Welcome to HSLC GURU. This page provides a complete, exam-ready guide to ASSEB Class 12 Physics Chapter 3: Current Electricity in English medium. The chapter develops the microscopic and macroscopic pictures of electric current, builds Ohm’s law from drift velocity, and applies Kirchhoff’s laws to standard networks (Wheatstone bridge, meter bridge, potentiometer). All formulas are written using KaTeX, three SVG diagrams illustrate key circuits, and every textbook exercise is solved step by step. Use the sections below to revise concepts, practise numericals, and consolidate definitions before the ASSEB Higher Secondary final examination.

Chapter Summary

Electric current is the rate of flow of charge through a cross-section of a conductor. Inside a metallic conductor a constant electric field $E$ pushes free electrons to drift opposite to $E$ with a small drift velocity $v_d$. Although individual electrons move randomly with thermal speeds of $\sim 10^5$ m/s, the net drift speed is only of the order of $10^{-4}$ m/s, yet it is enough to set up a steady macroscopic current $I$.

Ohm’s law $V = IR$ is an empirical relation valid for ohmic materials at constant temperature. The resistance $R$ depends on geometry and material through $R = \rho L/A$, where $\rho$ is the resistivity. The temperature dependence $\rho_T = \rho_0[1 + \alpha(T-T_0)]$ explains why metals heat up and conduct slightly worse, while semiconductors conduct better with rise in temperature ($\alpha < 0$).

Real cells supply an EMF $\epsilon$ but lose some voltage to their internal resistance $r$, so the terminal voltage is $V = \epsilon – Ir$. Kirchhoff’s junction rule (KCL) expresses charge conservation, $\sum I = 0$ at any node, while Kirchhoff’s loop rule (KVL) expresses energy conservation, $\sum \epsilon = \sum IR$ around any closed loop. These two rules let us solve any DC network. The Wheatstone bridge, meter bridge and potentiometer are direct applications and form the foundation of precision DC measurements.

Key Formulas

Quantity	Formula	SI Unit
Electric current	$I = \dfrac{dq}{dt}$	ampere (A)
Drift velocity	$v_d = -\dfrac{eE\tau}{m}$	m s$^{-1}$
Mobility	$\mu = \dfrac{|v_d|}{E} = \dfrac{e\tau}{m}$	m$^2$ V$^{-1}$ s$^{-1}$
Current density	$\vec{J} = ne\vec{v_d} = \sigma\vec{E}$	A m$^{-2}$
Ohm’s law	$V = IR$	volt (V)
Resistance	$R = \dfrac{\rho L}{A}$	ohm ($\Omega$)
Conductivity	$\sigma = \dfrac{1}{\rho} = \dfrac{ne^2\tau}{m}$	S m$^{-1}$
Temperature variation	$\rho_T = \rho_0[1 + \alpha(T-T_0)]$	$\Omega$ m
Series resistance	$R_s = R_1 + R_2 + R_3 + \ldots$	$\Omega$
Parallel resistance	$\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots$	$\Omega$
Terminal voltage	$V = \epsilon – Ir$	volt
Wheatstone balance	$\dfrac{P}{Q} = \dfrac{R}{S}$	—
Electric power	$P = VI = I^2R = \dfrac{V^2}{R}$	watt (W)

The microscopic form of Ohm’s law links current density $\vec{J}$ to the applied field through conductivity:

$$\vec{J} = \sigma \vec{E}, \qquad \sigma = \dfrac{ne^2\tau}{m}.$$

For $N$ identical cells (each EMF $\epsilon$, internal resistance $r$) connected in series across an external resistance $R$:

$$I = \dfrac{N\epsilon}{R + Nr}.$$

For $N$ identical cells in parallel:

$$I = \dfrac{\epsilon}{R + r/N}.$$

Circuit Diagrams

Figure 3.1 — Simple closed circuit showing a cell of EMF $\epsilon$ with internal resistance $r$ driving current $I$ through an external resistor $R$.

Figure 3.2 — Wheatstone bridge. When the galvanometer reads zero, no current flows through it and the balance condition $P/Q = R/S$ holds.

Figure 3.3 — Potentiometer. A long resistance wire AB carries a steady current from a driver cell. The unknown EMF is balanced against the potential drop along the wire to the jockey J.

Conceptual Foundations — Detailed Notes

1. Electric current and conventional direction. In a metallic conductor connected to a battery, free electrons drift from the lower potential (negative terminal of the cell) to the higher potential (positive terminal). However, by historical convention the direction of the electric current is taken as the direction in which positive charges would move — i.e. opposite to the electron drift. Inside the cell, the EMF drives positive charges from the negative to the positive terminal; outside the cell (in the external circuit) the current flows from the positive terminal through the load and back to the negative terminal.

If a charge $\Delta q$ crosses a section of conductor in time $\Delta t$, the average current is $I_{\text{av}} = \Delta q/\Delta t$. The instantaneous current is the limit:

$$I = \lim_{\Delta t \to 0}\dfrac{\Delta q}{\Delta t} = \dfrac{dq}{dt}.$$

Although direction is essential, electric current itself is a scalar quantity because it does not obey the parallelogram law of vector addition; for the same reason, currents at a junction are added algebraically (with sign indicating direction).

2. The microscopic picture — drift of free electrons. In the absence of an electric field, free electrons in a metal move randomly in all directions with thermal speeds of the order of $10^5$ m/s. The average velocity over the entire population is zero, hence no net current. When an electric field $\vec{E}$ is applied, each electron experiences a force $-e\vec{E}$ and is uniformly accelerated until it suffers a collision with a positive ion. The cumulative effect of many collisions is a small directed velocity opposite to $\vec{E}$, called the drift velocity $\vec{v_d}$. If $\tau$ is the average time between collisions (relaxation time), then

$$\vec{v_d} = -\dfrac{e\vec{E}}{m}\tau,$$

where $m$ is the electron mass. Even though $|\vec{v_d}|$ is only of the order of $10^{-4}$ m/s, the immense number density $n \sim 10^{28}$–$10^{29}$ m$^{-3}$ ensures macroscopically observable currents.

3. From drift velocity to Ohm’s law. Consider a conductor of length $L$, area $A$ and free-electron density $n$. The number of electrons crossing any cross-section in unit time is $n A v_d$, carrying charge $I = n e A v_d$. Substituting $v_d = eE\tau/m$ and $E = V/L$,

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

Rearranging, $V = IR$ where $R = mL/(ne^2 A\tau) = \rho L/A$, with resistivity $\rho = m/(ne^2\tau)$. This is the microscopic justification of Ohm’s law: $V$ and $I$ are directly proportional as long as $n$, $\tau$ and physical conditions remain constant.

4. Limitations of Ohm’s law. Ohm’s law is not a fundamental law of nature; it is an empirical observation that holds for many metallic conductors at moderate fields and steady temperatures. It fails for: (i) non-ohmic devices like semiconductor diodes (different resistance for forward and reverse bias), (ii) high-current regimes where heating changes resistance significantly, (iii) gas discharges where charge carriers are produced by ionisation, and (iv) electrolytes where the V–I curve is non-linear and shows back EMF.

5. Resistors in series. When several resistors $R_1, R_2, \ldots, R_n$ are joined end to end so that the same current $I$ flows through each:

$$V = V_1 + V_2 + \ldots + V_n = I(R_1 + R_2 + \ldots + R_n) \implies R_s = \sum_i R_i.$$

The equivalent resistance is greater than each individual resistance.

6. Resistors in parallel. When resistors are connected so that the same potential difference $V$ exists across each, the total current splits:

$$I = I_1 + I_2 + \ldots = V\left(\dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots\right) \implies \dfrac{1}{R_p} = \sum_i \dfrac{1}{R_i}.$$

The equivalent resistance is less than each individual resistance — adding a resistor in parallel always increases the conductance.

7. EMF and internal resistance. A real electrochemical cell has internal resistance $r$ due to the electrolyte, electrode polarisation, and contact resistance. When current $I$ is drawn:

$$V_{\text{terminal}} = \epsilon – Ir.$$

The cell delivers true EMF only at zero current (open circuit). For a short circuit, $V \to 0$ and $I_{\max} = \epsilon/r$ — a useful but destructive limit since the cell rapidly heats and degrades.

8. Combinations of cells. For $n$ identical cells in series across external $R$:

$$I_{\text{series}} = \dfrac{n\epsilon}{R + nr}.$$

For $m$ identical cells in parallel:

$$I_{\text{parallel}} = \dfrac{\epsilon}{R + r/m}.$$

For a mixed grouping (m rows of n cells each in parallel, total $N = mn$):

$$I = \dfrac{n\epsilon}{R + nr/m}.$$

Maximum current is delivered when the external resistance equals the equivalent internal resistance of the combination.

9. Kirchhoff’s laws — strategy of application. (i) Mark each branch with an assumed current direction; if the algebraic answer is negative, the actual direction is reverse. (ii) Apply the junction rule to as many nodes as needed (one less than the total number of nodes gives independent equations). (iii) Apply the loop rule to enough independent loops to match the unknowns. (iv) When traversing a loop, take a battery EMF positive if traversed from $-$ to $+$ inside the source; take an IR drop positive if traversed in the direction of assumed current.

10. Wheatstone bridge — derivation. Assume the bridge is balanced. Let $I_1$ flow through P then Q, and $I_2$ flow through R then S, with no current through the galvanometer. Equality of potentials at B and D gives

$$I_1 P = I_2 R, \qquad I_1 Q = I_2 S.$$

Dividing yields the balance condition $P/Q = R/S$. The arrangement is independent of cell EMF and galvanometer resistance, making it ideal for null-method measurements of unknown resistances against known standards.

11. Meter bridge. The meter bridge is a practical Wheatstone bridge using a 1 m uniform resistance wire (typically of constantan or manganin). Two of the four arms are formed by lengths $l$ and $(100 – l)$ cm of this wire, the third arm contains the unknown resistance X, and the fourth arm contains a known resistance Y. At balance,

$$\dfrac{X}{Y} = \dfrac{l}{100 – l}.$$

Sources of error: (i) non-uniformity of the wire; (ii) end-corrections at the terminals; (iii) heating of the wire if cell current is too high. Best accuracy is obtained when the balance length is near 50 cm.

12. Potentiometer. A potentiometer is essentially a uniform resistance wire AB across which a steady “driver” cell maintains a current $I_0$, producing a uniform potential gradient $k = I_0\rho/A$ along the wire. Any test EMF $\epsilon$ smaller than the total potential drop across AB can be balanced by sliding the jockey J along the wire to the point where the potential drop from A to J equals $\epsilon$ — at that instant the galvanometer shows zero deflection and no current is drawn from the test cell. The EMF is then $\epsilon = kl$, where $l$ is the balance length. By comparing two cells through their balance lengths $l_1$ and $l_2$:

$$\dfrac{\epsilon_1}{\epsilon_2} = \dfrac{l_1}{l_2}.$$

The same arrangement can also measure the internal resistance of a cell: balance the cell first on open circuit (length $l_1$), then with a known shunt $R$ across it (length $l_2$). Then $r = R(l_1 – l_2)/l_2$.

13. Electric power and energy. A current $I$ driven by a potential difference $V$ delivers power

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

Over time $t$, the energy dissipated is $W = Pt$ joules. The commercial unit of electrical energy is the kilowatt-hour: $1$ kWh $= 3.6 \times 10^6$ J. Domestic appliances are rated by their power consumption at the standard supply voltage (usually 220 V in India).

Textbook Exercise — Solutions

Q1. The storage battery of a car has an EMF of 12 V. If the internal resistance of the battery is 0.4 $\Omega$, what is the maximum current that can be drawn from the battery?

Answer: Maximum current is drawn when the external resistance is zero (short circuit):

$$I_{\max} = \dfrac{\epsilon}{r} = \dfrac{12}{0.4} = 30 \text{ A}.$$

Q2. A battery of EMF 10 V and internal resistance 3 $\Omega$ is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer: Using $\epsilon = I(R + r)$:

$$R = \dfrac{\epsilon}{I} – r = \dfrac{10}{0.5} – 3 = 20 – 3 = 17 \, \Omega.$$

Terminal voltage $V = \epsilon – Ir = 10 – 0.5 \times 3 = 8.5$ V.

Q3. (a) Three resistors 1, 2 and 3 $\Omega$ are combined in series. What is the total resistance? (b) If the combination is connected to a battery of EMF 12 V and negligible internal resistance, find the potential drop across each resistor.

Answer: (a) $R_s = 1 + 2 + 3 = 6 \, \Omega$. (b) Current $I = 12/6 = 2$ A. Voltages: $V_1 = 2 \times 1 = 2$ V, $V_2 = 2 \times 2 = 4$ V, $V_3 = 2 \times 3 = 6$ V (sum = 12 V, as expected).

Q4. (a) Three resistors 2, 4 and 5 $\Omega$ are combined in parallel. What is the total resistance? (b) If the combination is connected to a battery of EMF 20 V and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery.

Answer: (a) $\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{5} = \dfrac{10 + 5 + 4}{20} = \dfrac{19}{20}$, so $R_p = 20/19 \approx 1.05 \, \Omega$. (b) Currents: $I_1 = 20/2 = 10$ A, $I_2 = 20/4 = 5$ A, $I_3 = 20/5 = 4$ A. Total $I = 10 + 5 + 4 = 19$ A.

Q5. At room temperature (27.0 °C) the resistance of a heating element is 100 $\Omega$. What is the temperature of the element if the resistance is found to be 117 $\Omega$, given that the temperature coefficient of the material of the resistor is $1.70 \times 10^{-4}$ °C$^{-1}$?

Answer: Using $R_T = R_0[1 + \alpha(T – T_0)]$:

$$T – T_0 = \dfrac{R_T – R_0}{\alpha R_0} = \dfrac{117 – 100}{1.70 \times 10^{-4} \times 100} = \dfrac{17}{0.017} = 1000 \, °C.$$

Therefore $T = 1000 + 27 = 1027 \, °C$.

Q6. A negligibly small current is passed through a wire of length 15 m and uniform cross-section $6.0 \times 10^{-7}$ m$^2$, and its resistance is measured to be 5.0 $\Omega$. What is the resistivity of the material at the temperature of the experiment?

Answer: From $R = \rho L/A$:

$$\rho = \dfrac{RA}{L} = \dfrac{5.0 \times 6.0 \times 10^{-7}}{15} = 2.0 \times 10^{-7} \, \Omega\,\text{m}.$$

Q7. A silver wire has a resistance of 2.1 $\Omega$ at 27.5 °C, and a resistance of 2.7 $\Omega$ at 100 °C. Determine the temperature coefficient of resistivity of silver.

Answer: Using $R_2 = R_1[1 + \alpha(T_2 – T_1)]$:

$$\alpha = \dfrac{R_2 – R_1}{R_1(T_2 – T_1)} = \dfrac{2.7 – 2.1}{2.1 \times (100 – 27.5)} = \dfrac{0.6}{152.25} \approx 3.94 \times 10^{-3} \, °C^{-1}.$$

Q8. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A, which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if room temperature is 27 °C? Take temperature coefficient of resistance of nichrome as $1.70 \times 10^{-4}$ °C$^{-1}$.

Answer: Initial resistance $R_1 = 230/3.2 = 71.875 \, \Omega$ at 27 °C. Steady resistance $R_2 = 230/2.8 = 82.143 \, \Omega$.

$$T_2 – T_1 = \dfrac{R_2 – R_1}{\alpha R_1} = \dfrac{82.143 – 71.875}{1.70 \times 10^{-4} \times 71.875} = \dfrac{10.268}{0.01222} \approx 840 \, °C.$$

Hence steady temperature $T_2 = 840 + 27 = 867 \, °C$.

Q9. Determine the current in each branch of the network shown: a network with EMF 10 V and resistors 5 $\Omega$, 5 $\Omega$, 10 $\Omega$, 10 $\Omega$, 5 $\Omega$ arranged in a Wheatstone-like configuration.

Answer: Apply Kirchhoff’s loop rule to the two independent loops. Solving the simultaneous equations gives the standard NCERT result:

Branch	Current
AB	$\tfrac{4}{17}$ A $\approx 0.235$ A
BC	$\tfrac{6}{17}$ A $\approx 0.353$ A
CD	$-\tfrac{4}{17}$ A (reverse)
AD	$\tfrac{6}{17}$ A
BD	$-\tfrac{2}{17}$ A
Total from cell	$\tfrac{10}{17}$ A $\approx 0.588$ A

Q10. (a) In a meter bridge, the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 $\Omega$. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips? (b) Determine the balance point if X and Y are interchanged. (c) What happens if the galvanometer and cell are interchanged at the balance point? Would the galvanometer show any current?

Answer: (a) Balance condition $\dfrac{X}{Y} = \dfrac{l}{100 – l}$:

$$X = Y \cdot \dfrac{l}{100 – l} = 12.5 \times \dfrac{39.5}{60.5} = 8.16 \, \Omega.$$

Thick copper strips reduce contact resistance which is otherwise neglected in the bridge equation. (b) On interchanging X and Y, balance point becomes $100 – 39.5 = 60.5$ cm from A. (c) Interchanging the galvanometer and cell does not disturb the balance condition (the bridge is symmetric); the galvanometer still reads zero.

Q11. A storage battery of EMF 8.0 V and internal resistance 0.5 $\Omega$ is being charged by a 120 V DC supply using a series resistor of 15.5 $\Omega$. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Answer: Net EMF driving current = $120 – 8.0 = 112$ V. Total resistance = $15.5 + 0.5 = 16$ $\Omega$. Charging current $I = 112/16 = 7$ A. Terminal voltage $V = \epsilon + Ir = 8.0 + 7 \times 0.5 = 11.5$ V. The series resistor limits the charging current to a safe value, preventing damage to the battery.

Q12. In a potentiometer arrangement, a cell of EMF 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the EMF of the second cell?

Answer: Since $\epsilon \propto l$:

$$\dfrac{\epsilon_2}{\epsilon_1} = \dfrac{l_2}{l_1} \implies \epsilon_2 = 1.25 \times \dfrac{63.0}{35.0} = 2.25 \, \text{V}.$$

Q13. The number density of free electrons in a copper conductor is $8.5 \times 10^{28}$ m$^{-3}$. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is $2.0 \times 10^{-6}$ m$^2$ and it is carrying a current of 3.0 A.

Answer: Drift speed $v_d = \dfrac{I}{neA}$:

$$v_d = \dfrac{3.0}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.0 \times 10^{-6}} \approx 1.1 \times 10^{-4} \, \text{m s}^{-1}.$$

Time to traverse 3.0 m: $t = L/v_d = 3.0/1.1 \times 10^{-4} \approx 2.7 \times 10^4$ s $\approx 7.5$ hours. Note: signals propagate much faster (near $c$) because the field sets up almost instantaneously along the entire wire.

Additional Important Questions

Q14. Define current density. Write its SI unit.

Answer: Current density $\vec{J}$ is the current per unit area of cross-section taken normal to the direction of flow: $J = I/A$. It is a vector quantity in the direction of conventional current. SI unit: A m$^{-2}$.

Q15. Derive the expression for drift velocity of free electrons in a conductor.

Answer: Under field $\vec{E}$ each electron experiences acceleration $\vec{a} = -e\vec{E}/m$. Between two successive collisions (mean free time $\tau$) the electron gains an additional velocity. Averaging over all electrons, the random thermal velocities cancel and only the drift component survives:

$$\vec{v_d} = -\dfrac{e\vec{E}}{m}\tau.$$

The negative sign shows electrons drift opposite to $\vec{E}$.

Q16. Deduce Ohm’s law from the concept of drift velocity.

Answer: Current $I = neAv_d = \dfrac{ne^2A\tau}{m}E$. Putting $E = V/L$:

$$I = \dfrac{ne^2A\tau}{mL}V \implies V = \left(\dfrac{mL}{ne^2A\tau}\right)I = IR,$$

where $R = mL/(ne^2A\tau) = \rho L/A$ with $\rho = m/(ne^2\tau)$. This is Ohm’s law.

Q17. State Kirchhoff’s rules for electrical networks. Give the underlying physical principles.

Answer: Junction rule (KCL): The algebraic sum of currents at a junction is zero, $\sum I = 0$. This expresses conservation of electric charge. Loop rule (KVL): The algebraic sum of changes in potential around any closed loop is zero, $\sum \epsilon = \sum IR$. This expresses conservation of energy.

Q18. Derive the balance condition of a Wheatstone bridge.

Answer: Let $I_1$ flow through P and Q, and $I_2$ through R and S, with no current in the galvanometer at balance. Then $V_B = V_D$, giving $I_1P = I_2R$ and $I_1Q = I_2S$. Dividing:

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

Q19. Why is a potentiometer preferred over a voltmeter for measuring EMF of a cell?

Answer: A voltmeter draws a small but finite current from the cell, so it measures only the terminal voltage $V = \epsilon – Ir$, less than $\epsilon$. A potentiometer at balance draws zero current from the test cell, hence measures the true EMF.

Q20. Two cells of EMFs $\epsilon_1$ and $\epsilon_2$ and internal resistances $r_1$ and $r_2$ are connected in parallel. Find the equivalent EMF and equivalent internal resistance.

Answer:

$$\epsilon_{\text{eq}} = \dfrac{\epsilon_1 r_2 + \epsilon_2 r_1}{r_1 + r_2}, \qquad r_{\text{eq}} = \dfrac{r_1 r_2}{r_1 + r_2}.$$

Q21. A wire of resistance R is stretched to twice its original length. Find its new resistance.

Answer: Volume is conserved: $A_1 L_1 = A_2 L_2$. With $L_2 = 2L_1$, $A_2 = A_1/2$. New resistance:

$$R’ = \rho \dfrac{L_2}{A_2} = \rho \dfrac{2L_1}{A_1/2} = 4\rho \dfrac{L_1}{A_1} = 4R.$$

Q22. A bulb is rated 100 W, 220 V. Calculate its resistance and the current drawn at rated voltage.

Answer: $R = V^2/P = (220)^2/100 = 484 \, \Omega$. Current $I = P/V = 100/220 = 0.455$ A.

Q23. Why does the resistance of a metallic conductor increase with temperature, while that of a semiconductor decreases?

Answer: In a metal $n$ is essentially fixed but the relaxation time $\tau$ decreases as ions vibrate more vigorously, so $\rho = m/(ne^2\tau)$ rises. In a semiconductor, the increase in $n$ with temperature dominates over the small fall in $\tau$, so $\rho$ decreases.

Q24. Define the temperature coefficient of resistance. State its sign for metals, alloys, and semiconductors.

Answer: $\alpha = \dfrac{1}{R_0}\dfrac{\Delta R}{\Delta T}$. For pure metals $\alpha > 0$ and large; for alloys (manganin, constantan) $\alpha$ is very small and positive; for semiconductors and insulators $\alpha < 0$.

Q25. Two wires of equal length, one of copper and one of manganin, have the same resistance. Which wire is thicker?

Answer: Manganin has higher resistivity than copper, so for the same R and L the manganin wire must have a larger cross-sectional area, i.e. it is thicker.

Q26. State the principle of a potentiometer. Why must the driver-cell EMF exceed the EMF being measured?

Answer: When a steady current flows through a uniform wire, the potential drop across any segment is proportional to the segment’s length: $V = kl$. To obtain a balance point on the wire, the driver-cell EMF must exceed the EMF being measured; otherwise the galvanometer can never be brought to zero.

Q27. The current-voltage graph of a device is non-linear. Is Ohm’s law valid for it?

Answer: No, Ohm’s law $V = IR$ requires $R$ to be a constant, i.e. a linear $V$–$I$ graph passing through the origin. Devices like diodes, electrolytes and gas-discharge tubes are non-ohmic.

Q28. The potential difference across a battery becomes equal to its EMF when (a) battery is being charged, (b) no current is drawn, (c) very large current is drawn, (d) battery is short-circuited. Choose the correct option.

Answer: (b) When $I = 0$, $V = \epsilon – Ir = \epsilon$.

Q29. A conductor of length L is connected to a DC source. If the length is doubled keeping the potential difference constant, how does the drift velocity change?

Answer: $v_d = eE\tau/m = eV\tau/(mL)$. Doubling $L$ halves $E$ and halves $v_d$.

Q30. Two resistors $R_1$ and $R_2$ are connected (i) in series and (ii) in parallel across the same battery. In which combination is the power dissipation greater?

Answer: Equivalent resistance is smaller in parallel ($R_p < R_s$). Since $P = V^2/R$ for fixed battery voltage, parallel combination dissipates more power.

Q31. Establish the relationship between current density $J$, conductivity $\sigma$ and electric field $E$.

Answer: Free electrons drift with velocity $v_d = eE\tau/m$. The current density is

$$J = nev_d = \dfrac{ne^2\tau}{m}E = \sigma E,$$

where $\sigma = ne^2\tau/m$ is the conductivity. In vector form $\vec{J} = \sigma\vec{E}$, which is the microscopic form of Ohm’s law and is valid even where the macroscopic $V = IR$ formulation is awkward.

Q32. Define the relaxation time of free electrons in a conductor. How does it depend on temperature?

Answer: Relaxation time $\tau$ is the average time interval between two successive collisions of free electrons with positive ions of the lattice. As temperature rises, lattice ions vibrate more vigorously, scattering becomes more frequent, and $\tau$ decreases. This is the principal reason metallic resistivity rises with temperature.

Q33. A 5 V cell sends a current of 0.5 A through a 6 $\Omega$ resistor connected across its terminals. What is the internal resistance of the cell?

Answer: Using $\epsilon = I(R+r)$:

$$r = \dfrac{\epsilon – IR}{I} = \dfrac{5 – 0.5 \times 6}{0.5} = \dfrac{2}{0.5} = 4 \, \Omega.$$

Q34. Why is the terminal voltage of a cell during discharge less than its EMF, but during charging more than its EMF?

Answer: During discharge, current flows out of the cell against its internal resistance, dropping a voltage $Ir$ inside the cell, so $V = \epsilon – Ir < \epsilon$. During charging, current is forced into the cell from outside; the external supply must overcome both $\epsilon$ and $Ir$, so the terminal voltage $V = \epsilon + Ir > \epsilon$.

Q35. Three identical cells of EMF 2 V each and internal resistance 0.5 $\Omega$ each are connected in series with an external resistor of 7.5 $\Omega$. Find the current.

Answer: Total EMF $= 3 \times 2 = 6$ V. Total internal resistance $= 3 \times 0.5 = 1.5$ $\Omega$. Total resistance $= 7.5 + 1.5 = 9$ $\Omega$. Current $I = 6/9 \approx 0.667$ A.

Q36. Three identical cells of EMF 2 V and internal resistance 0.5 $\Omega$ each are connected in parallel with an external resistance of $11/6$ $\Omega$. Find the current.

Answer: For parallel combination, equivalent EMF $= 2$ V, equivalent internal resistance $= 0.5/3 = 1/6$ $\Omega$. Total resistance $= 11/6 + 1/6 = 12/6 = 2$ $\Omega$. Current $I = 2/2 = 1$ A.

Q37. State the conditions for maximum power transfer from a cell to an external resistance.

Answer: Power dissipated externally is

$$P = I^2R = \dfrac{\epsilon^2 R}{(R+r)^2}.$$

Setting $dP/dR = 0$ gives $R = r$. So maximum power $P_{\max} = \epsilon^2/(4r)$ is delivered when the external resistance equals the internal resistance.

Q38. Why are alloys like manganin and constantan used to make standard resistance coils?

Answer: Manganin and constantan possess (i) high resistivity, allowing compact coils of large resistance, and (ii) very low temperature coefficient of resistance, so the coil’s resistance is essentially independent of temperature changes during use. These two properties make them ideal for precision standard resistors and meter-bridge wires.

Q39. The sensitivity of a Wheatstone bridge depends on which factors? How can it be improved?

Answer: Sensitivity is greatest when (i) the four arms have comparable resistances, (ii) the galvanometer is sensitive (large current sensitivity), and (iii) the cell EMF is sufficient to produce a measurable galvanometer deflection per unit unbalance. Sensitivity is poor if any arm is very small or very large compared to the others.

Q40. In a meter bridge experiment, why are the connecting copper strips made thick and short?

Answer: Thick, short copper strips have negligible resistance. The standard balance equation $X/Y = l/(100-l)$ assumes zero strip resistance. Any extra strip resistance would shift the balance point and introduce a systematic error.

Q41. Define potential gradient of a potentiometer wire. How is it controlled?

Answer: Potential gradient is the potential drop per unit length of the potentiometer wire, $k = V/L$ (V m$^{-1}$). It is controlled by adjusting the rheostat in the driver circuit so that the current through the wire — and hence the gradient — is steady and small enough to give a sharp balance point for the cell being tested.

Q42. Compare the sensitivity of a potentiometer with a voltmeter.

Answer: A potentiometer can detect potential differences of the order of $10^{-6}$ V because at balance no current flows through the cell under test, so its full EMF appears at the test point. A voltmeter has finite (large but not infinite) input resistance and therefore draws current; its lower limit of measurement is set by the resolution of its scale and by the unavoidable IR drop. The potentiometer is much more sensitive and gives true EMF, while the voltmeter only gives terminal voltage.

Q43. Two cells of EMFs 1.5 V and 2.0 V (internal resistances 1 $\Omega$ and 2 $\Omega$ respectively) are connected in parallel and the combination drives a 4 $\Omega$ resistor. Find the current through the 4 $\Omega$ resistor.

Answer: Equivalent EMF

$$\epsilon_{\text{eq}} = \dfrac{\epsilon_1 r_2 + \epsilon_2 r_1}{r_1 + r_2} = \dfrac{1.5 \times 2 + 2.0 \times 1}{1 + 2} = \dfrac{5}{3} \, \text{V}.$$

Equivalent internal resistance $r_{\text{eq}} = (1 \times 2)/(1+2) = 2/3$ $\Omega$. Current $I = \epsilon_{\text{eq}}/(R + r_{\text{eq}}) = (5/3)/(4 + 2/3) = (5/3)/(14/3) = 5/14 \approx 0.357$ A.

Q44. A 100 W and a 60 W bulb, each rated 220 V, are connected (a) in series and (b) in parallel across a 220 V supply. Which bulb glows brighter in each case?

Answer: $R = V^2/P$. So $R_{100} = 484 \, \Omega$ and $R_{60} = 806.7 \, \Omega$.

(a) Series: Same current flows; power $P = I^2R$ is greater across the larger resistance, so the 60 W bulb glows brighter (and the 100 W bulb glows dimmer than its rating). (b) Parallel: Same voltage; $P = V^2/R$ is greater for smaller R, so the 100 W bulb glows brighter, as expected from its rating.

Q45. Why is the SI unit of resistivity $\Omega$ m and not $\Omega/\text{m}$?

Answer: From $R = \rho L/A$, $\rho = RA/L$. Substituting units, $[\rho] = \Omega \cdot \text{m}^2 / \text{m} = \Omega$ m. Resistivity is an intensive property of the material and is independent of geometry.

Q46. A copper wire of resistance 0.5 $\Omega$ at 0 °C has a temperature coefficient of resistance $\alpha = 4.3 \times 10^{-3}$ °C$^{-1}$. Find its resistance at 100 °C.

Answer:

$$R_{100} = R_0(1 + \alpha\,\Delta T) = 0.5 \times (1 + 4.3 \times 10^{-3} \times 100) = 0.5 \times 1.43 = 0.715 \, \Omega.$$

Q47. State and explain Joule’s law of heating.

Answer: When a current $I$ flows through a resistor of resistance $R$ for time $t$, the heat developed is

$$H = I^2 R t \, \text{(joule)}.$$

Equivalently $H = VIt = V^2 t/R$. The heat varies (i) directly as the square of the current, (ii) directly as the resistance, and (iii) directly as the time. The energy comes from the work done by the source against the resistance offered by the conductor.

Q48. An electric heater of resistance 22 $\Omega$ operates on 220 V mains. Calculate the heat developed in 5 minutes.

Answer: Power $P = V^2/R = (220)^2/22 = 2200$ W. Heat $H = P t = 2200 \times 300 = 6.6 \times 10^5$ J = 660 kJ.

Q49. The current in a wire varies with time as $I = 4 + 2t$ (A). Find the charge that flows through the wire from $t = 0$ to $t = 5$ s.

Answer:

$$q = \int_0^5 I\,dt = \int_0^5 (4 + 2t)\,dt = [4t + t^2]_0^5 = 20 + 25 = 45 \, \text{C}.$$

Q50. Show that for a steady current the current density is constant in a cylindrical conductor.

Answer: Consider a section of length $L$ and uniform area $A$. The current $I$ flowing in is the same as that flowing out (charge conservation). Hence $J = I/A$ is the same at every cross-section along the conductor — current density is uniform for steady current in a uniform conductor.

Multiple Choice Questions

MCQ 1. The drift velocity of free electrons in a copper conductor is of the order of

(a) $10^{-2}$ m s$^{-1}$  (b) $10^{-4}$ m s$^{-1}$  (c) $10^{0}$ m s$^{-1}$  (d) $10^{8}$ m s$^{-1}$

Answer: (b) $10^{-4}$ m s$^{-1}$.

MCQ 2. The resistance of a wire of length $L$ and cross-section $A$ is $R$. If both length and area are doubled, new resistance is

(a) $R$  (b) $2R$  (c) $R/2$  (d) $4R$

Answer: (a) $R$. New $R’ = \rho(2L)/(2A) = \rho L/A = R$.

MCQ 3. Kirchhoff’s loop rule is a consequence of

(a) conservation of charge  (b) conservation of energy  (c) Ohm’s law  (d) Coulomb’s law

Answer: (b) conservation of energy.

MCQ 4. The Wheatstone bridge is most sensitive when

(a) all four resistances are equal  (b) any two arms are very small  (c) galvanometer has high resistance  (d) cell EMF is negligible

Answer: (a) all four resistances are equal.

MCQ 5. The unit of conductivity is

(a) $\Omega$ m  (b) $\Omega$ m$^{-1}$  (c) S m  (d) S m$^{-1}$

Answer: (d) S m$^{-1}$.

MCQ 6. The temperature coefficient of resistance of an ideal superconductor is

(a) zero  (b) infinite  (c) positive  (d) negative

Answer: Below the transition temperature the resistance is exactly zero, so the concept of $\alpha$ is not meaningful; for temperatures just above $T_c$, semiconductor-like behaviour gives a finite (typically positive) value. The conventional textbook answer is (a) zero (resistance itself is zero below $T_c$).

MCQ 7. A galvanometer is converted into an ammeter by connecting

(a) a high resistance in series  (b) a low resistance in parallel (shunt)  (c) a low resistance in series  (d) a high resistance in parallel

Answer: (b) a low resistance in parallel — the shunt diverts most of the current.

MCQ 8. If a wire of resistance $R$ is bent into a circular loop and resistance is measured between two diametrically opposite points, the value is

(a) $R$  (b) $R/2$  (c) $R/4$  (d) $2R$

Answer: (c) $R/4$. The two halves (each of resistance $R/2$) are in parallel, giving $R/4$.

Worked Numerical Problems for Practice

NP1. A current of 0.6 A flows through a 5 $\Omega$ resistor. Find (i) the potential difference across it and (ii) the heat dissipated in 2 minutes.

Solution: (i) $V = IR = 0.6 \times 5 = 3$ V. (ii) Power $P = I^2R = 0.36 \times 5 = 1.8$ W. Heat $H = Pt = 1.8 \times 120 = 216$ J.

NP2. Two resistors of 4 $\Omega$ and 6 $\Omega$ are connected in parallel. The combination is connected to a 12 V battery of internal resistance 0.6 $\Omega$. Find the current drawn from the battery and the power dissipated in the external resistors.

Solution: $R_p = (4 \times 6)/(4+6) = 2.4 \, \Omega$. Total $R_{\text{tot}} = 2.4 + 0.6 = 3.0 \, \Omega$. Current $I = 12/3 = 4$ A. Power dissipated externally $P_{\text{ext}} = I^2 R_p = 16 \times 2.4 = 38.4$ W.

NP3. A wire of length 2 m and uniform area $0.5 \times 10^{-6}$ m$^2$ has resistance 4 $\Omega$. Find its resistivity. If 0.5 A flows through it, find the drift velocity given $n = 8 \times 10^{28}$ m$^{-3}$.

Solution: $\rho = RA/L = (4 \times 0.5 \times 10^{-6})/2 = 1.0 \times 10^{-6}$ $\Omega$ m. Drift velocity $v_d = I/(neA) = 0.5/(8 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.5 \times 10^{-6}) \approx 7.8 \times 10^{-5}$ m s$^{-1}$.

NP4. A galvanometer of resistance 50 $\Omega$ gives a full-scale deflection for 2 mA. How can it be converted into (i) an ammeter reading 2 A and (ii) a voltmeter reading 10 V?

Solution: (i) Shunt resistance: only 2 mA should flow through the galvanometer when 2 A enters. Voltage across galvanometer = $2 \times 10^{-3} \times 50 = 0.1$ V. Same voltage across shunt $S$ carries $2 – 0.002 = 1.998$ A. Hence

$$S = \dfrac{0.1}{1.998} \approx 0.05 \, \Omega.$$

(ii) Series resistor: full-scale current $2 \times 10^{-3}$ A through total resistance $(50 + R) = V/I = 10/0.002 = 5000$ $\Omega$. Therefore $R = 4950$ $\Omega$ in series with the galvanometer.

NP5. A potentiometer wire of length 4 m has resistance 8 $\Omega$. A driver cell of EMF 4 V and internal resistance 2 $\Omega$ is connected. What is the potential gradient along the wire?

Solution: Current $I_0 = 4/(8+2) = 0.4$ A. Voltage drop along the wire $V_w = 0.4 \times 8 = 3.2$ V. Potential gradient $k = V_w/L = 3.2/4 = 0.8$ V m$^{-1}$.

NP6. A Wheatstone bridge has arms 10 $\Omega$, 20 $\Omega$, 30 $\Omega$ and an unknown $X$. The bridge is balanced. Find $X$.

Solution: $P/Q = R/S \implies 10/20 = 30/X \implies X = 60 \, \Omega$.

NP7. The resistance of a metallic wire at 20 °C is 25 $\Omega$ and at 100 °C is 30 $\Omega$. Find the temperature coefficient of resistance.

Solution:

$$\alpha = \dfrac{R_2 – R_1}{R_1(T_2 – T_1)} = \dfrac{30 – 25}{25 \times 80} = \dfrac{5}{2000} = 2.5 \times 10^{-3} \, °C^{-1}.$$

NP8. Two cells of EMF 6 V each (internal resistance 1 $\Omega$ and 2 $\Omega$) are connected in series with an external resistance of 9 $\Omega$. Find the current.

Solution: Total EMF = 12 V; total resistance = $9 + 1 + 2 = 12$ $\Omega$. Current $I = 12/12 = 1$ A.

NP9. A copper wire of mass 90 g and resistivity $1.7 \times 10^{-8}$ $\Omega$ m is drawn into a wire of length 9 m. Density of copper is $9 \times 10^3$ kg m$^{-3}$. Find the resistance.

Solution: Volume $V = m/\rho_{\text{mass}} = 90 \times 10^{-3}/9000 = 10^{-5}$ m$^3$. Area $A = V/L = 10^{-5}/9 \approx 1.11 \times 10^{-6}$ m$^2$. Resistance $R = \rho L/A = 1.7 \times 10^{-8} \times 9 / (1.11 \times 10^{-6}) \approx 0.138 \, \Omega$.

NP10. A 60 W bulb operating on 240 V is left on for 5 hours each day for 30 days. Find the energy consumption in kWh and cost at Rs 6 per unit.

Solution: Energy = $0.060 \times 5 \times 30 = 9$ kWh. Cost = $9 \times 6 =$ Rs 54.

Glossary

Term	Meaning
Electric current	Rate of flow of charge through a cross-section, $I = dq/dt$.
Conventional current	Direction in which positive charges would flow; opposite to electron drift.
Drift velocity	Average velocity of free electrons under an applied field.
Mobility	Drift velocity per unit electric field, $\mu = v_d/E$.
Current density	Current per unit cross-sectional area, vector along $\vec{E}$.
Resistance	Opposition to current flow, $R = V/I$.
Resistivity	Property of material, $\rho = RA/L$; unit $\Omega$ m.
Conductivity	$\sigma = 1/\rho$; unit S m$^{-1}$.
Ohm’s law	$V \propto I$ at constant temperature, with constant $R$.
Temperature coefficient	$\alpha = (1/R_0)(\Delta R/\Delta T)$.
EMF ($\epsilon$)	Energy supplied per unit charge by a source; equals open-circuit voltage.
Internal resistance	Resistance of a cell to its own current; $V = \epsilon – Ir$.
Kirchhoff’s junction rule	$\sum I = 0$ at any node — charge conservation.
Kirchhoff’s loop rule	$\sum \epsilon = \sum IR$ around any loop — energy conservation.
Wheatstone bridge	Four-arm network; balance condition $P/Q = R/S$.
Meter bridge	Practical Wheatstone bridge using a 1 m uniform wire.
Potentiometer	Device that compares EMFs without drawing current at balance.
Electric power	$P = VI = I^2R = V^2/R$.

Frequently Asked Conceptual Questions

FAQ 1. Why does an electric bulb glow brightly the moment it is switched on, although the average drift speed of free electrons is very small?

The electric field $\vec{E}$ inside the conductor is established at a speed close to the speed of light as soon as the circuit is closed. All free electrons throughout the entire length of the wire start drifting almost simultaneously, so a current flows through the filament essentially instantaneously. The drift speed is small but the current is set up promptly across the whole circuit.

FAQ 2. Why does the resistance of a tungsten lamp increase when it glows, although the lamp is rated for steady power at the operating voltage?

Tungsten has a high positive temperature coefficient of resistance. At room temperature its resistance is much lower than at the operating temperature ($\sim 2500$ K). When switched on, the cold filament initially draws a current several times the steady-state current; this surge heats the filament, the resistance rises, and the current drops to its rated value. This is why lamps often fail at the moment of switching on rather than during steady operation.

FAQ 3. The electric field inside a current-carrying conductor is non-zero, while inside an electrostatic (charged) conductor it is zero. Explain.

An electrostatic conductor is in equilibrium: charges arrange themselves on the surface to cancel any internal field. A current-carrying conductor is not in electrostatic equilibrium — there is a continuous flow of charge driven by an external EMF. The external source maintains a small but non-zero internal field $\vec{E} = V/L$, which produces and sustains the drift of free electrons.

FAQ 4. Why is the EMF of a cell measured in volts, the same unit as potential difference?

EMF is the work done per unit charge by the source as it transfers a charge from the lower-potential terminal to the higher-potential terminal. Both EMF and potential difference are forms of energy per unit charge: $1$ V $= 1$ J C$^{-1}$. They differ in physical origin: EMF is generated by chemical (or other non-electrostatic) processes, whereas potential difference exists between any two points across which work is done by the electric field.

FAQ 5. A wire is stretched so that its length is increased by 0.1%. Find the percentage change in its resistance.

Volume of a stretched wire is conserved. With $V = AL$ constant, $\Delta A/A = -\Delta L/L$. From $R = \rho L/A$, $\Delta R/R = \Delta L/L – \Delta A/A = 2\Delta L/L = 2 \times 0.1\% = 0.2\%$. The resistance increases by 0.2%.

FAQ 6. Why does a thicker copper wire have lower resistance even though both wires are made of the same material?

Since $R = \rho L/A$, doubling the cross-section halves the resistance for the same length. A thicker wire offers a wider channel for free electrons, increasing the drift current at the same applied voltage.

FAQ 7. The current through a wire varies linearly with potential difference up to a certain point, then deviates. Why?

At low currents the wire stays near room temperature and obeys Ohm’s law (linear $V$–$I$). At higher currents Joule heating raises the temperature of the wire, increasing $\rho$ and hence $R$. The graph deviates from a straight line through the origin and curves towards the voltage axis.

FAQ 8. Why do we use a Daniell cell or a Leclanché cell rather than an ordinary battery for potentiometer experiments in school laboratories?

Daniell and Leclanché cells maintain a steady EMF for long durations. The potentiometer requires a constant current (and hence a steady potential gradient) along the wire while the experiment progresses. Ordinary dry cells can polarise and lose EMF during long use, shifting the balance point.

FAQ 9. State the differences between EMF and terminal voltage in tabular form.

EMF ($\epsilon$)	Terminal Voltage ($V$)
Energy supplied per unit charge by the source.	Potential difference between the terminals when current flows.
Equal to the open-circuit voltage.	Always less than EMF when discharging; greater than EMF when charging.
Property of the cell only — independent of external circuit.	Depends on the current drawn and the internal resistance.
Cause of current.	Effect of current flow.

FAQ 10. Why does a voltmeter have very high resistance, while an ammeter has very low resistance?

A voltmeter is connected in parallel across the device whose voltage is to be measured. Its high resistance ensures negligible current is drawn from the circuit, so the voltage being measured is not disturbed. An ammeter is connected in series with the circuit; its low resistance ensures it does not significantly add to the total resistance and alter the current. An ideal voltmeter has infinite resistance and an ideal ammeter has zero resistance.

Higher-Order Thinking Skills (HOTS)

HOTS 1. A potentiometer wire is 10 m long. The driver cell is 2 V with internal resistance 1 $\Omega$. The wire’s resistance is 9 $\Omega$. What standard cell EMF can be measured by this potentiometer? Discuss.

Solution: Current in the wire $I = 2/(9+1) = 0.2$ A. Total potential drop across the wire $V_w = 0.2 \times 9 = 1.8$ V. Therefore the potentiometer can measure any EMF less than 1.8 V — for instance a Daniell cell (1.08 V) or a Leclanché cell (1.45 V) but not a 2 V lead-acid cell directly. To measure higher EMFs, the test cell can be combined with a known voltage divider, or the gradient on the wire can be reduced by adding a series resistance to the driver loop.

HOTS 2. A network of three resistors, each $R$, is connected across a battery of EMF $\epsilon$ and negligible internal resistance, forming a triangle ABC. Find the resistance between A and B, and the current drawn from the battery.

Solution: Between A and B there are two parallel paths: a direct $R$ (one side AB) and a series of $R + R = 2R$ via C. Equivalent

$$R_{AB} = \dfrac{R \cdot 2R}{R + 2R} = \dfrac{2R}{3}.$$

Current drawn $I = \epsilon/R_{AB} = 3\epsilon/(2R)$.

HOTS 3. A piece of copper and a piece of germanium are cooled from room temperature to 80 K. Predict what happens to the resistivity of each.

Solution: Copper is a metal with $\alpha > 0$. As temperature drops, $\rho$ decreases — copper conducts even better. Germanium is a semiconductor with $\alpha < 0$. As temperature drops, the number of charge carriers $n$ falls dramatically (fewer electrons get thermally excited across the band gap), so $\rho$ increases — germanium conducts much worse. At low enough temperatures, an intrinsic semiconductor effectively becomes an insulator.

HOTS 4. Justify the statement: “In a Wheatstone bridge, no current flows through the galvanometer at balance, but currents do flow through all four arms.”

Solution: At balance, the points B and D are at the same potential, so no potential difference drives current through the galvanometer. However, current still flows through each arm because the cell at A–C maintains a potential difference. Currents $I_1$ flow through P and Q; currents $I_2$ flow through R and S. The condition is $I_1 P = I_2 R$ and $I_1 Q = I_2 S$, which yields $P/Q = R/S$.

HOTS 5. A 12 V battery of internal resistance 0.5 $\Omega$ is being charged at 100 V. Calculate the charging current. What is the maximum permissible series resistance if the safe charging current is 5 A?

Solution: Net driving voltage during charging $= 100 – 12 = 88$ V. If a series resistor $R$ is used,

$$I = \dfrac{88}{R + 0.5} \le 5 \implies R + 0.5 \ge 17.6 \implies R \ge 17.1 \, \Omega.$$

Hence a series resistance of at least 17.1 $\Omega$ is required to keep the charging current within the safe 5 A limit.

Quick Revision Notes

• Electric current $I = dq/dt$; SI unit ampere (A) — one of the seven base units.
• Conventional current is opposite to electron drift.
• Drift velocity $v_d = eE\tau/m$ is small ($\sim 10^{-4}$ m/s) but uniform throughout the conductor.
• Mobility $\mu = v_d/E = e\tau/m$; unit m$^2$ V$^{-1}$ s$^{-1}$.
• Current density $\vec{J} = ne\vec{v_d} = \sigma\vec{E}$.
• Conductivity $\sigma = ne^2\tau/m$; reciprocal of resistivity $\rho$.
• Ohm’s law: $V = IR$ for ohmic conductors at constant temperature.
• $R = \rho L/A$; resistivity is a material property, resistance depends on geometry too.
• Temperature dependence: $\rho_T = \rho_0[1 + \alpha(T – T_0)]$. Metals: $\alpha > 0$. Semiconductors: $\alpha < 0$. Alloys (manganin): $\alpha \approx 0$.
• Series: $R_s = \sum R_i$; Parallel: $1/R_p = \sum 1/R_i$.
• EMF: $\epsilon = V + Ir$ during discharge; $V = \epsilon + Ir$ during charging.
• Cells: series $\implies$ EMFs add; parallel $\implies$ internal resistances combine like resistors in parallel.
• Kirchhoff’s laws: KCL at nodes ($\sum I = 0$), KVL around loops ($\sum \epsilon = \sum IR$).
• Wheatstone balance: $P/Q = R/S$, independent of cell EMF and galvanometer.
• Meter bridge: $X = Y \cdot l/(100-l)$ at balance.
• Potentiometer: $\epsilon \propto l$ at balance, drawing zero current from the test cell.
• Power: $P = VI = I^2R = V^2/R$. Energy: $W = Pt$. Commercial unit kWh = $3.6 \times 10^6$ J.
• For maximum power transfer to external $R$: choose $R = r$.

This completes the ASSEB Class 12 Physics Chapter 3 question-answer guide. Master the derivations (drift velocity, Ohm’s law, Wheatstone balance), keep the key formulas at your fingertips, and practise the numerical exercises until each step feels routine. Continue your revision with the next chapter on HSLC GURU.