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The Vault
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The Vault
CBSE Class 12 Physics · 11 questions · 26 marks
Current electricity is where the abstract field ideas turn into wires, cells and switches you can actually hold. This chapter follows one electron as it drifts through a copper lattice, then scales that picture up into Ohm's law, internal resistance, and networks that Kirchhoff's two rules can crack in minutes. A large slice of the board paper hides in these numericals, so practise the arithmetic until the units stop tripping you.
A copper wire of cross-sectional area 2.0×10⁻⁶ m² carries a steady current of 5.0 A. If the free electron density in copper is 8.5×10²⁸ m⁻³, the drift speed of the electrons is nearest to:
Answer
1.8×10⁻⁴ m/s — from I = nAev_d, v_d = I/(nAe) = 5.0/(8.5×10²⁸ × 2.0×10⁻⁶ × 1.6×10⁻¹⁹) = 5.0/(2.72×10⁴) = 1.84×10⁻⁴ m/s. Drift speeds are tiny; the bulb lights instantly because the electric field, not the electrons, travels through the circuit almost at the speed of light.
A uniform metal wire of resistance R is stretched, without loss of material, until its length is doubled. Its new resistance is:
Answer
4R — stretching keeps the volume Al constant, so doubling l halves A. Since R = ρl/A, the resistance changes by a factor (2)/(1/2) = 4. In general, for a wire stretched n times its original length, R' = n²R.
Three resistors of 6 Ω, 3 Ω and 2 Ω are joined in parallel across a battery. The equivalent resistance of the combination is:
Answer
1 Ω — for a parallel combination 1/R_p = 1/6 + 1/3 + 1/2 = (1 + 2 + 3)/6 = 1, so R_p = 1 Ω. Note that the equivalent resistance in parallel is always smaller than the smallest individual resistance, which rules out the 6 Ω and 11 Ω options at a glance.
A cell of emf 2.0 V and internal resistance 0.50 Ω is connected across an external resistance of 3.5 Ω. The terminal potential difference of the cell is:
Answer
1.75 V — the current is I = ε/(R + r) = 2.0/(3.5 + 0.50) = 0.50 A. The terminal voltage is V = ε - Ir = 2.0 - (0.50 × 0.50) = 2.0 - 0.25 = 1.75 V. Equivalently V = IR = 0.50 × 3.5 = 1.75 V, which is a useful cross-check.
Assertion (A): The resistivity of a semiconductor such as germanium decreases as its temperature is raised, whereas that of a metal increases. Reason (R): Raising the temperature of a semiconductor breaks many covalent bonds and increases the number density n of charge carriers far more rapidly than it reduces the relaxation time τ.
Answer
Both A and R are true and R is the correct explanation of A — resistivity is ρ = m/(ne²τ). In a metal n is essentially fixed, so hotter lattice vibrations shorten τ and ρ rises. In a semiconductor τ also falls, but n climbs so steeply with temperature that the product neτ increases overall and ρ drops.
Define drift velocity of free electrons in a conductor and derive the relation I = nAev_d connecting it with the current.
Answer
Drift velocity is the small constant average velocity that free electrons acquire along a conductor under an applied electric field, superposed on their large random thermal motion. Its magnitude is v_d = eEτ/m, where τ is the average relaxation time between collisions. Derivation: Consider a conductor of cross-sectional area A carrying current I, with n free electrons per unit volume, each of charge magnitude e, drifting with speed v_d. In a small time t, every electron moves a distance v_d t, so all electrons lying within a cylinder of length v_d t and area A cross the chosen section. Volume of that cylinder = A v_d t, so the number of electrons crossing = n A v_d t. Charge crossing the section, q = (n A v_d t) e. Therefore I = q/t = nAev_d. The current density is j = I/A = nev_d.
A coil of wire is 200 m long and has a uniform cross-sectional area of 0.50 mm². Its resistance is measured to be 6.8 Ω at room temperature. Calculate the resistivity of the material and suggest what the metal is likely to be.
Answer
From R = ρl/A, the resistivity is ρ = RA/l. Here R = 6.8 Ω, A = 0.50 mm² = 0.50×10⁻⁶ m², l = 200 m. ρ = 6.8 × 0.50×10⁻⁶/200 ρ = 3.4×10⁻⁶/200 ρ = 1.7×10⁻⁸ Ω m. This value is characteristic of copper, one of the lowest resistivities among common conductors, which is why copper is used for domestic wiring.
State Kirchhoff's junction rule and loop rule, and name the conservation principle on which each is based.
Answer
Junction rule: at any junction in an electrical network, the algebraic sum of the currents is zero, that is the total current entering equals the total current leaving (ΣI = 0). It follows from conservation of electric charge, since charge cannot pile up at a point in a steady current. Loop rule: around any closed loop of a network, the algebraic sum of the changes in potential is zero (ΣΔV = 0), counting emfs as rises and IR drops as falls. It follows from conservation of energy, because the electrostatic field is conservative and a charge returning to its starting point must have zero net work done on it.
Starting from the motion of a free electron between collisions, derive an expression for the drift velocity in a metal and hence show that resistivity is ρ = m/(ne²τ). Using this, estimate the relaxation time for copper, for which n = 8.5×10²⁸ m⁻³ and ρ = 1.7×10⁻⁸ Ω m. (mass of electron = 9.1×10⁻³¹ kg)
Answer
Derivation of drift velocity: A potential difference V across a conductor of length l sets up a uniform field E = V/l inside it. Each free electron of charge -e experiences a force F = -eE and therefore an acceleration a = -eE/m. Between two successive collisions with the lattice ions an electron accelerates freely, but every collision randomises its velocity, so the average of the initial thermal velocities is zero. If τ is the average time between collisions (relaxation time), the average velocity gained is v_d = a τ = -eEτ/m, so the drift speed is v_d = eEτ/m, directed opposite to E. Derivation of resistivity: The current is I = nAev_d = nAe × eEτ/m = (ne²τ/m) A E. Putting E = V/l: I = (ne²τ/m) × A V/l Rearranging, V/I = (m/ne²τ) × (l/A). Comparing with R = ρl/A gives ρ = m/(ne²τ), and conductivity σ = 1/ρ = ne²τ/m. Notice that this also proves Ohm's law: at fixed temperature n, τ and m are constants, so V is directly proportional to I. Numerical: τ = m/(ne²ρ). Denominator: e² = (1.6×10⁻¹⁹)² = 2.56×10⁻³⁸ C². ne² = 8.5×10²⁸ × 2.56×10⁻³⁸ = 2.176×10⁻⁹. ne²ρ = 2.176×10⁻⁹ × 1.7×10⁻⁸ = 3.70×10⁻¹⁷. τ = 9.1×10⁻³¹/3.70×10⁻¹⁷ = 2.5×10⁻¹⁴ s. So a conduction electron in copper travels freely for only about 25 femtoseconds before its next collision.
(a) Draw the arrangement in words and, using Kirchhoff's rules, derive the balance condition of a Wheatstone bridge. (b) In such a bridge P = 12 Ω, Q = 18 Ω and R = 20 Ω. Find the value of S for balance, and the current drawn from a 6.0 V battery of negligible internal resistance connected across the bridge when it is balanced.
Answer
(a) Arrangement: Four resistors P, Q, R and S form a quadrilateral ABCD. P lies in arm AB and Q in arm BC; R lies in arm AD and S in arm DC. A cell is connected across the diagonal AC and a sensitive galvanometer across the other diagonal BD. Let the current from the cell divide at A into I₁ through P and I₂ through R, and let I_g be the current through the galvanometer from B to D. At balance the galvanometer shows no deflection, so I_g = 0. Then by the junction rule the current through P also flows through Q, and the current through R also flows through S. Applying the loop rule to loop ABDA (with I_g = 0, so no potential drop across the galvanometer): I₁P - I₂R = 0, hence I₁P = I₂R ... (1) Applying the loop rule to loop BCDB: I₁Q - I₂S = 0, hence I₁Q = I₂S ... (2) Dividing (1) by (2): P/Q = R/S. This is the balance condition of the Wheatstone bridge. Because it involves only a ratio, an unknown resistance can be measured accurately without knowing the emf of the cell or the resistance of the galvanometer. (b) At balance, S = QR/P = (18 × 20)/12 = 360/12 = 30 Ω. With I_g = 0, the branch P + Q is simply in series: 12 + 18 = 30 Ω. The branch R + S is in series: 20 + 30 = 50 Ω. These two branches are in parallel across the battery: R_eq = (30 × 50)/(30 + 50) = 1500/80 = 18.75 Ω. Current drawn from the battery, I = V/R_eq = 6.0/18.75 = 0.32 A.
Read the passage and answer the questions that follow: A small workshop runs a soldering station from a rechargeable 12 V battery pack. When new, the pack has an internal resistance of 0.40 Ω. The technician, Ravi, first connects a single heating coil of resistance 10 Ω across the pack and notes that the voltmeter across the terminals no longer reads the full 12 V. Later he adds a second, identical 10 Ω coil in parallel with the first. (a) Calculate the current drawn when only one coil is connected. (b) Find the terminal potential difference of the battery pack in that case. (c) Calculate the power dissipated in the coil and the power wasted inside the battery pack. (d) When the second coil is added in parallel, find the new current and terminal voltage, and explain why the reading of the voltmeter falls further.
Answer
(a) Total circuit resistance = R + r = 10 + 0.40 = 10.4 Ω. I = ε/(R + r) = 12/10.4 = 1.15 A. (b) Terminal voltage V = ε - Ir = 12 - (1.15 × 0.40) = 12 - 0.46 = 11.54 V. Check: V = IR = 1.15 × 10 = 11.5 V, agreeing to rounding. (c) Power in the coil: P = I²R = (1.15)² × 10 = 1.323 × 10 = 13.2 W. Power wasted inside the pack: P_r = I²r = 1.323 × 0.40 = 0.53 W. The total power delivered by the battery is εI = 12 × 1.15 = 13.8 W, which matches 13.2 + 0.53 W. (d) Two 10 Ω coils in parallel give R' = 10/2 = 5.0 Ω. New current I' = 12/(5.0 + 0.40) = 12/5.4 = 2.22 A. New terminal voltage V' = 12 - (2.22 × 0.40) = 12 - 0.89 = 11.11 V. The voltmeter reads less because the terminal voltage is ε - Ir: lowering the external resistance raises the current, so a larger share of the emf is used up across the internal resistance of the pack. As the pack ages and r grows, this drop becomes worse for the same load.
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