Chapter 3 · Current Electricity

3.2 Resistivity, Its Temperature Dependence and Electrical Power

Not everything obeys Ohm's law, resistivity spans more than twenty powers of ten from silver to glass, metals resist more when hot while semiconductors resist less — and every resistor turns electrical energy into heat at the rate $P=VI=I^2R$.

Measure the resistance of a 60 W filament bulb with a multimeter and you get a few tens of ohms. Yet at its working voltage of $230$ V it draws only about a quarter of an ampere, which means a resistance near $900\ \Omega$. Nothing is wrong with the meter: the filament at $2500\ ^\circ\text{C}$ is a very different conductor from the same filament at room temperature. The same physics makes a toaster element glow, lets a platinum wire act as a precise thermometer in a laboratory or a hospital, and makes the thermistor in a phone charger cut the current when the battery warms up.

In Section 3.1 we met Ohm's law and the resistivity $\rho$ of a material, and explained both with the drift of free electrons. This section asks where Ohm's law fails, compares the resistivities of real materials, studies how resistivity changes with temperature (with the temperature coefficient of resistivity $\alpha$), and finally works out the electrical power delivered to a resistor — the energy behind every heater, every electricity bill and every power line.

Where Ohm's Law Fails

Ohm's law is not a law of nature in the way that conservation of charge is. It is an observed property of many materials — metals at a steady temperature above all — and it is so useful that we name devices after whether they obey it.

📘 Definition 3.6 — Ohmic and non-ohmic devices
A device is ohmic if, under fixed physical conditions, the current through it is proportional to the potential difference across it, so that its $I$–$V$ graph is a straight line through the origin with the same slope for both directions of current. A device whose $I$–$V$ graph is not such a line is non-ohmic. For any device, the ratio $R=V/I$ at a particular operating point is called its (static) resistance there.

Devices depart from Ohm's law in three typical ways:

  1. $V$ is not proportional to $I$. A filament lamp: as the current grows, the filament heats up and its resistance rises, so the $I$–$V$ curve bends towards the $V$-axis. A thermistor does the opposite: as it warms up its resistance falls and the curve bends upwards.
  2. The relation depends on the sign of $V$. A semiconductor diode (Chapter 14) lets a large current through in one direction and almost none in the other.
  3. The relation is not unique. In some devices (gallium arsenide, for instance) the same current flows at more than one voltage: over part of the graph, the current falls as the voltage rises.

Even a perfectly ohmic metal wire becomes "non-ohmic" if it is allowed to heat up. Ohm's law is a statement about a material at a fixed temperature; the filament lamp is a metal whose temperature changes with the current.

✏️ Example 1 — Is a filament lamp ohmic?
A $12$ V car lamp draws $0.20$ A when $2.0$ V is applied across it and $0.50$ A at $12$ V. Its resistance measured with a very small current at room temperature is $7.2\ \Omega$. (a) Find its resistance at $2.0$ V and at $12$ V. Is it ohmic? (b) Find its power at $12$ V. (c) Estimate the current at the instant it is switched on at $12$ V, and explain why lamps usually fail at switch-on.
(a) Resistance $R=V/I$: $\dfrac{2.0}{0.20}=10\ \Omega$ and $\dfrac{12}{0.50}=24\ \Omega$. Six times the voltage gives only $2.5$ times the current: not ohmic. The filament is hotter at $12$ V.
(b) Power $P=VI=(12)(0.50)=6.0$ W (the formula is justified later in this section).
(c) Switch-on At the first instant the filament is still cold: $I\approx\dfrac{12}{7.2}\approx1.7$ A — over three times the working current. The cold filament gets a burst of power, heats unevenly and suffers thermal stress, so filaments usually burn out just as they are switched on.
Check Choose "filament lamp" in Interactive 1: the readouts at $2.0$ V and $12$ V give $V/I=10\ \Omega$ and $24\ \Omega$.
Interactive 1 — I–V Characteristics: Ohmic and Non-Ohmic

Choose a device and sweep the applied potential difference (press Play, or drag the slider). The dot runs along the device's $I$–$V$ curve. Compare the static resistance $V/I$ with the slope of the curve — they agree at every point only for the ohmic resistor. The dashed line joins the origin to the dot (its slope is $I/V$, so it measures the static resistance); the short red line is the tangent (its slope measures $\Delta I/\Delta V$).

2.0
$V$—
$I$—
static resistance $V/I$—
slope $\Delta V/\Delta I$—

Resistivity of Various Materials

Resistivity varies enormously from one material to another — far more than almost any other physical property. A centimetre cube of silver has a resistance of about $10^{-6}\ \Omega$; a centimetre cube of glass can have $10^{12}\ \Omega$ or more. Materials fall into three broad groups:

Material$\rho$ at $20\ ^\circ\text{C}$ ($\Omega\ \text{m}$)$\alpha$ ($^\circ\text{C}^{-1}$)Typical use
silver$1.6\times10^{-8}$$3.8\times10^{-3}$contacts, conductive inks
copper$1.7\times10^{-8}$$3.9\times10^{-3}$house wiring, motor windings
aluminium$2.7\times10^{-8}$$4.0\times10^{-3}$overhead power lines
tungsten$5.6\times10^{-8}$$4.5\times10^{-3}$lamp filaments
platinum$1.1\times10^{-7}$$3.9\times10^{-3}$resistance thermometers
manganin (alloy)$4.4\times10^{-7}$$\approx2\times10^{-6}$standard resistors
nichrome (alloy)$1.1\times10^{-6}$$4\times10^{-4}$heating elements
silicon (pure)$\approx2\times10^{3}$negativechips (after doping)

Approximate values; real samples vary with purity and treatment. Two alloys stand out. Nichrome (nickel–chromium) has a resistivity about $65$ times that of copper and does not oxidise when red hot — ideal for heater elements. Manganin and constantan have resistivities that hardly change with temperature, so resistors made from them keep their value when they warm up; they are used for standard resistors and precision instruments.

✏️ Example 2 — Why power lines are made of aluminium
An overhead line must have a given resistance per kilometre. Compare an aluminium conductor with a copper one of the same length and resistance: find (a) the ratio of their cross-sectional areas, (b) the ratio of their masses. Densities: copper $8.9\times10^3\ \text{kg/m}^3$, aluminium $2.7\times10^3\ \text{kg/m}^3$.
(a) Areas Equal $R$ and $l$ in $R=\rho l/A$ need $A\propto\rho$: $\dfrac{A_\text{Al}}{A_\text{Cu}}=\dfrac{2.7\times10^{-8}}{1.7\times10^{-8}}\approx1.59$. The aluminium wire is thicker.
(b) Masses Mass $=$ density $\times\,Al$, so $\dfrac{m_\text{Al}}{m_\text{Cu}}=1.59\times\dfrac{2.7}{8.9}\approx0.48$.
Conclusion For the same resistance, aluminium weighs less than half as much as copper (and costs much less), so the pylons can be lighter and further apart. Copper wins inside buildings and machines, where space matters more than weight.

Temperature Dependence of Resistivity

Heat a copper wire from room temperature to $100\ ^\circ\text{C}$ and its resistance rises by about $30\%$. Over a moderate range of temperature the increase is very nearly linear, which suggests describing it by the fractional change of resistivity per degree.

📘 Definition 3.7 — Temperature coefficient of resistivity
If a material has resistivity $\rho_0$ at a reference temperature $T_0$, then over a limited range of temperature $$\rho_T=\rho_0\left[1+\alpha\,(T-T_0)\right],$$ where $\alpha=\dfrac{\rho_T-\rho_0}{\rho_0\,(T-T_0)}$ is the temperature coefficient of resistivity, the fractional change in resistivity per degree, in $^\circ\text{C}^{-1}$ (or $\text{K}^{-1}$, the same size of degree). Because the length and area of a wire change very little with temperature, the same law holds for its resistance: $R_T=R_0\left[1+\alpha\,(T-T_0)\right]$.
📐 Result 3.3 — How resistivity changes with temperature

Metals: $\alpha\gt0$, of the order of $4\times10^{-3}\ ^\circ\text{C}^{-1}$; the resistivity rises nearly linearly with temperature over a wide range (it curves at very low temperatures).

Alloys such as nichrome, manganin and constantan: $\alpha\gt0$ but very small, so their resistivity is almost independent of temperature.

Semiconductors: the resistivity decreases rapidly as the temperature rises ($\alpha\lt0$, and not constant), and insulators behave in the same way.

All three follow from $\rho=\dfrac{m}{ne^2\tau}$ (Result 3.2).

Why metals resist more when hot. In a metal the number $n$ of free electrons per unit volume hardly changes with temperature. But the ions of the lattice vibrate more vigorously as the metal warms up, so the electrons collide with them more often: the relaxation time $\tau$ falls, and $\rho\propto1/\tau$ rises. Why semiconductors resist less when hot. In a semiconductor $n$ is small at room temperature, because most electrons are bound. Heating frees more of them — $n$ grows very steeply with temperature, far outweighing the small fall in $\tau$ — so $\rho\propto1/n$ falls. A device built on this effect is a thermistor with a negative temperature coefficient (an NTC thermistor), the most common temperature sensor in electronics.

🧭 Problem-solving strategy — temperature and resistance
  1. Identify the reference temperature $T_0$ at which $R_0$ (or $\rho_0$) and $\alpha$ are quoted — often $0\ ^\circ\text{C}$ or $20\ ^\circ\text{C}$.
  2. For a single unknown use $R_T=R_0[1+\alpha(T-T_0)]$ directly, or its rearrangement $T-T_0=\dfrac{R_T-R_0}{\alpha R_0}$.
  3. If two readings $R_1$ at $T_1$ and $R_2$ at $T_2$ are given and $\alpha$ is referred to $0\ ^\circ\text{C}$, write both as $R=R_0(1+\alpha T)$ and solve the pair; do not divide by $R_1$ as if it were $R_0$.
  4. Check the sign: metals get hotter → larger $R$; semiconductors → smaller $R$.
✏️ Example 3 — A platinum resistance thermometer
A platinum thermometer has resistance $R_0=100.0\ \Omega$ at $0\ ^\circ\text{C}$, and for this grade of platinum $\alpha=3.85\times10^{-3}\ ^\circ\text{C}^{-1}$. (a) Its resistance in a bath is $138.5\ \Omega$. Find the temperature of the bath. (b) What does it read in a patient's mouth at $37.0\ ^\circ\text{C}$? (c) By how many ohms does its resistance change per degree?
(a) Temperature $$T=\frac{R_T-R_0}{\alpha R_0}=\frac{138.5-100.0}{(3.85\times10^{-3})(100.0)}=\frac{38.5}{0.385}=100\ ^\circ\text{C}$$ — a bath of boiling water.
(b) Body temperature $R=100.0\left[1+(3.85\times10^{-3})(37.0)\right]\approx114.2\ \Omega$.
(c) Sensitivity $\alpha R_0=0.385\ \Omega$ per $^\circ\text{C}$. A meter that resolves $0.004\ \Omega$ therefore resolves about $0.01\ ^\circ\text{C}$.
Note Over hundreds of degrees platinum is not exactly linear; the Research–Teaching Nexus at the end of this section measures how much that matters.
✏️ Example 4 — How hot is the motor winding?
The copper winding of a motor has a resistance of $2.00\ \Omega$ at $20\ ^\circ\text{C}$, when the motor is cold. After a long run its resistance is $2.30\ \Omega$. Find the average temperature of the winding. ($\alpha_\text{Cu}=3.9\times10^{-3}\ ^\circ\text{C}^{-1}$ at $20\ ^\circ\text{C}$.)
Concept Engineers use exactly this method to monitor windings they cannot reach with a thermometer: the winding is its own thermometer.
Rise $T-20=\dfrac{R_T-R_{20}}{\alpha R_{20}}=\dfrac{2.30-2.00}{(3.9\times10^{-3})(2.00)}\approx38.5\ ^\circ\text{C}$.
Answer $T\approx58\ ^\circ\text{C}$ (an average over the whole winding — hot spots are hotter).
Check In Interactive 2 choose copper and set $T=58\ ^\circ\text{C}$: the ratio $R_T/R_{20}$ reads $1.148\approx1.15$, and $2.00\ \Omega\times1.15=2.30\ \Omega$ ✓.
Interactive 2 — Resistance of a Wire: Material, Shape and Temperature

Pick a material and set the length and diameter of the wire: the drawing and $R=\rho l/A$ update. Then change the temperature (or press Play to heat and cool the wire): the marker moves along the straight line $R_T=R_{20}[1+\alpha(T-20)]$. Compare the steep line of a pure metal with the almost flat line of manganin.

201.0020
$\rho$ at $T$ (Ω m)—
area $A$ (m²)—
$R$ at 20 °C (Ω)—
$R$ at $T$ (Ω)—
$R_T/R_{20}$—

Electrical Energy and Power

Consider a conductor $AB$ carrying a steady current $I$ from $A$ to $B$, with a potential difference $V=V_A-V_B\gt0$ across it. In a time $\Delta t$ a charge $\Delta q=I\Delta t$ enters at $A$ and the same charge leaves at $B$. Its electric potential energy has fallen by $$\Delta U=\Delta q\,(V_A-V_B)=IV\,\Delta t.$$ If the charges moved freely this energy would become kinetic energy. But inside the conductor they keep colliding with the ions, so their average kinetic energy (their drift) stays the same; the energy is handed to the lattice at every collision and the ions vibrate more strongly — the conductor heats up. This is Joule heating. In a motor some of the energy becomes mechanical work instead, and in a battery being charged some is stored as chemical energy; but the rate at which electrical energy is delivered is always $IV$.

📐 Result 3.4 — Electrical power

The power delivered to any device carrying a current $I$ with a potential difference $V$ across it is $$P=VI.$$ For a resistor of resistance $R$ (where $V=IR$) all of it becomes heat at the rate $$P=I^2R=\frac{V^2}{R}.$$ The energy delivered in time $t$ at a steady rate is $W=Pt=VIt$. The SI unit of power is the watt ($1\ \text{W}=1\ \text{J/s}=1\ \text{V A}$); electricity is sold by the kilowatt hour, $1\ \text{kWh}=3.6\times10^{6}$ J.

Which form to use? $I^2R$ is convenient when the current is fixed — resistors in series, a transmission line carrying a given current. $V^2/R$ is convenient when the potential difference is fixed — appliances in parallel on the mains. At fixed $V$ a smaller resistance draws more power, which is why a $2$ kW heater element has a lower resistance than a $1$ kW one.

Power transmission. A power station must send a power $P$ along lines of total resistance $R_c$. The current is $I=P/V$, where $V$ is the transmission voltage, so the power wasted in the lines is $$P_\text{loss}=I^2R_c=\frac{P^2R_c}{V^2}.$$ Raising $V$ ten times cuts the loss a hundred times. That is why electricity is transmitted at hundreds of kilovolts and stepped down near homes with transformers (Chapter 7).

✏️ Example 5 — A room heater and the electricity bill
A heater is rated $1000$ W at $230$ V. (a) Find its resistance and the current it draws. (b) In the evening the mains voltage falls to $207$ V. Find the power now, assuming the resistance stays the same. (c) The heater is used for $2$ hours a day for $30$ days. How many units (kWh) is that, and what does it cost at ₹ 8 per unit?
(a) Rating $R=\dfrac{V^2}{P}=\dfrac{230^2}{1000}=52.9\ \Omega$; $I=\dfrac PV=\dfrac{1000}{230}\approx4.35$ A.
(b) Low voltage $P=\dfrac{207^2}{52.9}\approx810$ W. A $10\%$ fall in voltage costs $19\%$ of the power, because $P\propto V^2$.
(c) Energy $W=(1.0\ \text{kW})(2\ \text{h})(30)=60$ kWh. Cost: $60\times8=$ ₹ 480.
Check $60\ \text{kWh}=60\times3.6\times10^6=2.16\times10^{8}$ J — enough to lift a one-tonne car about $22$ km.
✏️ Example 6 — Why transmit at high voltage?
A power station sends $1.0$ MW to a town through lines of total resistance $5.0\ \Omega$. Find the current, the power lost in the lines and the percentage lost if the transmission voltage is (a) $10$ kV, (b) $100$ kV.
Concept $I=P/V$; the loss is $I^2R_c$ (the line's own current through its own resistance).
(a) 10 kV $I=\dfrac{1.0\times10^6}{1.0\times10^4}=100$ A; $P_\text{loss}=(100)^2(5.0)=5.0\times10^4$ W $=50$ kW, i.e. $5\%$.
(b) 100 kV $I=10$ A; $P_\text{loss}=(10)^2(5.0)=500$ W, only $0.05\%$.
Trap Do not use $V^2/R_c$ with the transmission voltage — that $V$ is not the potential difference across the line wires (most of it is across the town's load).
⚠️ Common mistakes with temperature and power

1. Treating $\alpha$ as if it were the same for every reference temperature. Strictly $\alpha$ referred to $0\ ^\circ\text{C}$ and $\alpha$ referred to $20\ ^\circ\text{C}$ differ slightly. For small ranges it hardly matters; in two-reading problems use the strategy box above.

2. Expecting every material to resist more when hot. Semiconductors and insulators do the opposite.

3. Using the rated power at a different voltage. "$1000$ W, $230$ V" fixes the resistance $V^2/P$. At any other voltage recompute $P=V^2/R$.

4. Confusing kW and kWh. The kilowatt is a rate (power); the kilowatt hour is an amount (energy).

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — conductors whose cross-section changes

Not required by the CBSE syllabus. Resistance and resistivity are listed by JEE Main, NEET and JEE Advanced; conductors of varying cross-section are a JEE-style application (it uses integration) that no syllabus names.

$R=\rho l/A$ needs a uniform cross-section. If the area varies along the current, slice the conductor into thin slabs of thickness $dx$ perpendicular to the current. The slabs are in series, so their resistances $\rho\,dx/A(x)$ add (uses integration): $$R=\int_0^L\frac{\rho\,dx}{A(x)}.$$ Truncated cone (radii $a$ and $b$, length $L$): the radius grows linearly and the integral gives $R=\dfrac{\rho L}{\pi ab}$ — reducing to $\rho L/\pi a^2$ when $a=b$.

Radial leakage between two coaxial cylinders (radii $a\lt b$, length $L$) filled with a poor conductor: the current flows outwards through cylindrical shells of area $2\pi rL$, so $$R=\int_a^b\frac{\rho\,dr}{2\pi rL}=\frac{\rho}{2\pi L}\ln\frac ba.$$ This gives the leakage resistance of a coaxial cable's insulation. The board examination uses only uniform conductors.

Exercises

I
Ohmic and non-ohmic devices; materials. Use the resistivities in the table above.
  • 1 A device carries $0.10$ A at $1.0$ V and $0.40$ A at $2.0$ V. Find $V/I$ at each point. Is it ohmic?
  • 2 Describe three different ways in which a device can fail to obey Ohm's law, with an example of each.
  • 3 Give two reasons why heater elements are made of nichrome rather than copper.
  • 4 Why are standard resistors made of manganin?
  • 5 A nichrome wire and a copper wire have the same length and diameter. Find the ratio of their resistances.
  • 6 Arrange silver, glass, pure silicon and nichrome in order of increasing resistivity.
  • 7 An aluminium wire is to replace a copper wire of diameter $2.0$ mm, with the same length and resistance. What diameter must it have?
  • 8 A $12$ V lamp takes $3.0$ A when lit; its cold resistance is $0.40\ \Omega$. Find its resistance when lit and the current at the instant of switching on.
1 $V/I=10\ \Omega$ and $5.0\ \Omega$. The ratio is not constant, so the device is non-ohmic (its resistance falls as the current rises — like a thermistor).
2 (i) $V$ not proportional to $I$ — filament lamp, thermistor. (ii) The relation depends on the direction of $V$ — diode. (iii) The relation is not unique (the same $I$ at more than one $V$) — gallium arsenide device.
3 Nichrome has a much higher resistivity, so a short, robust coil has the required resistance; and it does not oxidise (burn) at red heat. Its small $\alpha$ also keeps the power steady as it warms up.
4 Its temperature coefficient is tiny ($\approx2\times10^{-6}\ ^\circ\text{C}^{-1}$), so the resistance hardly changes when the room or the resistor warms up.
5 $\dfrac{R_\text{n}}{R_\text{c}}=\dfrac{\rho_\text{n}}{\rho_\text{c}}=\dfrac{1.1\times10^{-6}}{1.7\times10^{-8}}\approx65$.
6 Silver, nichrome, pure silicon, glass.
7 $A\propto\rho$, so $d\propto\sqrt\rho$: $d_\text{Al}=2.0\sqrt{\dfrac{2.7}{1.7}}\approx2.5$ mm.
8 Lit: $R=12/3.0=4.0\ \Omega$ — ten times the cold value. At switch-on $I\approx12/0.40=30$ A (in practice limited by the battery and wiring).
II
Temperature dependence. Take $\alpha_\text{Cu}=3.9\times10^{-3}\ ^\circ\text{C}^{-1}$ unless told otherwise.
  • 1 A copper coil has a resistance of $5.00\ \Omega$ at $20\ ^\circ\text{C}$. Find its resistance at $100\ ^\circ\text{C}$.
  • 2 A platinum thermometer has $R_0=100\ \Omega$ at $0\ ^\circ\text{C}$ and $\alpha=3.9\times10^{-3}\ ^\circ\text{C}^{-1}$. Find the temperature at which $R=125\ \Omega$.
  • 3 A nichrome element ($\alpha=4.0\times10^{-4}\ ^\circ\text{C}^{-1}$) has $R=50.0\ \Omega$ at $20\ ^\circ\text{C}$. Find its resistance at $820\ ^\circ\text{C}$.
  • 4 A tungsten filament ($\alpha=4.5\times10^{-3}\ ^\circ\text{C}^{-1}$) has $R=20.0\ \Omega$ at $20\ ^\circ\text{C}$ and $200\ \Omega$ when glowing. Estimate its working temperature. Why is the answer only an estimate?
  • 5 A wire has resistance $4.00\ \Omega$ at $20\ ^\circ\text{C}$ and $4.80\ \Omega$ at $70\ ^\circ\text{C}$. Find $\alpha$ referred to $20\ ^\circ\text{C}$.
  • 6 Use the electron model to explain why the resistivity of a metal increases with temperature.
  • 7 Explain why the resistivity of a semiconductor decreases with temperature.
  • 8 A carbon resistor ($\alpha=-5.0\times10^{-4}\ ^\circ\text{C}^{-1}$) and a copper resistor are joined in series. Their total resistance is $10.0\ \Omega$ and is to be independent of small changes of temperature. Find each resistance.
1 $R=5.00\left[1+(3.9\times10^{-3})(80)\right]=5.00(1.312)\approx6.56\ \Omega$.
2 $T=\dfrac{125-100}{(3.9\times10^{-3})(100)}=\dfrac{25}{0.39}\approx64\ ^\circ\text{C}$.
3 $R=50.0\left[1+(4.0\times10^{-4})(800)\right]=50.0(1.32)=66.0\ \Omega$ — only $32\%$ more after $800$ degrees.
4 $T-20=\dfrac{200-20.0}{(4.5\times10^{-3})(20.0)}=2000$, so $T\approx2.0\times10^{3}\ ^\circ\text{C}$. The linear law is only reliable over a limited range; over two thousand degrees tungsten's resistivity is noticeably non-linear (and the filament also expands), so this is a rough estimate.
5 $\alpha=\dfrac{4.80-4.00}{(4.00)(50)}=4.0\times10^{-3}\ ^\circ\text{C}^{-1}$.
6 $\rho=m/(ne^2\tau)$. In a metal $n$ is fixed, but hotter ions vibrate with larger amplitude, so collisions are more frequent: $\tau$ falls and $\rho$ rises.
7 In a semiconductor few electrons are free at room temperature; heating frees many more, so $n$ rises steeply. This outweighs the fall in $\tau$, so $\rho$ falls.
8 The total changes by $(R_\text{C}\alpha_\text{C}+R_\text{Cu}\alpha_\text{Cu})\Delta T$, which must vanish: $R_\text{Cu}(3.9\times10^{-3})=R_\text{C}(5.0\times10^{-4})$, so $R_\text{Cu}=0.128R_\text{C}$. With $R_\text{C}+R_\text{Cu}=10.0$: $R_\text{C}\approx8.86\ \Omega$, $R_\text{Cu}\approx1.14\ \Omega$.
III
Electrical power and energy. Assume resistances stay constant unless told otherwise.
  • 1 Find the resistance of a lamp rated $60$ W, $230$ V, and the current it draws.
  • 2 The same lamp is connected to a $115$ V supply. Find the power.
  • 3 A $2.0$ kW kettle is used for $15$ minutes a day for $30$ days. Find the energy used in kWh and the cost at ₹ 7 per unit.
  • 4 A $10\ \Omega$ resistor carries $2.0$ A. Find the power and the heat produced in $5.0$ minutes.
  • 5 Bulbs rated $100$ W and $60$ W (both for $220$ V) are joined in series across $220$ V. Which glows brighter? Find the power of each.
  • 6 The same two bulbs are joined in parallel across $220$ V. Which glows brighter now?
  • 7 Two heater coils work on the same mains supply. Coil $A$ has the lower resistance. Which coil gives out more heat per second, and why?
  • 8 A circuit has a $5$ A fuse on the $230$ V mains. Can a $1.5$ kW iron and a $100$ W lamp be used on it at the same time?
1 $R=\dfrac{V^2}{P}=\dfrac{230^2}{60}\approx882\ \Omega$; $I=\dfrac{60}{230}\approx0.26$ A.
2 Half the voltage gives a quarter of the power: $P=\dfrac{115^2}{882}=15$ W.
3 $W=(2.0\ \text{kW})(0.25\ \text{h})(30)=15$ kWh; cost $15\times7=$ ₹ 105.
4 $P=I^2R=(2.0)^2(10)=40$ W; $Q=Pt=(40)(300)=1.2\times10^{4}$ J.
5 $R_{100}=\dfrac{220^2}{100}=484\ \Omega$, $R_{60}=\dfrac{220^2}{60}\approx807\ \Omega$. In series $I=\dfrac{220}{1291}\approx0.170$ A, so $P_{100}=I^2(484)\approx14.1$ W and $P_{60}=I^2(807)\approx23.4$ W. The 60 W bulb glows brighter: same current, larger resistance, $P=I^2R$.
6 In parallel each gets the full $220$ V and its rated power: $100$ W and $60$ W. The 100 W bulb is brighter ($P=V^2/R$, smaller $R$).
7 Coil $A$. Both have the same $V$, and $P=V^2/R$ is larger for the smaller $R$.
8 Total $1600$ W, so $I=\dfrac{1600}{230}\approx7.0$ A $\gt5$ A. No — the fuse would blow.
IV
Applications and modelling. Copper: $\rho=1.7\times10^{-8}\ \Omega\ \text{m}$, $\alpha=3.9\times10^{-3}\ ^\circ\text{C}^{-1}$; water: $c=4200\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}$.
  • 1 $2.0$ MW is transmitted through lines of total resistance $4.0\ \Omega$. Find the power lost, and the percentage, at $20$ kV and at $200$ kV.
  • 2 A heater coil is rated $1000$ W at $230$ V. One-tenth of its length is cut off and the rest is used on the same supply. Find the new power. Is this a safe repair?
  • 3 Modelling. A $1.5$ kW kettle heats $1.0$ kg of water from $25\ ^\circ\text{C}$ to $100\ ^\circ\text{C}$. Find the least time it can take. Why is the real time longer?
  • 4 The copper winding of a fan has a resistance of $0.500\ \Omega$ at $20\ ^\circ\text{C}$ and $0.580\ \Omega$ after an hour's running. Find its temperature.
  • 5 Modelling. An electric car charges at $32$ A through a copper cable with two conductors, each $5.0$ m long and of cross-section $6.0\ \text{mm}^2$. Find the resistance of the conductors and the power wasted in the cable.
  • 6 A $12$ V battery is rated at $60$ ampere hours. How much energy can it deliver, in joules?
1 At $20$ kV: $I=100$ A, $P_\text{loss}=(100)^2(4.0)=40$ kW $=2\%$. At $200$ kV: $I=10$ A, $P_\text{loss}=400$ W $=0.02\%$.
2 $R'=0.9R$, so $P'=\dfrac{V^2}{0.9R}=\dfrac{1000}{0.9}\approx1.1\times10^{3}$ W. Not safe: the coil runs hotter than designed and the current rises by $11\%$.
3 $Q=mc\Delta T=(1.0)(4200)(75)=3.15\times10^{5}$ J; $t=\dfrac{Q}{P}=\dfrac{3.15\times10^5}{1500}=210$ s $=3.5$ min. In practice heat is also lost to the kettle body and the air.
4 $T-20=\dfrac{0.580-0.500}{(3.9\times10^{-3})(0.500)}\approx41$, so $T\approx61\ ^\circ\text{C}$.
5 Total length $10$ m: $R=\dfrac{(1.7\times10^{-8})(10)}{6.0\times10^{-6}}\approx0.028\ \Omega$; $P=I^2R=(32)^2(0.0283)\approx29$ W.
6 Charge $q=(60\ \text{A})(3600\ \text{s})=2.16\times10^5$ C; $W=qV=(2.16\times10^5)(12)\approx2.6\times10^{6}$ J ($0.72$ kWh).
🎯 Exam corner — resistivity, temperature and power BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Sketch, on one set of axes, how the resistivity varies with temperature for (i) copper, (ii) nichrome, (iii) a semiconductor such as silicon. Explain the difference between (i) and (iii).

Sketch (i) Copper: a rising, nearly straight line over ordinary temperatures (curving towards a small constant value near absolute zero). (ii) Nichrome: a much higher but almost flat line — very small positive slope. (iii) Silicon: a curve falling steeply as $T$ increases.
Explanation $\rho=m/(ne^2\tau)$. Copper: $n$ is constant; $\tau$ falls as lattice vibrations increase, so $\rho$ rises. Silicon: $n$ increases rapidly as more electrons are freed by heating, so $\rho$ falls.
Exam tip Label the axes ($\rho$ against $T$) and mark which curve is which — the labels carry the marks.

E2. Board · 3 marks Two bulbs rated $220$ V, $100$ W and $220$ V, $40$ W are connected in series across a $220$ V supply. Find the current and the power consumed by each bulb. Which glows brighter? Assume the resistances do not change.

Resistances $R=V^2/P$: $R_{100}=484\ \Omega$, $R_{40}=1210\ \Omega$; in series $R=1694\ \Omega$.
Current $I=\dfrac{220}{1694}\approx0.130$ A.
Powers $P_{100}=I^2R_{100}\approx8.2$ W, $P_{40}=I^2R_{40}\approx20.4$ W. The 40 W bulb glows brighter. Exam tip: in series compare $I^2R$ — the larger resistance (the lower-rated bulb) wins.

E3. JEE Main The resistance of a wire is $3.0\ \Omega$ at $20\ ^\circ\text{C}$ and $3.6\ \Omega$ at $80\ ^\circ\text{C}$. Taking $R=R_0(1+\alpha T)$ with $T$ in $^\circ\text{C}$, the temperature coefficient $\alpha$ is about (A) $3.3\times10^{-3}\ ^\circ\text{C}^{-1}$ (B) $3.6\times10^{-3}\ ^\circ\text{C}^{-1}$ (C) $4.0\times10^{-3}\ ^\circ\text{C}^{-1}$ (D) $2.8\times10^{-3}\ ^\circ\text{C}^{-1}$.

Equations $3.0=R_0(1+20\alpha)$ and $3.6=R_0(1+80\alpha)$. Dividing: $1.2(1+20\alpha)=1+80\alpha$, so $0.2=56\alpha$.
Solve $\alpha=\dfrac{0.2}{56}\approx3.57\times10^{-3}\ ^\circ\text{C}^{-1}$ (and $R_0=2.8\ \Omega$).
Answer (B). Exam tip: (A) is $\dfrac{0.6}{3.0\times60}$ — it treats the $20\ ^\circ\text{C}$ value as $R_0$, which the question's form of the law does not allow.

E4. NEET A heater coil is cut into two equal halves and only one half is used on the same supply. The heat produced per second becomes (A) half (B) the same (C) double (D) four times.

Concept Fixed $V$: $P=V^2/R$. Half the length means half the resistance.
Answer (C) double. Exam tip: (D) would be right if both halves were joined in parallel ($R/4$).

E5. JEE Advanced (One or more correct.) Resistors $R_1$ and $R_2$ have temperature coefficients $\alpha_1=4.0\times10^{-3}\ ^\circ\text{C}^{-1}$ and $\alpha_2=-1.0\times10^{-3}\ ^\circ\text{C}^{-1}$ (at the working temperature). For small temperature changes: (A) in series, the combination's resistance does not change if $R_1:R_2=1:4$ (B) in parallel, it does not change if $R_1:R_2=4:1$ (C) in series with $R_1=R_2$, the combination has $\alpha=1.5\times10^{-3}\ ^\circ\text{C}^{-1}$ (D) $R_2$ is most likely a pure metal.

Series $\Delta R=(R_1\alpha_1+R_2\alpha_2)\Delta T=0$ needs $4R_1=R_2$: (A) true. With $R_1=R_2=R$: $\alpha=\dfrac{R\alpha_1+R\alpha_2}{2R}=1.5\times10^{-3}$: (C) true.
Parallel $\dfrac1R=\dfrac1{R_1}+\dfrac1{R_2}$; $\Delta\left(\dfrac1{R_i}\right)=-\dfrac{\alpha_i\Delta T}{R_i}$. Zero total change needs $\dfrac{\alpha_1}{R_1}+\dfrac{\alpha_2}{R_2}=0$, i.e. $R_1=4R_2$: (B) true.
Answer (A), (B), (C). (D) is false: a negative $\alpha$ indicates a semiconductor such as carbon. Exam tip: series combinations add $R\alpha$; parallel combinations add $\alpha/R$.

E6. JEE Main NEET A generating station supplies $1.1$ MW at $11$ kV through lines of total resistance $5.0\ \Omega$. The power lost in the lines is (A) $5.0$ kW (B) $50$ kW (C) $24$ MW (D) $0.50$ kW.

Current $I=\dfrac PV=\dfrac{1.1\times10^6}{1.1\times10^4}=100$ A.
Loss $I^2R=(100)^2(5.0)=5.0\times10^4$ W — about $4.5\%$ of the power sent.
Answer (B). Exam tip: (C) is $V^2/R$ with the transmission voltage — more than the station produces, a sure sign of the wrong formula.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How large an error does the straight-line law $R=R_0(1+\alpha T)$ make when a platinum thermometer is used far above $100\ ^\circ\text{C}$?

Why it matters. Platinum resistance thermometers ("Pt100" sensors, $100\ \Omega$ at $0\ ^\circ\text{C}$) monitor temperatures in pharmaceutical factories, food sterilisers, power stations and hospital laboratories worldwide. An international standard describes their resistance with a small quadratic correction to the linear law. Knowing when the correction matters is a real calibration question.

What to do.

  1. For $T\ge0\ ^\circ\text{C}$ the standard sensor follows $R=R_0\left(1+AT+BT^2\right)$ with $R_0=100\ \Omega$, $A=3.9083\times10^{-3}\ ^\circ\text{C}^{-1}$ and $B=-5.775\times10^{-7}\ ^\circ\text{C}^{-2}$. Compute $R$ at $0,\ 100,\ 200,\ 300$ and $400\ ^\circ\text{C}$.
  2. Find the "average" coefficient between $0$ and $100\ ^\circ\text{C}$: $\alpha=\dfrac{R_{100}-R_0}{100R_0}$.
  3. Pretend you only know this $\alpha$. Use the linear law to convert each true resistance back into a temperature. Tabulate the error (linear reading minus true temperature).
  4. Plot the error against temperature. Over what range is the linear law good to within $1\ ^\circ\text{C}$?

Tools: Spreadsheet or calculator; graph paper or plotting software.

Hand in: Your table, the error graph, and a recommendation on when the quadratic term must be used.

Going further. The quadratic term makes $dR/dT$ fall as $T$ rises. What does that mean for the thermometer's sensitivity (ohms per degree) at $400\ ^\circ\text{C}$ compared with $0\ ^\circ\text{C}$?

Instructor note: $R=100.00,\ 138.51,\ 175.86,\ 212.05,\ 247.09\ \Omega$; $\alpha=3.8505\times10^{-3}\ ^\circ\text{C}^{-1}$; linear readings $\approx0,\ 100,\ 197,\ 291,\ 382\ ^\circ\text{C}$, so the errors are $0,\ 0,\ -3,\ -9,\ -18\ ^\circ\text{C}$. The error is $\dfrac{B\,T(T-100)}{A+100B}$, zero at $0$ and $100\ ^\circ\text{C}$, and stays within $1\ ^\circ\text{C}$ from $0$ up to about $145\ ^\circ\text{C}$. Sensitivity: $dR/dT=R_0(A+2BT)$ falls from $0.391$ to $0.345\ \Omega/^\circ\text{C}$ at $400\ ^\circ\text{C}$.

🧪 Self-check quiz

0/5 correct

Q1. As the temperature of pure silicon rises, its resistivity

Q2. Manganin is used for standard resistors mainly because it has

Q3. A heater coil is replaced by one of half the resistance, on the same mains supply. The power

Q4. One kilowatt hour is

Q5. The $I$–$V$ graph of an ohmic conductor at constant temperature is

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