Board Previous Year Questions

Board Exam question bank

This section is your Board-exam workout room for Current Electricity. It is a curated bank of 35 questions in the exact format that appears on CBSE and state-board papers:

• Very Short Answer (VSA): 1 mark — one-line definitions or facts.
• Short Answer (SA-I): 2 marks — 2–3 sentence explanations or small numericals.
• Short Answer (SA-II): 3 marks — derivations of intermediate length, or detailed explanations.
• Long Answer (LA): 5 marks — full derivations with circuit diagrams, or multi-part conceptual questions.

How to use this section

Board examiners love specific derivations and definitions. Memorise the structure of each answer — not just the result. A 5-mark derivation loses 1–2 marks if you skip labelling vectors, state Kirchhoff's rules, or forget to give units. Treat this as a writing exercise, not just a reading one: re-derive each one on paper.

Legend used below:

• [Important] — these show up year after year. Prioritise these if you're short on time.
• [CBSE] — a question that has appeared in a past CBSE paper. The year is noted only when certain.
• [Diagram required] — drawing the circuit is worth marks. Skip the diagram, lose marks.

Every answer follows the Board-graded format: statement / law / derivation steps / final boxed result / one-line physical meaning. Reading one answer at a time, then closing the book and re-writing it from memory, is the single best study technique before the Board exam. Good luck!

VSA — 1 mark

Q1. Define drift velocity. [Important]

The drift velocity is the average velocity with which free electrons in a conductor drift along the direction opposite to the applied electric field, superimposed on their random thermal motion.

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

Its SI unit is m/s, and its typical magnitude in a current-carrying copper wire is $\sim 10^{-4}$ m/s.

Q2. State Ohm's law.

At constant temperature (and subject to no chemical or physical changes), the current $I$ flowing through an ohmic conductor is directly proportional to the potential difference $V$ across its ends:

$$V = IR$$

where $R$ is the resistance (a constant for the conductor at fixed temperature).

Q3. Define electrical resistivity.

Resistivity (or specific resistance) $\rho$ of a material is the resistance offered by a specimen of unit length and unit cross-sectional area. It relates the resistance to the geometry:

$$R = \rho \frac{L}{A}$$

SI unit: ohm-metre (Ω·m). It depends on the material and temperature but not on the size or shape of the specimen.

Q4. Define EMF of a cell.

The electromotive force (EMF) $\varepsilon$ of a cell is the work done per unit charge by the non-electrostatic force inside the cell in moving a unit positive charge from its negative terminal to its positive terminal:

$$\varepsilon = \frac{W}{q}$$

SI unit: volt (V). Equivalently, $\varepsilon$ is the terminal voltage of the cell when no current is drawn from it (open circuit).

Q5. State the two Kirchhoff's rules. [Important]

Junction rule (KCL): At any junction in a circuit, the algebraic sum of currents is zero (consequence of charge conservation): $$\sum I = 0$$

Loop rule (KVL): The algebraic sum of potential differences around any closed loop is zero (consequence of energy conservation): $$\sum \Delta V = 0$$

Q6. Define mobility of a charge carrier.

Mobility $\mu$ is defined as the drift velocity acquired per unit applied electric field:

$$\mu = \frac{|v_d|}{E}$$

SI unit: m²/(V·s) (or cm²/(V·s) in practice). A higher mobility means the charge carriers respond more strongly to a given electric field.

Q7. What is the SI unit of potential gradient on a potentiometer wire?

The potential gradient $k$ is the fall of potential per unit length along the potentiometer wire:

$$k = \frac{V}{\ell}$$

SI unit: volt per metre (V/m) — or more commonly volt per centimetre (V/cm) or millivolt per centimetre (mV/cm) in practice.

SA-I — 2 marks

Q8. Distinguish between EMF and terminal voltage of a cell. [Important]

EMF ($\varepsilon$)	Terminal voltage ($V$)
Property of the cell (depends on chemistry).	Depends on the external circuit.
Measured when no current flows ($I = 0$).	Measured when current is flowing.
Always equal to maximum possible voltage.	$V = \varepsilon - Ir$ while discharging; $V = \varepsilon + Ir$ while charging.

They are equal only in open circuit ($I = 0$). When a cell is delivering current, $V < \varepsilon$; when it's being charged, $V > \varepsilon$.

Q9. Why is the current flowing through a conductor constant even though the drift velocity is very small ($\sim 10^{-4}$ m/s)?

The free-electron number density $n$ in a metallic conductor is extremely large — about $10^{29}$ electrons per cubic metre. Even with a tiny drift velocity, the product $I = neAv_d$ gives a macroscopic current. Further, an electric field sets up almost instantaneously throughout the conductor, so electrons all along the wire begin to drift together. Thus the current is steady even though individual electrons move slowly.

Q10. Why is manganin or constantan used for making standard resistors?

Manganin (Cu-Mn-Ni alloy) and constantan (Cu-Ni alloy) are preferred for standard resistors because:

• They have a very low temperature coefficient of resistance ($\alpha \approx 10^{-5}$ /°C), so their resistance barely changes with temperature. This ensures the standard value is stable.
• They have a high resistivity, so a small coil of wire gives a reasonable resistance value.

A pure copper wire, in contrast, has $\alpha \approx 4 \times 10^{-3}$ /°C — its resistance can change by several percent over the range of room temperatures.

Q11. Why does the resistance of a conductor increase with temperature, while that of a semiconductor decreases?

In a metal, increased temperature causes more frequent collisions between free electrons and lattice ions, shortening the relaxation time $\tau$. Since $\rho = m/(ne^2\tau)$ and $n$ is essentially constant, $\rho$ (and hence $R$) increases.

In a semiconductor, increased temperature frees many more charge carriers, so $n$ increases sharply. This increase in $n$ far outweighs any reduction in $\tau$, so $\rho$ decreases. Hence semiconductors have a negative temperature coefficient of resistance.

Q12. State and briefly explain the principle of the potentiometer. [Important]

Principle: The potential difference across any portion of a uniform wire carrying a steady current is directly proportional to the length of that portion.

$$V \propto \ell, \quad V = k\ell$$

where $k$ is the potential gradient. Because the setup is a null method (galvanometer reads zero at balance, so no current flows from the test cell), the reading equals the true EMF — not a loaded terminal voltage — making the potentiometer ideal for precise EMF measurements.

Q13. A voltmeter of finite resistance, when connected across a cell, reads a voltage slightly less than the true EMF of the cell. Why?

A voltmeter has finite resistance $R_V$. When connected across a cell of EMF $\varepsilon$ and internal resistance $r$, it draws a small current $I = \varepsilon/(R_V + r)$. Its reading is the terminal voltage:

$$V = IR_V = \varepsilon \cdot \frac{R_V}{R_V + r} < \varepsilon$$

To minimise this loading error, use a voltmeter with $R_V \gg r$ — or use a potentiometer, which draws zero current at balance.

SA-II — 3 marks

Q14. Derive the expression $I = neAv_d$ relating current to drift velocity. [Important] [Diagram required] [CBSE recurring]

Setup. Consider a conductor of uniform cross-section $A$, with free-electron number density $n$. In steady state, electrons drift with velocity $v_d$ under an applied field.

Derivation. In time $\Delta t$, each electron moves a distance $v_d \Delta t$. Consider a cylindrical volume of length $v_d \Delta t$ and cross-sectional area $A$ adjacent to a cross-section of the wire. The number of electrons in this volume is:

$$N = n \cdot (A \cdot v_d \Delta t) = n A v_d \Delta t$$

Each electron carries charge $e$. So the total charge crossing any cross-section in time $\Delta t$:

$$\Delta Q = (n A v_d \Delta t)e$$

Current:

$$\boxed{\, I = \frac{\Delta Q}{\Delta t} = n e A v_d \, }$$

Physical meaning. Current is set by the number density $n$, charge per carrier $e$, cross-section $A$, and drift speed $v_d$ — not by the speed of individual electrons' random motion.

Q15. Derive the expression for drift velocity $v_d = eE\tau/m$. [Important] [CBSE recurring]

Setup. An electron (charge $-e$, mass $m$) in a conductor experiences an electric field $\vec{E}$, giving it an acceleration $\vec{a} = -e\vec{E}/m$.

Derivation. The electron's velocity $\vec{v}_i$ just after its $i$-th collision is essentially random, so on averaging over many electrons, $\langle \vec{v}_i \rangle = 0$. Between two successive collisions, the electron accelerates for a time $t_i$ with average value $\tau$.

$$\vec{v} = \vec{v}_i + \vec{a}t_i$$

Averaging over all electrons:

$$\langle \vec{v} \rangle = \langle \vec{v}_i \rangle + \vec{a}\langle t_i \rangle = 0 + \vec{a}\tau = \frac{-e\vec{E}\tau}{m}$$

This average is the drift velocity. Taking magnitudes:

$$\boxed{\, v_d = \frac{eE\tau}{m} \, }$$

Physical meaning. Shorter relaxation time means electrons are interrupted more frequently, giving smaller drift velocity and larger resistance.

Q16. Using the drift velocity formula, derive Ohm's law $V = IR$ and the expression $\rho = m/(ne^2\tau)$ for resistivity. [Important] [CBSE]

Derivation. Start with $I = neAv_d$ and $v_d = eE\tau/m$:

$$I = neA \cdot \frac{eE\tau}{m} = \frac{n e^2 A \tau}{m}E$$

For a conductor of length $L$, $E = V/L$. Therefore,

$$I = \frac{n e^2 A \tau}{mL}V$$

This has the form $I = V/R$, so:

$$R = \frac{mL}{n e^2 A \tau}$$

Since $R = \rho L/A$,

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

Physical meaning. Resistivity depends on microscopic quantities: carrier density $n$ and relaxation time $\tau$.

Q17. Derive the expression $V = \varepsilon - Ir$ for the terminal voltage of a cell delivering current $I$ through an external resistance $R$. [Important] [Diagram required]

Setup. A cell of EMF $\varepsilon$ and internal resistance $r$ is connected across an external resistance $R$. Current $I$ flows in the circuit.

Derivation. Apply Kirchhoff's loop rule:

$$+\varepsilon - Ir - IR = 0$$

Hence, $$I = \frac{\varepsilon}{R + r}$$

The terminal voltage $V$ equals the potential difference across the external resistor:

$$V = IR$$

So,

$$\boxed{\, V = \varepsilon - Ir \, }$$

Physical meaning. The terminal voltage is less than the EMF by the lost volts $Ir$ while the cell is discharging.

Q18. Derive the equivalent EMF and internal resistance of two cells connected in series (aiding). [Diagram required]

Setup. Two cells with EMFs $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ are connected in series aiding each other.

Derivation. If current $I$ flows through external resistance $R$, then:

$$\varepsilon_1 + \varepsilon_2 - I(r_1 + r_2) - IR = 0$$

Comparing with an equivalent single-cell form,

$$\varepsilon_{\text{eq}} - Ir_{\text{eq}} - IR = 0$$

we get:

$$\boxed{\, \varepsilon_{\text{eq}} = \varepsilon_1 + \varepsilon_2, \quad r_{\text{eq}} = r_1 + r_2 \, }$$

If the cells oppose, EMFs subtract algebraically but internal resistances still add.

Q19. Derive the equivalent EMF and internal resistance of two cells connected in parallel. [Important] [Diagram required] [CBSE]

Setup. Two cells with EMFs $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ are connected in parallel.

Let terminal voltage be $V$. Then,

$$V = \varepsilon_1 - I_1r_1 = \varepsilon_2 - I_2r_2$$

So,

$$I_1 = \frac{\varepsilon_1 - V}{r_1}, \qquad I_2 = \frac{\varepsilon_2 - V}{r_2}$$

Total current:

$$I = I_1 + I_2 = \frac{\varepsilon_1 - V}{r_1} + \frac{\varepsilon_2 - V}{r_2}$$

Rearranging into equivalent-cell form gives:

$$\boxed{\, \varepsilon_{\text{eq}} = \frac{\varepsilon_1/r_1 + \varepsilon_2/r_2}{1/r_1 + 1/r_2}, \quad \frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2} \, }$$

Physical meaning. The parallel combination has a weighted-average EMF and a smaller effective internal resistance.

Q20. Derive the balance condition of a Wheatstone bridge using Kirchhoff's rules. [Important] [CBSE]

Setup. Four resistors $P, Q, R, S$ form a diamond; a battery is connected across one diagonal and a galvanometer across the other.

At balance, galvanometer current is zero, so current $I_1$ flows through $P$ and $Q$, and current $I_2$ flows through $R$ and $S$.

Since the potentials at the galvanometer junctions are equal,

$$I_1P = I_2R$$

and

$$I_1Q = I_2S$$

Dividing,

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

Physical meaning. The bridge balance depends only on the resistance ratios, not on the battery EMF or galvanometer resistance.

Q21. Derive the meter bridge formula for an unknown resistance. [Diagram required]

Setup. A 1-metre-long uniform wire is used as two ratio arms. A known resistance $R$ is in one gap and an unknown resistance $X$ in the other. Let the balance length from the left end be $\ell$ cm.

Since the wire is uniform, resistance is proportional to length. Therefore, the ratio arms are:

$$P \propto \ell, \qquad Q \propto 100 - \ell$$

Using the Wheatstone balance condition:

$$\frac{\ell}{100 - \ell} = \frac{R}{X}$$

Hence,

$$\boxed{\, X = R \cdot \frac{100 - \ell}{\ell} \, }$$

Physical meaning. The absolute resistance per unit length of the wire cancels out, leaving only the measurable length ratio.

Q22. Derive the formula $r = R(\ell_1 - \ell_2)/\ell_2$ for measuring the internal resistance of a cell using a potentiometer. [Important] [Diagram required] [CBSE]

Step 1 — Key open. No current is drawn from the cell, so the terminal voltage equals its EMF:

$$\varepsilon = k\ell_1$$

Step 2 — Key closed. A resistance $R$ is connected across the cell. The terminal voltage becomes

$$V = \varepsilon \cdot \frac{R}{R+r}$$

and this balances at length $\ell_2$:

$$V = k\ell_2$$

Divide the two equations:

$$\frac{\ell_1}{\ell_2} = \frac{\varepsilon}{V} = \frac{R+r}{R}$$

Therefore,

$$\boxed{\, r = R \cdot \frac{\ell_1 - \ell_2}{\ell_2} \, }$$

Physical meaning. Internal resistance is obtained from a known resistor and two balance lengths only.

LA — 5 marks

Q23. With a neat circuit diagram, describe the working of a potentiometer for comparing EMFs of two cells. Derive the relevant formula. [Important] [Diagram required] [CBSE]

Principle. A steady current through a uniform wire produces a potential drop directly proportional to length:

$$V = k\ell$$

where $k$ is the potential gradient.

Working. Let two cells of EMFs $\varepsilon_1$ and $\varepsilon_2$ balance at lengths $\ell_1$ and $\ell_2$ respectively, with the same potentiometer setting.

Then,

$$\varepsilon_1 = k\ell_1, \qquad \varepsilon_2 = k\ell_2$$

Dividing,

$$\boxed{\, \frac{\varepsilon_1}{\varepsilon_2} = \frac{\ell_1}{\ell_2} \, }$$

Why it is accurate:

• It is a null method.
• No current is drawn from the test cells at balance.
• Galvanometer resistance does not matter.
• The gradient cancels while comparing EMFs.

Precautions:

• Driver cell must have EMF greater than the test cells.
• Positive terminals must be connected correctly.
• The potentiometer current must remain steady.

Q24. With a neat circuit diagram, describe the construction and working of a meter bridge to determine an unknown resistance. Derive the balance formula and state any two precautions. [Important] [Diagram required] [CBSE]

Construction.

• A 1-metre-long uniform resistance wire is stretched along a wooden scale.
• The two end gaps hold the known resistance $R$ and the unknown resistance $X$.
• A battery, key, and rheostat drive current through the wire.
• A galvanometer and jockey are used to locate the null point.

Working. At the null point, let the length from the left end be $\ell$ cm. Then the two wire segments have resistances proportional to $\ell$ and $100-\ell$.

Applying the Wheatstone bridge condition:

$$\frac{\ell}{100-\ell} = \frac{R}{X}$$

Hence,

$$\boxed{\, X = R \cdot \frac{100-\ell}{\ell} \, }$$

Precautions:

1. The null point should be near the middle for best accuracy.
2. The jockey should be pressed lightly.
3. The wire should not be overheated.
4. Connections should be tight and clean.

Q25. Draw a labelled circuit diagram of a potentiometer for measuring the internal resistance of a cell. Derive the expression for internal resistance. [Diagram required] [CBSE]

Working. With the shunt key open, the balance length is $\ell_1$:

$$\varepsilon = k\ell_1$$

With a known resistor $R$ connected across the cell, the balance length becomes $\ell_2$ and terminal voltage is:

$$V = k\ell_2 = \varepsilon \cdot \frac{R}{R+r}$$

Dividing,

$$\frac{\ell_1}{\ell_2} = \frac{R+r}{R}$$

Therefore,

$$\boxed{\, r = R \cdot \frac{\ell_1 - \ell_2}{\ell_2} \, }$$

Precautions:

1. Use a suitable resistor $R$.
2. Close the key only briefly.
3. Keep the driver current steady.

Q26. Describe an experiment to verify Ohm's law for a metallic conductor, including circuit diagram, procedure, observations, and graph. [Diagram required]

Aim. To verify Ohm's law for a metallic conductor.

Apparatus. Resistance wire, battery, ammeter, voltmeter, rheostat, key, connecting wires.

Procedure.

1. Connect the ammeter in series and voltmeter in parallel with the conductor.
2. Vary the current using the rheostat.
3. Record corresponding values of current and voltage.
4. Plot the graph of $V$ versus $I$.

Observation. The graph is a straight line through the origin.

Therefore,

$$V \propto I \Rightarrow V = IR$$

The slope of the graph gives the resistance.

Precautions:

1. Avoid heating the conductor.
2. Make proper circuit connections.
3. Take several observations.

Q27. State Kirchhoff's rules. Using these rules, derive the balance condition of a Wheatstone bridge. [Important] [Diagram required] [CBSE]

Kirchhoff's rules:

• Junction rule: algebraic sum of currents at a junction is zero.
• Loop rule: algebraic sum of potential differences in a closed loop is zero.

For a Wheatstone bridge at balance, the galvanometer current is zero. Hence the currents through opposite arms satisfy:

$$I_1P = I_2R$$ $$I_1Q = I_2S$$

Dividing,

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

Physical significance. The bridge balance depends only on resistance ratios and is therefore highly accurate.

Numericals

Q28. A wire is stretched to double its length, keeping the volume constant. How does its resistance change? [Important]

Since volume $AL$ remains constant, doubling the length halves the area.

Original resistance: $$R = \rho \frac{L}{A}$$

New resistance: $$R' = \rho \frac{2L}{A/2} = 4R$$

Answer: The resistance becomes 4 times the original.

Key idea: When volume is constant, resistance varies as the square of the length.

Q29. Three resistors of resistance $R$ each are connected first in series and then in parallel. Find the ratio of power dissipated in the two combinations (for the same applied voltage $V$).

For series: $$R_s = 3R, \qquad P_s = \frac{V^2}{3R}$$

For parallel: $$R_p = \frac{R}{3}, \qquad P_p = \frac{V^2}{R/3} = \frac{3V^2}{R}$$

Hence,

$$\frac{P_s}{P_p} = \frac{1}{9}$$

Answer: $P_{series} : P_{parallel} = 1 : 9$.

Q30. Two cells of EMFs 1.5 V and 2 V with internal resistances 1 Ω and 2 Ω respectively are connected in parallel across a 10-Ω external resistor. Find: (a) the equivalent EMF, (b) the equivalent internal resistance, (c) the current through the 10-Ω resistor.

(a) Equivalent EMF: $$\varepsilon_{eq} = \frac{1.5/1 + 2/2}{1/1 + 1/2} = \frac{2.5}{1.5} \approx 1.67\text{ V}$$

(b) Equivalent internal resistance: $$r_{eq} = \frac{1 \times 2}{1+2} = \frac{2}{3} \approx 0.67\text{ Ω}$$

(c) Current: $$I = \frac{1.67}{10 + 0.67} \approx 0.156\text{ A}$$

Answer: $\varepsilon_{eq} \approx 1.67$ V, $r_{eq} \approx 0.67$ Ω, $I \approx 0.156$ A.

Q31. A battery of EMF 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, find (a) the resistance of the resistor, (b) the terminal voltage of the battery. [Important] [CBSE]

Total circuit resistance: $$R_{total} = \frac{\varepsilon}{I} = \frac{10}{0.5} = 20\text{ Ω}$$

Therefore, $$R = 20 - 3 = 17\text{ Ω}$$

Terminal voltage: $$V = \varepsilon - Ir = 10 - 0.5 \times 3 = 8.5\text{ V}$$

Answers: $R = 17$ Ω, $V = 8.5$ V.

Cross-check: $V = IR = 0.5 \times 17 = 8.5$ V.

Q32. A heating element of an electric heater is rated at 1000 W / 220 V. Find (i) the current drawn, (ii) the resistance of the element.

(i) Current: $$I = \frac{P}{V} = \frac{1000}{220} \approx 4.55\text{ A}$$

(ii) Resistance: $$R = \frac{V^2}{P} = \frac{220^2}{1000} = 48.4\text{ Ω}$$

Answers: $I \approx 4.55$ A, $R = 48.4$ Ω.

Cross-check: $V = IR$ and $P = I^2R$.

Q33. State the condition for maximum power transfer from a cell to an external resistance. Derive the expression for the maximum power. [Important] [CBSE]

Power delivered to load:

$$P_R = \frac{\varepsilon^2R}{(R+r)^2}$$

Differentiating with respect to $R$ and setting to zero gives:

$$R = r$$

So the condition for maximum power transfer is:

$$\boxed{R = r}$$

At this condition,

$$P_{max} = \frac{\varepsilon^2}{4r}$$

Physical meaning. At maximum power transfer, half the power is lost inside the cell, so efficiency is only 50%.

Q34. Using Kirchhoff's rules, find the current in a single loop containing a 12-V cell, a 2-Ω resistor, an opposing 8-V cell, and a 4-Ω resistor in series. [CBSE]

Applying KVL:

$$+12 - 2I - 8 - 4I = 0$$

$$4 = 6I \Rightarrow I = 0.67\text{ A}$$

The current is driven by the stronger 12-V cell, so the 8-V cell is being charged.

Answer: Current = 0.67 A.

Q35. In an experiment, the temperature coefficient of a resistance wire was found to be $4 \times 10^{-3}$ /°C. At 20 °C, the resistance is 50 Ω. What is the resistance at 120 °C?

Using $$R_T = R_0[1 + \alpha(T - T_0)]$$

$$R_{120} = 50[1 + 4 \times 10^{-3}(120 - 20)] = 50(1 + 0.4) = 70\text{ Ω}$$

Answer: $R = 70$ Ω.

Resistance increases by 40% over this 100 °C range, which is typical of metals like copper.