Current Electricity

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XII, Chapter 3.

You flip a switch and the bulb lights instantly. Do electrons really race through the wire?

They do not. In a household copper wire the electrons drift at a fraction of a millimetre per second, slower than a snail. Yet the bulb glows the moment you press the switch, even if the wire is a hundred metres long. The resolution of this puzzle is the heart of the chapter: the electric field travels through the wire at close to the speed of light and pushes every free electron at once, while each electron itself only creeps forward.

Once you see current as "a slow drift on top of a fast random motion, started everywhere by a field", the rest follows: Ohm's law, resistivity, why metals heat up, what a battery actually does, and how to solve any circuit.

Electric current: charge crossing a section per second

If a net charge q crosses a cross-section in time t, the steady current is

$I = q/t$, and in general $I = \lim (\Delta Q/\Delta t)$ as $\Delta t \to 0$

The SI unit is the ampere ($1\ \text{A} = 1\ \text{C s}^{-1}$). Nerve impulses carry microamperes, a household geyser a few amperes, and a lightning stroke tens of thousands of amperes.

By convention, current points in the direction a positive charge would move, so in a metal it is opposite to the electron flow. Although we draw an arrow on it, current is a scalar. Two currents meeting at a junction add as plain numbers (with signs for in and out), never by the parallelogram law.

Drift velocity: the slow march inside a fast mess

Inside a metal, the positive ions sit fixed in a lattice and the free electrons move randomly at high speed (hundreds of km s⁻¹ at room temperature) in all directions. Because the motion is random, the average velocity is zero, and so there is no net current without a field.

Apply an electric field E along the wire. Each electron now feels a force and accelerates with $a = eE/m$, opposite to E. But it does not accelerate for long: it collides with an ion and loses its "memory" of the earlier push. If τ is the relaxation time, the average time between collisions, then the average velocity gained is

$v_d = -(eE/m)\tau$, with magnitude $v_d = eE\tau/m$

Think of it as a crowd leaving a stadium: everyone jostles in random directions, but a slight bias towards the exit gives a slow net flow. The tiny net bias is the drift velocity.

To connect this to current, take a wire of cross-section A with n free electrons per unit volume. In one second, all electrons within a length vd of a cross-section cross it, so the charge crossing per second is

$I = neAv_d$, and the current density is $j = I/A = nev_d$

Current density j is a vector (unit A m⁻²) pointing along E. Dividing out the field gives the mobility: drift velocity per unit field.

$\mu = v_d/E = e\tau/m$, with SI unit m² V⁻¹ s⁻¹

Why it matters. Mobility is the number that tells you how easily a carrier drifts. Electrons in a semiconductor are more mobile than holes, which is one reason n-type devices switch faster; you will meet this in Semiconductor Electronics.

Beyond the textbook. For the exam: free electrons collide with ions and drift with $v_d = eE\tau/m$, giving Ohm's law. In reality: this is the classical free electron picture. Real conduction electrons obey quantum mechanics and move at the Fermi speed, around 10⁶ m s⁻¹ in copper, and in a perfect crystal they would not scatter at all. Resistance comes from lattice vibrations and impurities, which is why it rises with temperature. The drift formula still gives the right dependence on E, n and τ, which is why it is used at this level.

Worked example: drift speed in a household wire. A copper wire of cross-section 2.0 mm² carries 3.2 A. Take $n = 8.5 \times 10^{28}\ \text{m}^{-3}$ and $e = 1.6 \times 10^{-19}\ \text{C}$. Find the drift speed and how long an electron takes to travel 1 m along the wire.

Common wrong step: using the area as 2.0 instead of 2.0 × 10⁻⁶ m², which makes the answer a million times too small. Another trap is to conclude the bulb must wait hours to light. The bulb lights because the field acts on electrons already sitting in the filament; none of them has to travel from the switch.

Pause and predict: the same current flows through a thick wire and then a thin wire joined end to end (same material). In which part do the electrons drift faster?

Check your answer: in the thin wire. The current I is the same everywhere in a series path, n is the same, so $v_d = I/(neA)$ is larger where A is smaller. The thin wire also has the stronger field, a point you will use in the next section.

Ohm's law, resistance and resistivity

If you vary the potential difference V across many conductors, the current I rises in proportion. Ohm's law says $V \propto I$, written

$$V = IR$$

R is the resistance (SI unit ohm, $1\ \Omega = 1\ \text{V A}^{-1}$). The real content of Ohm's law is that R stays constant while V changes, so the V–I graph is a straight line through the origin. The relation $V = IR$ on its own is just the definition of R and holds even for devices that disobey Ohm's law.

Why R depends on size. Doubling the length doubles the number of obstacles an electron meets, so R doubles. Doubling the cross-section gives the current twice as many parallel lanes, so R halves. Hence

$$R = \dfrac{\rho l}{A}$$

where ρ is the resistivity, a property of the material (and temperature) alone, in Ω m. Its reciprocal is the conductivity $\sigma = 1/\rho$, in S m⁻¹ (siemens per metre).

The microscopic version follows from $j = nev_d$ and $v_d = eE\tau/m$:

$j = (ne^2\tau/m)E = \sigma E$, so $\sigma = ne^2\tau/m$ and $\rho = m/(ne^2\tau)$

This single formula explains almost every resistivity question: if n or τ changes, ρ changes. It also gives $\sigma = ne\mu$.

Worked example: stretching a wire. A wire of resistance 5 Ω is drawn uniformly until its radius becomes half. What are its new resistance and resistivity? If it is connected to a fixed voltage, what happens to the power?

Common wrong step: writing $R \propto 1/A$ only and getting 4R. That forgets that the same metal must also get longer, and R grows with length. The factor of 16 is 4 (longer) times 4 (thinner).

Limits of Ohm's law. Many devices disobey it, and NTA asks you to recognise them.

Beyond the textbook. For the exam: a bulb filament is a non-ohmic conductor because its resistance changes as it heats. In reality: a tungsten filament at its working temperature of roughly 2500 to 3000 K has a resistance roughly ten times its value at room temperature. That is why a bulb draws a big surge when switched on and why bulbs tend to fail at the moment of switching rather than during steady glow.

How resistivity changes with temperature

For a metal over a modest temperature range, resistivity grows almost linearly:

$$\rho_T = \rho_0 [1 + \alpha (T - T_0)]$$

α is the temperature coefficient of resistivity, with unit K⁻¹. It is positive for metals. The same relation works for resistance, $R = R_0 [1 + \alpha (T - T_0)]$, when the dimensions barely change.

Why a metal gets worse as it warms: in $\rho = m/(ne^2\tau)$, the number of free electrons n is almost fixed, but hotter ions vibrate harder and collide more often, so τ falls and ρ rises.

Why a semiconductor gets better as it warms: here n is small and heat frees many more carriers, so n rises sharply. That growth beats the small fall in τ, so ρ falls and α is negative. Insulators behave the same way.

Pause and predict: a metal wire and a semiconductor are each connected to a battery and heated. What happens to the current in each?

Check your answer: in the metal wire the current falls because ρ rises, and in the semiconductor the current rises because ρ falls. This is the exact claim behind the semiconductor "negative α" questions.

Beyond the textbook. For the exam: ρ rises linearly with T for metals, with α positive. In reality: the straight line holds only in a limited range. Near absolute zero, the resistivity of an ordinary metal does not vanish but levels off at a small "residual" value set by impurities. Some materials do much more: mercury drops to zero resistance at about 4.2 K, which was the first superconductor ever found.

Electrical energy and power

When a charge $\Delta Q = I\Delta t$ falls through a potential difference V, it gives up energy VΔQ. All of this becomes heat in a resistor, so

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

Which form to use depends on what is fixed. In series the current is common, so use $P = I^2R$: the larger resistance gets more power. In parallel the voltage is common, so use $P = V^2/R$: the smaller resistance gets more power. Commercial energy is billed in the kilowatt hour: $1\ \text{kWh} = 3.6 \times 10^{6}\ \text{J}$.

Why power companies transmit at high voltage. To deliver power P over cables of resistance Rc, the current is $P/V$, and the wasted power is $P_c = (P/V)^2 R_c = P^2 R_c/V^2$. Raising V ten times cuts the loss a hundred times.

Worked example: two bulbs in series. A 100 W, 220 V bulb and a 60 W, 220 V bulb are joined in series across a 220 V supply. Which glows brighter, and what power does each use?

Common wrong step: assuming the 100 W bulb must be brighter because it says 100 W. The rating is valid only at 220 V across that bulb. In series each bulb gets less than 220 V, and the one with the larger resistance takes the larger share. A quick check: the total power $14.1 + 23.4 = 37.5$ W equals $(100 \times 60)/(100 + 60) = 37.5$ W.

Resistors in series and parallel

A handy fact: $n$ identical resistors $R$ give $nR$ in series and $R/n$ in parallel. A 12 Ω wire bent into a circle has 3 Ω between two diametrically opposite points, because the two 6 Ω halves are in parallel.

A cell: emf, terminal voltage and internal resistance

A cell does not create charge. It keeps a potential difference between its terminals by doing chemical work on charges. The emf ε is the potential difference between the terminals when no current flows (open circuit). A real cell also has internal resistance r, the resistance of its electrolyte and electrodes. When a current I is drawn, a part Ir of the emf is lost inside, so the terminal voltage is

$V = \varepsilon - Ir$, and with an external resistance $R$, $I = \varepsilon/(R + r)$

The largest current comes from a short circuit (R = 0) and is $I_{\max} = \varepsilon/r$. During charging, current is forced into the cell, so the terminal voltage becomes $V = \varepsilon + Ir$. This is the one place students must flip the sign.

Worked example: a cell with internal resistance. A cell of emf 12 V and internal resistance 0.5 Ω is joined to a 5.5 Ω resistor. Find the current, the terminal voltage, the power in the resistor and the power lost inside the cell.

Common wrong step: using the full 12 V across the resistor. Only 11 V reaches it, because 1 V is lost across r.

Maximum power transfer. The power in R is $P = \varepsilon^2R/(R + r)^2$. It is largest when $R = r$, and then $P_{\max} = \varepsilon^2/4r$. At that point only half of the total power reaches the load, so the efficiency is 50%. If two different loads R₁ and R₂ draw the same power from one cell, setting the two powers equal gives $r = \sqrt{R_1R_2}$.

Beyond the textbook. For the exam: a cell has fixed ε and r. In reality: internal resistance is not constant. It rises as a cell discharges, and it changes with temperature, which is why an old torch battery with a "good" voltage on a voltmeter still cannot light a bulb. A voltmeter draws almost no current, so it reads close to ε, while the bulb draws a real current and drops a large Ir.

Combining cells

Kirchhoff's rules

Not every circuit can be reduced by series and parallel. Two rules solve any network.

Junction rule: at any junction, the sum of currents entering equals the sum leaving. This is conservation of charge, since charge cannot pile up at a point in a steady current.

Loop rule: around any closed loop, the algebraic sum of potential changes is zero. This is conservation of energy, since a charge returning to its start has the same potential energy.

Signs are the part that decides your marks. Choose a direction of travel around the loop, and apply this table.

Worked example: parallel cells, and who is charging whom. A cell (4 V, 1 Ω) and a cell (2 V, 1 Ω) are connected in parallel, with their positive terminals together, across a 2.5 Ω resistor. Find the current in each cell and in the resistor.

The negative sign means cell 2 carries current into its positive terminal, so it is being charged by cell 1. The terminal voltage across the resistor is 2.5 V, which is above the 2 V emf of cell 2.

Common wrong step: assuming both cells discharge and getting I₂ positive by guess. A negative answer is not an error; it only says the true direction is opposite to the one you assumed. The equivalent-cell shortcut agrees: $\varepsilon_{eq} = (4 + 2)/(1 + 1) = 3$ V and $r_{eq} = 0.5\ \Omega$, so $I = 3/(0.5 + 2.5) = 1$ A.

Wheatstone bridge and metre bridge

Four resistors P, Q, R, S form a diamond with a battery across one diagonal and a galvanometer across the other. If the galvanometer shows no deflection (balance), the two junctions on its sides are at the same potential, and

$$\dfrac{P}{Q} = \dfrac{R}{S}$$

Pair the arms so that P and Q are in the same branch before the galvanometer, and R and S in the other. At balance the galvanometer branch can be removed or ignored, which makes the rest of the circuit easy.

The metre bridge is a Wheatstone bridge made from a 100 cm uniform wire, whose resistance is proportional to length. An unknown resistance X is placed in one gap and a known resistance R in the other. If the null point is at a distance l from the end near X, then

$$\dfrac{X}{R} = \dfrac{l}{100 - l}$$

Worked example: metre bridge. In a metre bridge the unknown resistance X is in the left gap and $R = 10\ \Omega$ is in the right gap. The balance point is 40 cm from the left end. Find X.

Common wrong step: inverting the ratio and getting 15 Ω. Resistance is proportional to length, so the shorter wire segment sits on the side with the smaller resistance. Since 40 cm is less than 60 cm, X must be less than R.

Beyond the textbook. For the exam: a metre bridge finds an unknown resistance by balancing lengths. In reality: the bridge is most accurate when the null point lies near the middle of the wire. Near the ends, a small error of 1 mm in reading l causes a much larger fractional error in $l/(100 - l)$. That is why the known resistance R is chosen so that the balance point falls between about 40 cm and 60 cm.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

Current electricity applies the potential difference you built in Electrostatic Potential and Capacitance, and the field E driving drift comes from Electric Charges and Fields. Moving charges are the source of magnetic effects in Moving Charges and Magnetism. The idea of a varying current continues in Alternating Current, and the contrast between conductors and semiconductors returns in Semiconductor Electronics.