Chapter 3: Current Electricity Long Questions | physics Class 12 CBSE

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Chapter 3: Current Electricity Long Questions | physics Class 12 CBSE | ByteNotes.in

Long Answer

Hard difficulty • Structured explanation

Hard

Question 1

Long Form

Derive an expression for the drift velocity of electrons in a metallic conductor and hence establish the relation j = σE.

When an electric field E is applied, electrons are accelerated with acceleration a = –eE/m. The i-th electron, after its last collision at time ti ago, has velocity Vi = vi + (–eE/m)ti, where vi is its post-collision velocity.
Averaging over all N electrons: vd = average of Vi = average of vi – (eE/m) × average of ti. Since post-collision velocities are random, their average is zero. The average of ti is the relaxation time τ.
Therefore vd = –eEτ/m. Current density j = nevd (in magnitude), giving j = (ne²τ/m)E.
Comparing with j = σE, conductivity σ = ne²τ/m, and resistivity ρ = 1/σ = m/(ne²τ).
This derivation shows Ohm's law emerges naturally from the electron drift model, with conductivity depending on electron density, charge, mass, and relaxation time — all microscopic quantities.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 2

Long Form

Derive the balance condition for a Wheatstone bridge using Kirchhoff's rules and explain how it is used to measure an unknown resistance.

In a Wheatstone bridge with resistors R₁, R₂, R₃, R₄, a battery across A-C, and galvanometer across B-D, at balance Ig = 0. Kirchhoff's junction rule at D gives I₁ = I₃, and at B gives I₂ = I₄.
Applying Kirchhoff's loop rule to loop ADBA: –I₁R₁ + I₂R₂ = 0, giving I₁/I₂ = R₂/R₁.
For loop CBDC: I₂R₄ – I₁R₃ = 0 (using I₃ = I₁, I₄ = I₂), giving I₁/I₂ = R₄/R₃.
Equating both expressions: R₂/R₁ = R₄/R₃, or the balance condition R₁/R₂ = R₃/R₄.
To find unknown R₄: place it in the fourth arm, fix R₁ and R₂, vary R₃ until galvanometer reads zero, then R₄ = R₃(R₂/R₁). This null method is highly accurate as it does not depend on the galvanometer calibration.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 3

Long Form

Compare the resistivity of conductors, semiconductors, and insulators. Explain why semiconductor resistivity decreases with temperature while metal resistivity increases.

Metals have very low resistivities (10⁻⁸ to 10⁻⁶ Ωm); semiconductors occupy an intermediate range; insulators like rubber and ceramic have resistivities up to 10¹⁸ times greater than metals.
Resistivity is given by ρ = m/(ne²τ). In metals, electron density n is essentially constant with temperature, but rising temperature increases collision frequency, decreasing τ and hence increasing ρ.
In semiconductors, both n and τ change with temperature. As temperature rises, many more electrons are promoted to the conduction band — n increases significantly.
This large increase in n more than compensates the decrease in τ, resulting in a net decrease in ρ for semiconductors with increasing temperature.
This distinction is exploited in electronics: semiconductor resistivity can also be engineered by adding impurities (doping), enabling the design of transistors, diodes, and other devices.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 4

Long Form

Derive expressions for the equivalent EMF and equivalent internal resistance of n cells connected (i) in series and (ii) in parallel.

Series: For two cells in series, VAC = VAB + VBC = (ε₁ – Ir₁) + (ε₂ – Ir₂) = (ε₁ + ε₂) – I(r₁ + r₂). Comparing with VAC = εeq – Ireq: εeq = ε₁ + ε₂ and req = r₁ + r₂. For n cells: εeq = Σεi and req = Σri.
Parallel: For two cells in parallel, I = I₁ + I₂ = (ε₁ – V)/r₁ + (ε₂ – V)/r₂. Combining: V = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) – I(r₁r₂)/(r₁ + r₂). So εeq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) and req = r₁r₂/(r₁ + r₂).
In general for n parallel cells: 1/req = Σ(1/ri) and εeq/req = Σ(εi/ri).
Series combination is preferred when higher terminal voltage is needed; parallel combination is preferred when lower internal resistance (higher current capacity) is required.
If any cell in a series string is connected in reverse, its EMF enters with a negative sign: εeq = ε₁ – ε₂ (when ε₁ > ε₂), though internal resistances still add.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 5

Long Form

Explain the origin of resistance in a metallic conductor using the electron drift model. How does this model account for the temperature dependence of resistance in metals?

In a metal, conduction electrons undergo random thermal motion and collide with fixed positive ions. When an electric field is applied, electrons gain a drift velocity vd = eEτ/m in the direction opposite to E.
During drift, electrons frequently collide with ions (average time between collisions is τ), losing their gained velocity and re-randomising their direction. This opposition to directed motion constitutes electrical resistance.
From the drift model, ρ = m/(ne²τ). Resistance R = ρl/A = ml/(ne²τA). This is the microscopic origin of resistance.
As temperature increases, the thermal speeds of electrons increase, so they collide more frequently with ions. This reduces τ (shorter average time between collisions), increasing ρ and hence R.
Thus the model explains both the existence of resistance and its positive temperature coefficient for metals, while assuming n remains essentially constant with temperature in metals.

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Long Answer

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Medium

Question 6

Long Form

Analyse the power dissipated in a resistive circuit and apply it to explain why electrical power is transmitted at high voltage. Derive the relevant expressions.

When current I flows through a conductor with potential difference V, the rate of energy dissipation is P = IV. Using Ohm's law: P = I²R = V²/R (Joule's law).
For power transmission: a device requiring power P is connected via cables of resistance Rc. If V is the supply voltage, current I = P/V. Power wasted in cables: Pc = I²Rc = (P/V)²Rc = P²Rc/V².
This shows Pc ∝ 1/V²: doubling the transmission voltage reduces cable losses by a factor of four.
Since cables are hundreds of kilometres long and Rc is significant, transmitting at very high V (e.g., hundreds of kV) keeps Pc negligibly small compared to P.
Transformers step up voltage at the power station and step it down to safe levels (230 V) at homes. This makes high-voltage transmission both efficient and safe for consumers.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 7

Long Form

State Kirchhoff's rules and apply them to analyse a complex circuit with at least two loops, writing all equations systematically.

Junction Rule: At any junction, ΣI_in = ΣI_out (based on charge conservation). Loop Rule: For any closed loop, the algebraic sum of all potential changes is zero (based on energy conservation).
To apply: label all branch currents with arrows; use junction rule to minimise unknowns; write loop equations by traversing each loop, adding +IR for resistor traversed opposite to current, –IR for resistor traversed with current; add +ε for EMF traversed from – to +, and –ε for EMF traversed from + to –.
For a loop containing a 10V cell and resistors R₁, R₂: –I₁R₁ – I₂R₂ + 10 = 0.
Solve the simultaneous equations obtained from as many independent loops as needed to find all unknown currents.
Verify: substitute back into all equations; currents with negative values simply flow opposite to the assumed arrow direction, which is physically valid.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 8

Long Form

Distinguish between EMF and terminal voltage of a cell. Derive the relationship between them and explain the condition under which terminal voltage equals EMF.

EMF (ε) is the potential difference between the terminals when no current flows (open circuit); it represents the total work done per unit charge by the cell in maintaining current.
Terminal voltage V is the actual potential difference across the terminals when current I flows; it is less than EMF because some voltage is dropped across the internal resistance r of the cell.
Derivation: V = (potential at P) – (potential at N) = ε – Ir, where the term Ir is the voltage drop across the internal resistance.
The current in the circuit is I = ε/(R + r), so V = IR = εR/(R + r), which is always less than ε when I ≠ 0.
Terminal voltage equals EMF only when I = 0, i.e., in open circuit (R → ∞). This is the standard condition for measuring EMF of a cell without drawing current.

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Long Answer

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Hard

Question 9

Long Form

Analyse the variation of resistivity with temperature for metals, alloys, and semiconductors, and explain each case using the microscopic model.

For metals, ρT = ρ₀[1 + α(T – T₀)] where α > 0. The graph of ρ vs T is approximately linear over moderate ranges, deviating at very low temperatures. This occurs because increasing T decreases τ while n stays constant.
For alloys like Nichrome, Manganin, and Constantan, α is very small (near zero). Their resistivity barely changes with temperature, making them ideal for standard resistors. The complex alloy structure makes τ relatively insensitive to temperature.
For semiconductors, resistivity decreases with temperature (α effectively negative) as shown by the exponential-like decrease. As T rises, vastly more electrons move from valence to conduction band, increasing n dramatically.
In semiconductors, the increase in n far outweighs the decrease in τ in ρ = m/(ne²τ), giving a net decrease in ρ.
Adding impurities to semiconductors also increases n (doping), reducing ρ further — a technique exploited in manufacturing transistors and integrated circuits.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 10

Long Form

Explain why the drift velocity of electrons is very small (of the order of mm/s) yet large currents can flow in conductors. Compare drift speed with thermal speed and speed of electric field propagation.

Drift speed is small (vd ~ 1 mm/s for typical currents) because it is only a tiny superposition on the much larger random thermal velocities. However, current I = nevdA involves the enormous electron number density n ~ 10²⁸–10²⁹ m⁻³.
Even though vd is tiny, n is so large that the product nevd gives a substantial current density. A small drift of billions of electrons per second through every cross-section adds up to ampere-scale currents.
Thermal speed of copper atoms at 300 K is about 2 × 10² m/s — roughly 10⁵ times larger than vd. The random thermal motion does not contribute to current; only the net drift does.
The electric field propagates along the conductor at the speed of electromagnetic waves (~3 × 10⁸ m/s), approximately 10¹¹ times faster than vd.
This explains why current appears instantaneously upon closing a circuit: the field signal (not the electrons) travels at near light-speed to set every electron drifting locally.

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Chapter 3: Current Electricity | Long Questions

Derive an expression for the drift velocity of electrons in a metallic conductor and hence establish the relation j = σE.

Derive the balance condition for a Wheatstone bridge using Kirchhoff's rules and explain how it is used to measure an unknown resistance.

Compare the resistivity of conductors, semiconductors, and insulators. Explain why semiconductor resistivity decreases with temperature while metal resistivity increases.

Derive expressions for the equivalent EMF and equivalent internal resistance of n cells connected (i) in series and (ii) in parallel.

Explain the origin of resistance in a metallic conductor using the electron drift model. How does this model account for the temperature dependence of resistance in metals?

Analyse the power dissipated in a resistive circuit and apply it to explain why electrical power is transmitted at high voltage. Derive the relevant expressions.

State Kirchhoff's rules and apply them to analyse a complex circuit with at least two loops, writing all equations systematically.

Distinguish between EMF and terminal voltage of a cell. Derive the relationship between them and explain the condition under which terminal voltage equals EMF.

Analyse the variation of resistivity with temperature for metals, alloys, and semiconductors, and explain each case using the microscopic model.

Explain why the drift velocity of electrons is very small (of the order of mm/s) yet large currents can flow in conductors. Compare drift speed with thermal speed and speed of electric field propagation.

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