Current Electricity Class 12: Notes, Derivations and Numericals

What Current Electricity is about

Weightage: This chapter is part of Units I and II, Electrostatics; Current Electricity, which carry 16 of the 70 theory marks together with Electric Charges and Fields and Electrostatic Potential and Capacitance.Source: CBSE Physics syllabus 2026-27, Units I and II Electrostatics; Current Electricity

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper. See the questions

Key concepts

Electric current and current density

NCERT section 3.2 and 3.3

Current is the rate of flow of charge through a cross-section: $I=\frac{dq}{dt}$, or $I=\frac{q}{t}$ for a steady current. Its SI unit is the ampere ($1A=1C{s}^{-1}$). By convention its direction is the direction in which positive charge would flow, opposite to the motion of electrons. Current is a scalar (it obeys ordinary addition at a junction).

Current density $\vec{j}$ is current per unit area normal to the flow, $j=\frac{I}{A}$, unit A m⁻²; it is a vector along the flow of positive charge. Ohm's law in its microscopic form is $\vec{j}=\sigma \vec{E}$, where $\sigma =\frac{1}{\rho }$ is the conductivity.

In a metal with no field, free electrons move randomly in all directions, so there is no net flow. A field gives them a net drift and a current flows.

Worked example

A current of 0.8 A flows through a wire of cross-sectional area $2\times 10^{-6}{m}^2$ for 5 minutes. Find (a) the charge that passes, (b) the number of electrons, (c) the current density. [2 marks]

Try it first, then show the solutionSolution

1. (a) $q=It=0.8\times 300=240C$
2. (b) $n=\frac{q}{e}=\frac{240}{1.6\times 10^{-19}}=1.5\times 10^{21}$
3. (c) $j=\frac{I}{A}=\frac{0.8}{2\times 10^{-6}}=4\times 10^{5}A{m}^{-2}$

Answer: (a) 240 C (b) $1.5\times 10^{21}$ electrons (c) $4\times 10^{5}A{m}^{-2}$

Ohm's law, resistance and resistivity

NCERT section 3.4

Ohm's law: the current through a conductor is proportional to the potential difference across it, $V=IR$, provided physical conditions (such as temperature) stay the same. R is the resistance, unit ohm (Ω).

Resistance depends on the shape and the material: $R=\rho \frac{l}{A}$, where ρ is the resistivity, a property of the material only (unit Ω m). Doubling the length doubles R; doubling the area halves it.

A handy result: if a wire is stretched so its length becomes n times, its volume stays the same, so its area becomes $\frac{1}{n}$ times and the resistance becomes $n^2$ times.

Resistivity varies over a huge range: metals have low ρ (good conductors), insulators extremely high ρ, and semiconductors lie in between. Conductivity $\sigma =\frac{1}{\rho }$ (unit S m⁻¹).

Worked example

A wire of resistance 5 Ω is drawn out until its length is three times the original. Find the new resistance. What happens to its resistivity? [2 marks]

Try it first, then show the solutionSolution

1. Volume is constant: $Al=A^{\prime }{l}^{\prime }$; with $l^{\prime }=3l$, $A^{\prime }=\frac{A}{3}$
2. $R^{\prime }=\rho \frac{3l}{A/3}=9\rho \frac{l}{A}=9R$
3. $R^{\prime }=9\times 5=45\Omega$
4. Resistivity is a property of the material, so it does not change.

Answer: 45 Ω; the resistivity is unchanged.

Drift velocity and the origin of resistivity

NCERT section 3.5

In a field $\vec{E}$, each free electron accelerates as $\vec{a}=-\frac{e\vec{E}}{m}$ between collisions with the ions. After each collision its velocity is random again. Averaged over all electrons, the random parts cancel and the gain since the last collision gives the drift velocity:

${\vec{v}}_d=-\frac{e\vec{E}}{m}\tau$

where τ is the relaxation time, the average time between collisions. Drift speeds are tiny (of the order of $10^{-4}$ to $10^{-3}m{s}^{-1}$), yet a lamp lights at once because the field is set up throughout the wire almost instantly and all the electrons start drifting together.

Current and drift velocity. In time Δt, electrons in a length $v_d\Delta t$ of the wire cross a section. Their number is $nA{v}_d\Delta t$ (n = free electrons per unit volume), so $I=\frac{neA{v}_d\Delta t}{\Delta t}=neA{v}_d$.

Origin of resistivity. Substituting $v_d=\frac{eE\tau }{m}$: $j=\frac{I}{A}=\frac{n{e}^2\tau }{m}E$. Comparing with $j=\sigma E$:

$\sigma =\frac{n{e}^2\tau }{m},\rho =\frac{m}{n{e}^2\tau }$

So resistivity depends on n (how many carriers) and τ (how often they collide).

Worked example

A copper wire of cross-section $2.5\times 10^{-6}{m}^2$ carries a current of 3.4 A. Copper has $8.5\times 10^{28}$ free electrons per m³. Find the drift speed. [2 marks]

Try it first, then show the solutionSolution

1. $v_d=\frac{I}{neA}$
2. $neA=8.5\times 10^{28}\times 1.6\times 10^{-19}\times 2.5\times 10^{-6}=3.4\times 10^{4}C{m}^{-1}$
3. $v_d=\frac{3.4}{3.4\times 10^4}=1\times 10^{-4}m{s}^{-1}$

Answer: $1\times 10^{-4}m{s}^{-1}$ (0.1 mm per second)

Mobility

NCERT section 3.5.1

Mobility μ is the drift speed per unit field: $\mu =\frac{|v_d|}{E}$. Since $v_d=\frac{eE\tau }{m}$, $\mu =\frac{e\tau }{m}$. Its SI unit is $m^2{V}^{-1}{s}^{-1}$.

Mobility depends only on the material and its temperature (through τ), not on the applied field. So if the potential difference across a wire is doubled, the field and the drift velocity double, the current density doubles, but the mobility stays the same.

The current can be written $I=neA\mu E$, which shows the two things that make a good conductor: many carriers (large n) and high mobility (large τ).

Worked example

In a metal the relaxation time of free electrons is $2.5\times 10^{-14}s$. Find the mobility, and the drift speed in a field of $2V{m}^{-1}$. ($e=1.6\times 10^{-19}C$, $m=9.1\times 10^{-31}kg$) [2 marks]

Try it first, then show the solutionSolution

1. $\mu =\frac{e\tau }{m}=\frac{1.6\times 10^{-19}\times 2.5\times 10^{-14}}{9.1\times 10^{-31}}=\frac{4.0\times 10^{-33}}{9.1\times 10^{-31}}$
2. $\mu \approx 4.4\times 10^{-3}{m}^2{V}^{-1}{s}^{-1}$
3. $v_d=\mu E=4.4\times 10^{-3}\times 2=8.8\times 10^{-3}m{s}^{-1}$

Answer: $\mu \approx 4.4\times 10^{-3}{m}^2{V}^{-1}{s}^{-1}$; $v_d\approx 8.8\times 10^{-3}m{s}^{-1}$

Limitations of Ohm's law: linear and non-linear V-I characteristics

NCERT section 3.6

Ohm's law is not a universal law. Materials and devices that obey it give a straight-line V-I graph through the origin (linear, ohmic). Many do not; the deviations are of three kinds:

1. V stops being proportional to I: the graph curves (for example a filament lamp as it heats up).
2. The relation depends on the sign of V: reversing V does not just reverse I with the same size (for example a p-n junction diode, which conducts easily one way only).
3. V is not a single-valued function of I: the same current can occur at more than one voltage (for example gallium arsenide, whose graph turns back on itself).

For such devices you can still define $R=\frac{V}{I}$ at a point, but it is not constant.

Worked example

Give two limitations of Ohm's law, each with an example, and state the shape of the V-I graph in each case. [2 marks]

Try it first, then show the solutionSolution

1. Non-linear relation: for a filament lamp, I rises more slowly than V at high V because the filament heats up and R rises; the graph curves towards the V axis.
2. Direction-dependent relation: a p-n junction diode passes a large current in forward bias but almost none in reverse bias; the graph is very different in the two quadrants.

Answer: (1) Non-linear V-I (filament lamp, curved graph); (2) behaviour depends on the polarity of V (diode, graph not symmetric). A third: V not single-valued in I (GaAs).

Temperature dependence of resistivity

NCERT section 3.8

For metals, over a limited range, ${\rho }_T={\rho }_0[1+\alpha (T-T_0)]$, where α is the temperature coefficient of resistivity (unit K⁻¹ or °C⁻¹). The same form holds for resistance: $R_T=R_0[1+\alpha (T-T_0)]$.

Why metals: $\rho =\frac{m}{n{e}^2\tau }$. In a metal n hardly changes with temperature, but hotter ions vibrate more, collisions become more frequent, τ falls and ρ rises.

Semiconductors and insulators: heating frees many more carriers, so n rises sharply and ρ falls (negative α).

Alloys such as nichrome, manganin and constantan have a high ρ and a very small α, so they are used for heating elements and for standard resistors whose value must not drift.

Worked example

A wire has a resistance of 10 Ω at 20 °C and a temperature coefficient of $4.0\times 10^{-3}{}^{\circ }{C}^{-1}$. Find (a) its resistance at 70 °C, (b) the temperature at which its resistance is twice its value at 20 °C. [2 marks]

Try it first, then show the solutionSolution

1. (a) $R=10[1+4.0\times 10^{-3}\times (70-20)]=10\times 1.2=12\Omega$
2. (b) $2R_0=R_0[1+\alpha \Delta T]\Rightarrow \Delta T=\frac{1}{\alpha }=\frac{1}{4.0\times 10^{-3}}=250{}^{\circ }C$
3. $T=20+250=270{}^{\circ }C$

Answer: (a) 12 Ω (b) 270 °C

Electrical energy and power

NCERT section 3.9

When charge q moves through a potential drop V in a resistor, it loses energy qV, which appears as heat (Joule heating). The rate is the power:

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

In series (same I), use $P=I^{2}R$: the larger resistor dissipates more. In parallel (same V), use $P=\frac{V^2}{R}$: the smaller resistor dissipates more.

Power transmission. Delivering power P at voltage V needs current $I=\frac{P}{V}$, and the cables waste $P_c=I^2{R}_c=\frac{P^2{R}_c}{V^2}$. So power is sent at very high voltage to cut the current and the loss; transformers (Chapter 7) step the voltage up and down.

An appliance rated (P, V) has resistance $R=\frac{V^2}{P}$; on a different supply V', its power is $P^{\prime }=\frac{V^{\prime 2}}{R}$.

Worked example

A heater is rated 1000 W, 220 V. Find (a) its resistance, (b) the power it uses on a 110 V supply, assuming R does not change. [2 marks]

Try it first, then show the solutionSolution

1. (a) $R=\frac{V^2}{P}=\frac{220^2}{1000}=\frac{48400}{1000}=48.4\Omega$
2. (b) $P^{\prime }=\frac{110^2}{48.4}=\frac{12100}{48.4}=250W$
3. Halving V quarters the power.

Answer: (a) 48.4 Ω (b) 250 W

EMF, terminal voltage and internal resistance

NCERT section 3.10

A cell has two electrodes in an electrolyte. Its emf ε is the potential difference between its terminals when no current is drawn (open circuit); it is the work done by the cell per unit charge in taking charge round the circuit. The electrolyte resists the flow inside the cell: this is the internal resistance r.

With an external resistance R, $I=\frac{\epsilon }{R+r}$ and the terminal voltage is

$V=\epsilon -Ir=\frac{\epsilon R}{R+r}$

so $V<\epsilon$ while the cell supplies current. When the cell is being charged (current forced in the reverse direction), $V=\epsilon +Ir$. The maximum current a cell can give is $\frac{\epsilon }{r}$ (short circuit). Graphs: V against I is a straight line falling from ε (intercept) with slope −r; V against R rises from 0 and approaches ε.

Worked example

A battery of emf 12 V and internal resistance 0.5 Ω is connected to a 5.5 Ω resistor. Find the current, the terminal voltage and the power wasted inside the battery. [2 marks]

Try it first, then show the solutionSolution

1. $I=\frac{\epsilon }{R+r}=\frac{12}{5.5+0.5}=2A$
2. $V=\epsilon -Ir=12-2\times 0.5=11V$ (check: $IR=2\times 5.5=11V$)
3. Power wasted inside $=I^{2}r=4\times 0.5=2W$

Answer: 2 A; 11 V; 2 W

Cells in series and in parallel

NCERT section 3.11

Series (positive of one to negative of the next): emfs add and internal resistances add:

${\epsilon }_{eq}={\epsilon }_1+{\epsilon }_2,r_{eq}=r_1+r_2$

If one cell is connected the wrong way round, its emf is subtracted (its internal resistance still adds).

Parallel (positive terminals joined together, negative together): let the common terminal voltage be V and the currents $I_1,I_2$ with $I=I_1+I_2$. From $V={\epsilon }_1-I_1{r}_1$ and $V={\epsilon }_2-I_2{r}_2$: $I=\frac{{\epsilon }_1-V}{r_1}+\frac{{\epsilon }_2-V}{r_2}$. Rearranging into the form $V={\epsilon }_{eq}-I{r}_{eq}$:

${\epsilon }_{eq}=\frac{{\epsilon }_1{r}_2+{\epsilon }_2{r}_1}{r_1+r_2},r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$

or $\frac{{\epsilon }_{eq}}{r_{eq}}=\frac{{\epsilon }_1}{r_1}+\frac{{\epsilon }_2}{r_2}$ and $\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}$. For n identical cells in parallel the emf stays ε and the internal resistance becomes $\frac{r}{n}$.

Worked example

A cell of emf 2 V and internal resistance 1 Ω is connected in parallel with a cell of emf 4 V and internal resistance 2 Ω (like terminals together). Find the equivalent emf and internal resistance. [2 marks]

Try it first, then show the solutionSolution

1. ${\epsilon }_{eq}=\frac{{\epsilon }_1{r}_2+{\epsilon }_2{r}_1}{r_1+r_2}=\frac{2\times 2+4\times 1}{1+2}=\frac{8}{3}V\approx 2.67V$
2. $r_{eq}=\frac{r_1{r}_2}{r_1+r_2}=\frac{1\times 2}{3}=\frac{2}{3}\Omega \approx 0.67\Omega$

Answer: ${\epsilon }_{eq}=\frac{8}{3}V$, $r_{eq}=\frac{2}{3}\Omega$

Kirchhoff's rules

NCERT section 3.12

Junction rule (current law): at any junction, the sum of currents entering equals the sum leaving, $\sum I=0$. It follows from conservation of charge (charge does not pile up at a junction in steady state).

Loop rule (voltage law): round any closed loop, the algebraic sum of changes in potential is zero, $\sum \Delta V=0$, or $\sum \epsilon =\sum IR$. It follows from conservation of energy (the electrostatic force is conservative, so you return to the same potential).

Sign rules for the loop law: going through a resistor in the direction of the current, the potential falls by IR (write $-IR$); against the current, $+IR$. Going through a cell from − to +, the potential rises by ε ($+\epsilon$); from + to −, $-\epsilon$.

Method: mark a current in each branch (guess a direction; a negative answer just means the flow is the other way), use the junction rule to reduce the unknowns, then write one loop equation per independent loop and solve.

Worked example

Points A and B are joined by three branches: (i) a 10 V cell (positive terminal towards A) in series with 1 Ω; (ii) a 6 Ω resistor; (iii) a 4 V cell (positive terminal towards A) in series with 2 Ω. Cells have no internal resistance. Find the current in each branch. [3 marks]

Try it first, then show the solutionSolution

1. Let $V_A-V_B=V$. Currents towards A: branch (i) $I_1=\frac{10-V}{1}$, branch (iii) $I_2=\frac{4-V}{2}$; current from A to B through 6 Ω: $I_3=\frac{V}{6}$ (each from the loop rule for that branch)
2. Junction rule at A: $(10-V)+\frac{4-V}{2}=\frac{V}{6}$. Multiply by 6: $60-6V+12-3V=V\Rightarrow V=7.2V$
3. $I_1=10-7.2=2.8A$; $I_3=\frac{7.2}{6}=1.2A$
4. $I_2=\frac{4-7.2}{2}=-1.6A$: 1.6 A flows away from A through the 4 V branch, so the 4 V cell is being charged. Check: $2.8-1.6=1.2$

Answer: 10 V branch: 2.8 A towards A; 6 Ω: 1.2 A from A to B; 4 V branch: 1.6 A from A into the 4 V cell's positive terminal (it is charged).

Wheatstone bridge

NCERT section 3.13

Four resistances P (A to B), Q (B to C), R (A to D) and S (D to C) form a loop; a cell is connected across A and C and a galvanometer across B and D. The bridge is balanced when the galvanometer shows no current ($I_g=0$).

Derivation of the balance condition. With $I_g=0$, the junction rule at B and D gives: current $I_1$ flows through P and then Q, and current $I_2$ through R and then S. Loop ABDA (no current in BD): $-I_{1}P+I_{2}R=0$. Loop BCDB: $-I_{1}Q+I_{2}S=0$. Dividing the first by the second:

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

The balance does not depend on the emf of the cell, and swapping the positions of the cell and the galvanometer keeps the bridge balanced. Knowing three resistances, you can find the fourth. When the bridge is balanced, the galvanometer branch can be removed when finding the equivalent resistance.

Worked example

In a Wheatstone bridge, P = 10 Ω, Q = 20 Ω and R = 15 Ω. What value of S balances it? If S is then replaced by two equal resistors in parallel, what should each be? [2 marks]

Try it first, then show the solutionSolution

1. $\frac{P}{Q}=\frac{R}{S}\Rightarrow S=\frac{QR}{P}=\frac{20\times 15}{10}=30\Omega$
2. Two equal resistors x in parallel give $\frac{x}{2}=30\Omega$, so $x=60\Omega$ each

Answer: S = 30 Ω; two 60 Ω resistors in parallel

Physics in real life: Current Electricity

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Drift Velocity, Mobility and Resistivity
• EMF, Internal Resistance and Cells
• Kirchhoff's Rules and Wheatstone Bridge

Current Electricity formula sheet

Current

$I=\frac{dq}{dt},I=\frac{q}{t}$

Current density

$j=\frac{I}{A},\vec{j}=\sigma \vec{E}$

Ohm's law and resistance

$V=IR,R=\rho \frac{l}{A}$

Stretching to n times the length: $R\to n^{2}R$.

Drift velocity

${\vec{v}}_d=-\frac{e\vec{E}\tau }{m}$

Current and drift velocity

$I=neA{v}_d$

Mobility

$\mu =\frac{v_d}{E}=\frac{e\tau }{m}$

Unit m² V⁻¹ s⁻¹.

Resistivity and conductivity

$\rho =\frac{m}{n{e}^2\tau },\sigma =\frac{1}{\rho }=\frac{n{e}^2\tau }{m}$

Temperature dependence

${\rho }_T={\rho }_0[1+\alpha (T-T_0)],R_T=R_0[1+\alpha (T-T_0)]$

Power

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

Rated appliance: $R=V_{rated}^2/P_{rated}$.

Cell with internal resistance

$I=\frac{\epsilon }{R+r},V=\epsilon -Ir$

Charging: $V=\epsilon +Ir$.

Cells in series

${\epsilon }_{eq}={\epsilon }_1+{\epsilon }_2,r_{eq}=r_1+r_2$

Cells in parallel

${\epsilon }_{eq}=\frac{{\epsilon }_1{r}_2+{\epsilon }_2{r}_1}{r_1+r_2},r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$

Kirchhoff's rules

$\sum I_{in}=\sum I_{out},\sum \epsilon =\sum IR\text{(round a loop)}$

Wheatstone bridge balance

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

Series and parallel resistors (for circuit problems)

$R_s=R_1+R_2+\dots ,\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\dots$

Current Electricity: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Current Electricity

The question patterns to expect, and the method that works for each.

Tricks to remember Current Electricity

Short memory aids. Say each one aloud once, then use it on a question the same day.

• I = n e A v ("I never eat a vegetable")

  Read $I=neA{v}_d$ as the silly sentence 'I never eat a vegetable'.

  Use it when: Drift velocity numericals and derivations.

• ρ = m over n e² τ

  'Mass over, n-e-squared-tau': heavier carriers or fewer, rarer collisions change ρ. More collisions (smaller τ) mean more resistance.

  Use it when: Explaining temperature dependence.

• Junction = charge, loop = energy

  Kirchhoff's first rule conserves charge (count of electrons), the second conserves energy (height of the potential hill).

  Use it when: One-mark 'on what principle' questions.

• Bridge ratios across

  Top ratio equals bottom ratio: P/Q (upper arms) = R/S (lower arms). Cross-multiply: PS = QR (opposite arms multiply equal).

  Use it when: Checking balance quickly in a network.

• Series cells add everything, parallel cells average the emf

  In parallel, the equivalent emf is a weighted average of the emfs (weights 1/r), so it always lies between them.

  Use it when: Sanity check for cells-in-parallel answers.

• Dim bulb is the big resistor in series

  In series, the lower-wattage bulb (larger R) glows brighter, because P = I²R with the same I.

  Use it when: Bulb comparison MCQs.

Current Electricity: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Current Electricity

The slips examiners see most often, and what to write instead.

• ✗Mistake: Thinking mobility changes with the applied voltage

  Students link 'more voltage, faster electrons' to mobility.

  ✓Do this: $\mu =e\tau /m$ depends only on the material and temperature. V changes $v_d$ and j, not μ.

  Usually costs: 1 mark in 2-mark reasoning items.

• ✗Mistake: Forgetting the internal resistance when finding current

  Treating every cell as ideal.

  Wrong working

  $I=\frac{12}{5.5}=2.18A$

  Right working

  $I=\frac{12}{5.5+0.5}=2A$

  ✓Do this: If r is given, it goes in the denominator.

  Usually costs: 1-2 marks.

• ✗Mistake: Inconsistent signs in Kirchhoff's loop equations

  Changing the convention halfway round a loop.

  ✓Do this: Pick a direction to go round, then: along the current through R, write −IR; − to + through a cell, write +ε. Stick to it for the whole loop.

  Usually costs: Usually all the numerical marks after the error.

• ✗Mistake: Using $R=\rho l/A$ with diameter in place of radius

  Rushing the area.

  Wrong working

  $A=\pi d^2$

  Right working

  $A=\pi r^2=\frac{\pi d^2}{4}$

  ✓Do this: Write A in terms of d explicitly before substituting.

  Usually costs: 1 mark.

• ✗Mistake: Writing $P=I^{2}R$ to compare bulbs in parallel

  Not noticing which quantity is common.

  ✓Do this: Parallel: same V, use $V^2/R$ (the smaller R is brighter). Series: same I, use $I^{2}R$ (the larger R is brighter).

  Usually costs: 1 mark in MCQs; 2 in reasoning.

• ✗Mistake: Missing the drift-velocity direction and order of magnitude

  Writing only the formula.

  ✓Do this: State that ${\vec{v}}_d$ is opposite to $\vec{E}$ for electrons and that it is of the order of $10^{-4}m{s}^{-1}$ to $10^{-3}m{s}^{-1}$.

  Usually costs: ½ mark in derivations.

NCERT exercise guide: Current Electricity

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Current Electricity: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper; 29 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Current Electricity

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Current Electricity question, by year

Practice questions on Current Electricity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The SI unit of mobility is:
   (a) m s⁻¹
   (b) m² V⁻¹ s⁻¹
   (c) V m⁻¹
   (d) m V s⁻¹

   Hint

   μ = v_d / E.

   Answer

   (b) m² V⁻¹ s⁻¹

2. Q2 · 1 mark · MCQBoard level

   A heater rated 1000 W, 220 V is used on a day when the supply voltage has fallen by 10%. Assuming its resistance does not change, its power falls by about:
   (a) 10%
   (b) 19%
   (c) 20%
   (d) 81%

   Hint

   R is fixed, so $P\propto V^2$.

   Answer

   (b) 19%: $P^{\prime }=(0.9{)}^{2}P=0.81P$, so the power falls by 19% (to 810 W).

3. Q3 · 1 mark · Very short answerBoard level

   Two wires A and B are of the same material. B has twice the length and half the radius of A. If A has resistance R, what is the resistance of B?

   Hint

   $R\propto l/r^2$.

   Answer

   $R_B=R\times \frac{2}{(1/2{)}^2}=8R$

4. Q4 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): Although the drift speed of electrons is very small, a bulb glows almost as soon as it is switched on.
   Reason (R): The electric field is set up throughout the circuit almost instantly, so electrons everywhere start drifting together.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   Think about where the electrons already are.

   Answer

   (a)

5. Q5 · 2 marks · Very short answerHarder board level

   The resistance of a wire is 5.0 Ω at 50 °C and 6.0 Ω at 100 °C. Taking $R_T=R_0(1+\alpha T)$ with T in °C, find α and the resistance at 0 °C.

   Hint

   Write the equation at both temperatures and divide to remove $R_0$.

   Answer

   $5=R_0(1+50\alpha )$ and $6=R_0(1+100\alpha )$. Dividing: $6(1+50\alpha )=5(1+100\alpha )\Rightarrow 200\alpha =1\Rightarrow \alpha =5\times 10^{-3}{}^{\circ }{C}^{-1}$. Then $R_0=\frac{5}{1.25}=4.0\Omega$.

6. Q6 · 2 marks · Very short answerHarder board level

   The potential difference across a conductor is doubled while its temperature stays the same. How do (a) the drift velocity, (b) the current density, (c) the mobility, (d) the resistivity change?

   Hint

   E = V/l.

   Answer

   (a) doubles (b) doubles (c) unchanged (d) unchanged.

7. Q7 · 2 marks · Very short answerHarder board level

   The V-I graph of a cell is a straight line. When I = 0, V = 6 V; when I = 2 A, V = 5 V. Find the emf, internal resistance and the maximum current the cell can supply.

   Hint

   $V=\epsilon -Ir$.

   Answer

   ε = 6 V; $r=\frac{6-5}{2}=0.5\Omega$; $I_{max}=\frac{6}{0.5}=12A$.

8. Q8 · 5 marks · Long answerHarder board level

   (a) Two cells of emf ${\epsilon }_1$, ${\epsilon }_2$ and internal resistance $r_1$, $r_2$ are connected in parallel (like terminals together). Derive expressions for the equivalent emf and internal resistance. (b) Cells of 4 V (1 Ω) and 6 V (2 Ω) are connected this way across a 4 Ω resistor. Find the current in the resistor and in each cell.

   Hint

   Write $V={\epsilon }_1-I_1{r}_1={\epsilon }_2-I_2{r}_2$ and $I=I_1+I_2$.

   Answer

   (a) $I=\frac{{\epsilon }_1-V}{r_1}+\frac{{\epsilon }_2-V}{r_2}$, so $V=\frac{{\epsilon }_1{r}_2+{\epsilon }_2{r}_1}{r_1+r_2}-I\frac{r_1{r}_2}{r_1+r_2}$, giving ${\epsilon }_{eq}=\frac{{\epsilon }_1{r}_2+{\epsilon }_2{r}_1}{r_1+r_2}$ and $r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$. (b) ${\epsilon }_{eq}=\frac{4\times 2+6\times 1}{3}=\frac{14}{3}V$, $r_{eq}=\frac{2}{3}\Omega$; $I=\frac{14/3}{2/3+4}=1A$; terminal voltage 4 V; 4 V cell: $\frac{4-4}{1}=0$; 6 V cell: $\frac{6-4}{2}=1A$.

9. Q9 · 3 marks · Short answerHarder board level

   In a Wheatstone bridge P = 2 Ω, Q = 4 Ω and R = 3 Ω. (a) Find S for balance. (b) The cell and galvanometer are interchanged. Is the bridge still balanced? Explain.

   Hint

   Write the balance condition both ways.

   Answer

   (a) $S=\frac{QR}{P}=\frac{4\times 3}{2}=6\Omega$. (b) Yes. After interchanging, the arms meeting at the new galvanometer ends pair up as P with R and Q with S, and the balance condition becomes $P/R=Q/S$, which is the same as $PS=QR$ (12 = 12). So no current flows through the galvanometer.

10. Q10 · 2 marks · Very short answerHOTS / competency

    A wire of resistance 12 Ω is bent into a circle. Find the resistance between (a) two diametrically opposite points, (b) two points that subtend $90^{\circ }$ at the centre.

    Hint

    The two arcs are in parallel.

    Answer

    (a) Two 6 Ω halves in parallel: 3 Ω. (b) Arcs of 3 Ω and 9 Ω in parallel: $\frac{3\times 9}{12}=2.25\Omega$.

11. Q11 · 2 marks · Very short answerHOTS / competency

    A steady current flows through a metal wire whose cross-section is larger at one end than at the other. Compare the current, current density and drift speed at the two ends.

    Hint

    Charge cannot pile up.

    Answer

    The current is the same everywhere (steady state). Current density $j=I/A$ and drift speed $v_d=I/neA$ are larger at the narrow end and smaller at the wide end.

12. Q12 · 2 marks · Very short answerHarder board level

    Resistors of 6 Ω and 3 Ω are in parallel across a 12 V supply. Find the power in each and say which gets hotter.

    Hint

    Same V across both.

    Answer

    $P_6=\frac{144}{6}=24W$; $P_3=\frac{144}{3}=48W$. The 3 Ω resistor dissipates more.

13. Q13 · 3 marks · Short answerHarder board level

    A copper wire of diameter 1.0 mm and length 5.0 m carries a current of 4.0 A when a potential difference of 0.43 V is applied across it. Taking $n=8.5\times 10^{28}{m}^{-3}$, $e=1.6\times 10^{-19}C$ and $m=9.1\times 10^{-31}kg$, find (a) the drift speed, (b) the electric field in the wire, (c) the mobility of the electrons, (d) their relaxation time.

    Hint

    Area $=\pi d^2/4$; $v_d=I/neA$; $E=V/l$; $\mu =v_d/E=e\tau /m$.

    Answer

    (a) $A=\frac{\pi (1.0\times 10^{-3}{)}^2}{4}=7.85\times 10^{-7}{m}^2$; $v_d=\frac{4.0}{8.5\times 10^{28}\times 1.6\times 10^{-19}\times 7.85\times 10^{-7}}\approx 3.7\times 10^{-4}m{s}^{-1}$. (b) $E=\frac{0.43}{5.0}=0.086V{m}^{-1}$. (c) $\mu =\frac{3.74\times 10^{-4}}{0.086}\approx 4.4\times 10^{-3}{m}^2{V}^{-1}{s}^{-1}$. (d) $\tau =\frac{\mu m}{e}=\frac{4.35\times 10^{-3}\times 9.1\times 10^{-31}}{1.6\times 10^{-19}}\approx 2.5\times 10^{-14}s$.