rt$, distinguishing it clearly from electric power.
• Compute electric power using $P = VI = I^2R = V^2/R$, choosing the simplest form for a given problem.
• Convert joules to kilowatt hours and compute an appliance's running cost over a stated time and tariff.
• Explain, using resistivity and Joule heating, why filaments use tungsten, coils use nichrome, wires use copper, and fuses melt at a set current.

Ask the tutor: Want to work through a household wiring problem — say, sizing the correct fuse for a room with a fan, two tube lights and an iron all running in parallel — using exactly this chapter's formulas?