Current Electricity

A 60 W bulb and a 100 W bulb, both rated 220 V, are wired in series across 220 V. Which glows brighter?

Almost everyone says the 100 W bulb.

It is the 60 W bulb. Higher wattage at a fixed rating means lower resistance, $R=V^2/P$ — and in series the current is common, so $P=I^{2}R$ hands the power to the larger resistance.

$R_{60}=\frac{220^2}{60}=807\Omega ,R_{100}=\frac{220^2}{100}=484\Omega$

Three ideas carry the whole chapter, and none is a formula:

• Ohm's law is a statement about a material, not a law of nature. It holds for metals at fixed temperature and fails for diodes, lamps and thermistors.
• All circuit analysis is two conservation laws. Junction rule is charge; loop rule is energy. Series, parallel, Wheatstone and the metre bridge are special cases of those two.
• Object versus material. Resistance belongs to a piece of wire; resistivity belongs to copper. Stretching changes one and not the other.

Scope note. The 2023 NTA revision removed the potentiometer, the resistor colour code, and resistances of different materials from JEE Main, and they stay out for 2026. Kirchhoff's laws, the Wheatstone bridge and the metre bridge are all still in.

1. Electric Current and Current Density

$I=\frac{dq}{dt}[\text{ampere}],\vec{J}=\frac{I}{A}=\sigma \vec{E}$

Conventional current flows the way a positive charge would move — opposite to the actual electron motion in a metal. The convention predates the electron and has simply been kept.

Current is a scalar, despite having a direction attached to it. The test: currents meeting at a junction add arithmetically, not vectorially. Two 3 A currents arriving give 6 A leaving, whatever the angle between the wires.

Current density is a genuine vector, and unlike $I$ it varies from point to point inside a conductor.

Illustration 1

The current in a wire varies as $I=3t^2+2t$ amperes. Find the charge crossing a given section between $t=1$ s and $t=3$ s, and the average current over that interval.

Charge is accumulated current, so this is an integral and not a multiplication:

$q={\int }_1^3(3t^2+2t)dt={\left[t^3+t^2\right]}_1^3=(27+9)-(1+1)=34\text{C}$

$\bar{I}=\frac{q}{\Delta t}=\frac{34}{2}=17\text{A}$

Note what the average current is not. Evaluating $I$ at the midpoint $t=2$ gives 16 A, which is close but wrong. Sampling a non-linear quantity at the middle of an interval is the standard slip here; only the integral is safe.

2. Drift Velocity and the Microscopic Picture

Electrons in a metal already move at random thermal speeds around $10^5$ m s⁻¹ in all directions. With no field these cancel and there is no current. A field superimposes a tiny systematic drift on that chaos:

$v_d=\frac{eE\tau }{m},I=neA{v}_d,\mu =\frac{v_d}{E}$

$\tau$ is the relaxation time — the average gap between collisions with the lattice. $n$ is the free-electron number density.

Drift velocity is around $10^{-4}$ m s⁻¹. A single electron takes hours to cross a metre of wire.

Trap. So why does a lamp light instantly? Because the field is set up throughout the circuit at nearly the speed of light, and every electron in the wire starts drifting at once. Nothing travels from the switch to the lamp. A pipe already full of water delivers flow at the far end the moment you open the tap.

Illustration 2

A wire tapers from cross-section $2A$ at one end to $A$ at the other. Compare $I$, $J$ and $v_d$ at the two ends.

Charge cannot pile up in a steady state, so $I$ is the same at both ends. Then:

$J=\frac{I}{A}\Rightarrow J_{narrow}=2J_{wide},v_d=\frac{I}{neA}\Rightarrow v_{d,narrow}=2v_{d,wide}$

Current is conserved along the wire; current density and drift speed are not. This is exactly why a thin filament glows and the thick supply lead feeding it does not.

3. Ohm's Law and Its Limits

$V=IR,\text{microscopically }\vec{J}=\sigma \vec{E}$

Materials obeying it are called ohmic, and the class is smaller than students assume.

Even a metal is ohmic only at fixed temperature. A filament lamp is made of metal and is markedly non-ohmic — purely because the current heats it.

Illustration 3

A filament lamp draws 0.5 A at 4 V and 0.8 A at 10 V. Find its resistance at each point, and say what the slope of the chord joining them represents.

Resistance at an operating point is $V/I$ at that point, never the slope of the curve:

$R_1=\frac{4}{0.5}=8\Omega ,R_2=\frac{10}{0.8}=12.5\Omega$

The resistance rises by more than half between the two points, because the extra power has heated the filament.

The chord gives $\Delta V/\Delta I=6/0.3=20$ Ω, which is neither of those. That is the dynamic resistance, describing how the device answers a small change riding on top of a bias. It is the wrong quantity for finding the current at a given voltage, and reaching for it is what the graph is testing. Ohm's law is not being violated here; it simply never applied, because $R$ was never constant.

4. Resistivity, Conductivity and Temperature

$R=\frac{\rho L}{A},\rho =\frac{m}{n{e}^2\tau },{\rho }_T={\rho }_0(1+\alpha \Delta T)$

The middle expression is the most valuable equation in the chapter, because it explains rather than computes.

One equation, two opposite behaviours, and the difference is only which factor wins.

Trap. Resistance is a property of the object; resistivity is a property of the material. Stretching a wire changes $R$ and leaves $\rho$ exactly as it was.

Stretching conserves volume, which makes those problems fast: stretch to $n$ times the length and the area falls by $n$, so $R$ grows by $n^2$.

Below a critical temperature some materials become superconducting, with $\rho$ exactly zero — not merely small. A current once started persists indefinitely.

Illustration 4

Two wires of the same material: the second is twice as long and twice as thick. Find $R_2/R_1$.

Twice as thick means twice the diameter, so four times the area:

$\frac{R_2}{R_1}=\frac{L_2}{L_1}\cdot \frac{A_1}{A_2}=2\times \frac{1}{4}\Rightarrow \frac{R_2}{R_1}=\frac{1}{2}$

The thicker wire has less resistance despite being longer. Area goes as the square of the diameter, and that squaring is where the marks are lost.

5. Combinations of Resistors

Those last two checks catch most arithmetic errors on sight.

Trap. These are the reverse of the capacitor rules, and the two chapters sit next to each other. Resistors add in series; capacitors add in parallel.

Illustration 5

A wire of total resistance $R$ is bent into a circle. Find the resistance between two points a quarter of the way round.

The two points split the ring into two arcs, in parallel, of resistance $R/4$ and $3R/4$:

$R_{eq}=\frac{(R/4)(3R/4)}{R}\Rightarrow R_{eq}=\frac{3R}{16}$

Check against the rule: $3R/16$ is smaller than $R/4$, the smaller arm. Any answer larger than $R/4$ is wrong before you check the algebra.

6. EMF, Internal Resistance and Terminal Voltage

EMF is the work done per unit charge by the source in driving charge round the circuit. Despite the name it is an energy per charge, not a force.

$V_{terminal}=\epsilon -Ir\text{(discharging)},V=\epsilon +Ir\text{(charging)}$

Terminal voltage equals the emf only at zero current. This is why an ideal voltmeter must draw negligible current, and why a nearly dead battery still reads close to its rating on open circuit but collapses under load.

Power delivered to an external $R$ is maximum at $R=r$. At that point exactly half the power is wasted inside the cell:

$P_{ext}=P_{int}\Rightarrow \eta =50\%$

Trap. Maximum power and maximum efficiency are different and incompatible goals. Power stations keep source resistance far below load resistance, accepting less than maximum power for efficiency near 100 per cent.

Illustration 6

A battery of emf 6 V reads a terminal voltage of 5.4 V while supplying 3 A. Find $r$ and the short-circuit current.

$r=\frac{\epsilon -V}{I}=\frac{0.6}{3}=0.2\Omega \Rightarrow I_{sc}=\frac{\epsilon }{r}=\frac{6}{0.2}=30\text{A}$

A small internal resistance means a large short-circuit current, which is precisely why shorting a car battery is dangerous and shorting a torch cell is merely futile.

7. Combinations of Cells

$I_{series}=\frac{n\epsilon }{R+nr},I_{parallel}=\frac{\epsilon }{R+r/n}$

Series multiplies the driving emf and the internal loss, so it pays only when the internal loss is a small share of the total.

• Series wins when $R\gg r$ — a large external resistance.
• Parallel wins when $R\ll r$ — a small external resistance.

Illustration 7

Twelve cells, each 1.5 V, are connected in series, but two of them are inserted the wrong way round. Find the net emf.

A reversed cell does not merely fail to contribute — it opposes. Each one costs twice its emf:

${\epsilon }_{net}=(12-2\times 2)(1.5)=8\times 1.5\Rightarrow {\epsilon }_{net}=12\text{V}$

The total internal resistance is unchanged at $12r$, because resistance has no polarity.

8. Kirchhoff's Laws

Junction rule — current in equals current out. This is conservation of charge; nothing accumulates at a point in a steady circuit.

Loop rule — potential differences around any closed loop sum to zero. This is conservation of energy; returning to a point must return you to its potential.

Assume any direction for each unknown current. A wrong guess simply returns a negative number.

Trap. A negative current is information, not an error. It usually means a cell is being driven backwards and is charging. Never rework the problem to make the sign positive.

You need as many independent equations as unknown currents, taking one junction equation fewer than the number of junctions.

Ammeters and voltmeters in the circuit

Connecting an ammeter in parallel with a component is the classic laboratory accident: its near-zero resistance short-circuits the component and usually destroys the meter.

Illustration 8

A 10 V cell of internal resistance 1 Ω charges a 4 V cell of internal resistance 2 Ω through a 3 Ω resistor in series. Find the current and each cell's terminal voltage.

The cells oppose, so the net driving emf is the difference:

$I=\frac{10-4}{1+2+3}=1\text{A}$

$V_{10}=10-(1)(1)=9\text{V}(\text{discharging}),V_4=4+(1)(2)=6\text{V}(\text{charging})$

Check the loop closes: $9=6+(1)(3)$. The charging cell's terminal voltage sits above its emf, which is the sign that energy is flowing into it.

Illustration 9

A 200 Ω and a 300 Ω resistor sit in series across 100 V. A voltmeter of resistance 600 Ω is placed across the 300 Ω. What does it read, and what is the true value?

True value first, with the voltmeter absent:

$V_{300}=100\times \frac{300}{500}=60\text{V}$

Connected, the voltmeter sits in parallel with the 300 Ω and changes the circuit it is measuring:

$R_{parallel}=\frac{300\times 600}{900}=200\Omega \Rightarrow V_{reading}=100\times \frac{200}{400}=50\text{V}$

A 17 per cent error, produced purely by measuring. A real voltmeter always reads low, because drawing current lowers the very potential difference it is reporting, and the error grows as its resistance falls toward that of the component it straddles. Ten times the resistance here would have cut the error to about two per cent.

9. Wheatstone Bridge and Metre Bridge

$\frac{P}{Q}=\frac{R}{S}\text{at balance}$

At balance the galvanometer reads zero: B and D sit at the same potential, so no current crosses.

The balance condition contains neither the galvanometer's resistance nor the cell's emf. That is the entire point of the design — a null method needs no calibration, so it beats reading a deflection. Interchanging the cell and the galvanometer leaves the condition unchanged.

The metre bridge replaces two arms with a uniform wire, so their ratio is set by lengths alone:

$\frac{R}{S}=\frac{l}{100-l}$

Accuracy is best near the middle of the wire, where the fractional error in reading a length is smallest — which is why the known resistance is chosen comparable to the unknown.

Illustration 10

Five resistors form a bridge: 4 and 8 Ω in the upper arms, 6 and 12 Ω in the lower arms, and 7 Ω bridging the midpoints. Find the equivalent resistance across the supply.

Test for balance before touching anything else:

$\frac{4}{8}=\frac{1}{2},\frac{6}{12}=\frac{1}{2}\Rightarrow \text{balanced}$

The midpoints therefore sit at equal potential, no current flows through the 7 Ω, and it can be deleted from the diagram outright.

Upper branch in series gives $4+8=12$ Ω, lower gives $6+12=18$ Ω, and those two are in parallel:

$R_{eq}=\frac{12\times 18}{30}=7.2\Omega$

The bridging resistor's value never entered the answer. Had the two ratios differed, no shortcut would exist and Kirchhoff's laws would be the only route — which is why the balance test is the first thing to try on any five-resistor network.

Illustration 11

In a metre bridge with $R=20$ Ω in the left gap, the balance point is at 40 cm. Find $S$, and the new balance point if the two gaps are interchanged.

$\frac{20}{S}=\frac{40}{60}\Rightarrow S=30\Omega$

Interchanging swaps the ratio, so the balance point moves to $100-40=60$ cm. Taking the mean of the two readings cancels any end-resistance error in the bridge — which is why the swap is a standard experimental step, not just an exam question.

10. Electrical Energy and Power

$P=VI=I^{2}R=\frac{V^2}{R}$

All three are equivalent, but the right choice saves time:

• Series — current is common, so use $P=I^{2}R$. Power goes to the larger resistance.
• Parallel — voltage is common, so use $P=V^2/R$. Power goes to the smaller resistance.

A higher-wattage bulb has lower resistance at the same rating, since $R=V^2/P$. That resolves the opening question: in series the 60 W bulb, having the larger resistance, glows brighter; in parallel — which is how household wiring actually works — each runs at its rating and the 100 W bulb wins.

Commercial energy is billed in kilowatt-hours, one unit being $3.6\times 10^6$ J.

Illustration 12

A heater rated 1000 W at 220 V is connected to a 110 V supply. Find the power it consumes.

The rating is not a property of the heater; its resistance is:

$R=\frac{220^2}{1000}=48.4\Omega \Rightarrow P=\frac{110^2}{48.4}=250\text{W}$

Half the voltage gives a quarter of the power, because $P\propto V^2$ at fixed $R$. Assuming the heater still draws 1000 W, or even 500 W, is the standard error.

Summary

• Ohm's law is a property of materials, not a law of nature; even a metal obeys it only at fixed temperature.
• Current is a scalar, current density a vector. Junction currents add arithmetically.
• $v_d\sim 10^{-4}$ m s⁻¹, yet lamps light instantly — the field propagates at nearly $c$.
• $I$ is the same all along a wire; $J$ and $v_d$ are not.
• $\rho =m/n{e}^2\tau$ explains both signs of $\alpha$: in metals $\tau$ falls, in semiconductors $n$ rises and wins.
• Resistance belongs to the object, resistivity to the material. Stretch to $n\times$ length and $R$ grows $n^2\times$.
• Resistors add in series, add reciprocally in parallel — the reverse of capacitors.
• $V=\epsilon -Ir$ discharging, $\epsilon +Ir$ charging. Terminal voltage equals emf only at zero current.
• Maximum power at $R=r$, where efficiency is exactly 50 per cent — power and efficiency are incompatible goals.
• Cells in series suit large external $R$; in parallel, small $R$. A reversed cell costs twice its emf.
• Junction rule is charge conservation, loop rule is energy conservation, and everything else follows.
• A negative Kirchhoff current is information — usually a cell being charged.
• Ideal ammeter: zero resistance, in series. Ideal voltmeter: infinite resistance, in parallel; a real one reads low.
• Bridge balances at $P/Q=R/S$, independent of galvanometer and cell — the strength of a null method.
• Metre bridge reads best near the middle; interchanging the gaps and averaging cancels end errors.
• Use $I^{2}R$ in series, $V^2/R$ in parallel. Halving the supply voltage quarters the power.