Current Electricity Exemplar Solutions Class 12 - Free PDF

Strategic angle. The question is asking which agent supplies the direction of j at each point of a curved wire, not its magnitude. The magnitude follows from the EMF (it fixes |j| = I/A once I is set by the loop's total resistance); the direction needs a position-dependent field, which only surface charges can provide. This is one of the few places in introductory electrodynamics where surface-charge effects, usually treated as electrostatic curiosities, become physically essential to steady currents.

1. Microscopic Ohm's law. j = σ E inside the wire (Ohm's law in local form). σ is a scalar for an isotropic conductor, so the direction of j at every point equals the direction of E at the same point. The job reduces to: what fixes the direction of E?
2. Straight wire baseline. A straight wire connecting battery terminals has E parallel to its axis everywhere — the longitudinal field set up by the battery terminals already lies along the wire because the geometry is uniform. No surface charges are needed there beyond the end caps.
3. Curved-wire challenge. A curved wire cannot have a constant-direction E everywhere inside — it must bend with the axis. But the battery's emf alone produces a roughly uniform external field; it can't supply a position-dependent direction inside the conductor.
4. Surface-charge resolution. Surface charges on the wire's outside form a thin layer (positive on the convex side of a bend, negative on the concave side). Their own field, added to the longitudinal driving field, gives a total E that locally follows the wire's axis. They are the steering agent. Ruling out (a), (c), (d) is then immediate: the emf fixes magnitude not direction, bulk carrier distribution is uniform in steady state.

Option (b).

Strategic angle. Compute _eq directly and test whether it lies between ₁ and ₂. The convex-combination view turns a circuit-theory question into a one-line inequality — much faster than re-deriving via Kirchhoff each time. It also makes the limit cases transparent: when one internal resistance is much smaller than the other, that cell dominates.

1. Derive parallel-cell formula via Kirchhoff. Let I be the load current leaving the junction. Apply Kirchhoff to each branch: branch 1 contributes V = ₁ - I₁ r₁, branch 2 contributes V = ₂ - I₂ r₂, with I₁ + I₂ = I. Solving gives _eq = (₁ r₂ + ₂ r₁)/(r₁+r₂) and r_eq = r₁ r₂/(r₁+r₂).
2. Re-write as convex combination. _eq = w₁₁ + w₂₂ with w₁ = r₂/(r₁+r₂), w₂ = r₁/(r₁+r₂). Note w₁ + w₂ = 1, both ∈ (0,1) for finite positive r₁, r₂.
3. Apply convex-combination bound. A convex combination of two positive numbers ₁ < ₂ always lies strictly between them. Hence option (a). The bound is strict because both weights are positive (no r_i = 0).
4. Eliminate distractors. (b) violates the lower bound; (c) is the series formula not parallel; (d) ignores the explicit r₁, r₂-dependence of every weight.

Option (a).

Strategic angle. Read the rough value of R from the present measurement, then pick S to put the null near the midpoint. The sensitivity argument is purely geometric: a fixed absolute reading uncertainty δ l (a millimetre, say) becomes a smaller fractional error when divided by a larger l₁ or 100-l₁. Both factors are largest together near the centre.

1. First-pass estimate. R ≈ 100 · (2.9/97.1) ≈ 2.99 Ω ≈ 3 Ω.
2. Sensitivity argument. Differentiating R = S l₁/(100-l₁) gives δ R/R = δ l₁ · [1/l₁ + 1/(100-l₁)]. The bracket is minimised at l₁ = 50 cm. For best sensitivity, the null should sit near 50 cm, which requires R ≈ S. So S = 3 Ω matches our estimate.
3. Quantify the gain. At l₁ = 2.9 cm, the bracket is ≈ 1/2.9 + 1/97.1 ≈ 0.355/cm. At l₁ = 50 cm it is ≈ 0.040/cm — nearly a 9-fold precision gain for the same δ l₁.
4. Options (a), (b), (d) all fail this geometric requirement — (a) cannot beat the sensitivity handicap, (b) makes it worse, (d) discards a fixable problem.

Option (c).

Strategic angle. Two facts pin the answer: the driving battery must put more potential across the wire than the largest measured EMF (existence of balance), and the wire drop should be only slightly larger so balance lengths are long (precision). The question is essentially asking the student to pick the choice that satisfies both constraints simultaneously.

1. Existence condition. V_wire ≥ max(₁, ₂) = 10 V. An 8 V driver (option a) cannot meet this — the galvanometer will never null on the 10 V cell.
2. Precision condition. Balance length l ≈ ε L/V_wire. With V_wire much larger than 10 V (option c forces V_wire = 80 V), l shrinks to a small fraction of the wire — a 1 mm jockey uncertainty then becomes a much larger fractional error. Lousy precision.
3. Sweet spot. A 15 V driver with a series resistor tuned to keep the wire drop just above 10 V (option b) gives long balance lengths for both cells, with the 10 V cell's balance close to the far end. Best precision compatible with the existence requirement.
4. Option (d). Confuses the potentiometer with the Wheatstone bridge. The potentiometer's primary job is EMF (voltage) comparison — Wheatstone bridges compare resistances.

Option (b).

Quick reading. Compute L/A for each pair of faces and pick the largest. The trick is to remember that L is the distance the current must travel (the dimension perpendicular to the chosen pair of faces, not the dimension along it), and A is the face area (the area perpendicular to the current). Confusing the two flips the answer.

1. Identify L and A for each face pair. For the 1 × 12 cm faces, the current crosses these two faces moving along the 10 cm length — so L = 10 cm, A = 0.5 cm². Repeat for each pair.
2. Pair (a): L/A = 10/0.5 = 20 cm^-1.
3. Pair (b): L/A = 0.5/10 = 0.05 cm^-1.
4. Pair (c): L/A = 1/5 = 0.2 cm^-1.
5. Maximum-R orientation. Largest L/A is pair (a) — by a factor of ≈ 400 over the worst pair. R_max/R_min = 20/0.05 = 400. Same rod, same material — geometry alone changes R by 400×.

Option (a).

Strategic angle. I = nAev_d has v_d but not the thermal velocity. So only drift contributes to the current. The deeper point is that current is a net flow of charge, and nets only register the small bias on top of the thermal Maxwell distribution; the bulk of the distribution cancels itself out.

1. Equilibrium argument. In equilibrium with no field, thermal velocities of electrons average to zero by isotropy: the Maxwell-Boltzmann distribution has equal probability for +v and -v in any direction.
2. Drift as a small bias. Applying a field E tilts the distribution by a tiny v_d = -eEτ/m (electrons negatively charged so they drift opposite to E). It is this tilt that we measure as current.
3. Quantitative check. For copper at room temperature, v_th ∼ 10⁵ m/s while v_d ∼ 10^-4 m/s in a household wire. The ratio is ∼ 10^-9 — drift is a billion times smaller, yet it carries all the current.
4. Formula. I = nAev_d. Only v_d appears, so only v_d enters I.

Option (a).

Strategic angle. Two equivalent ways to state the junction rule: ``charge in = charge out'' (conservation of charge) and ``no charge piles up at a node'' (steady state). Pick those. The key insight is that (b) and (d) are not independent — they are two faces of the same physical statement, applied to a steady current.

1. Junction rule. ∑ I_in = ∑ I_out.
2. Microscopic basis. Integrate the continuity equation ∇·j + ∂ρ/∂ t = 0 over a small volume around the node. Charge conservation gives the continuity equation itself; steady state gives ∂ρ/∂ t = 0. Together: ∇·j = 0, which integrated over the node is the junction rule.
3. Read off the right options. (b) Conservation of charge: gives the continuity equation. (d) No accumulation at a junction: the steady-state assumption.
4. Reject (a), (c). (a) refers to a non-standard ``conservation of j'' — j is a vector field, not a conserved scalar. (c) confuses momentum with charge; momentum of a single charge is not conserved at a junction (the wire's lattice and electric field redirect it).

Options (b), (d).

Strategic angle. Compute R_∥, note that it stays near R across the entire range of R', and read off the implications. The double-inequality r ≪ R ≪ R₀ is the problem's punchline: it forces R_∥ to be essentially R regardless of R', which in turn freezes both V_AB and I but lets I_R' swing wildly.

1. Bound R_∥. R_∥ = R R'/(R+R'). Two regimes:
   • R' → ∞: R_∥ → R.
   • R' = R₀: R_∥ = R R₀/(R+R₀) ≈ R because R ≪ R₀ so R+R₀ ≈ R₀.
   Thus R_∥ ≈ R throughout.
2. Total current I. I = V/(r + R_∥) ≈ V/(r+R), nearly constant in R'. Not sensitive to R', so (c) is wrong. The bound R_∥ ≤ R gives I ≥ V/(r+R) always, with equality at R' = ∞. Option (d) is correct.
3. Voltage across AB. V_AB = IR_∥ = VR_∥/(r+R_∥) ≈ V · R/(r+R) ≈ V since r ≪ R. Nearly constant — option (a) is correct.
4. Current through R'. I_R' = V_AB/R' ≈ V/R'. As R' runs from R₀ to ∞, I_R' runs from V/R₀ to 0 — a huge variation. Not nearly constant. Option (b) is wrong.

Options (a), (d).

Structural observation. Write ρ = m/(n e² τ) and ask: which factors depend on T? Only n (semiconductors) and τ (metals). L and m do not — to within a tiny correction in case of L, and not at all in case of m. The signature piece of the question is that the answer is a pair: one factor for metals, one for semiconductors.

1. Drude formula. ρ = m/(ne²τ). The four candidate factors in the options map onto this formula: (a) → n, (b) → τ, (c) → L (geometric, not in ρ — but feeds into R = ρ L/A), (d) → m.
2. Semiconductors/insulators: n(T) dominates. n(T) ∝ T^3/2 e^{-E_g/(2k_B T)} near the band gap. Exponential in T — orders-of-magnitude rise — so ρ falls sharply with rising T. Option (a).
3. Metals: τ(T) dominates. At high T, phonon scattering gives τ ∝ 1/T, so ρ ∝ T. Option (b).
4. Rule out (c), (d). Thermal expansion gives Δ L/L ∼ 10^-5/K — six orders of magnitude smaller than the carrier-density or relaxation-time variations. Electron mass is a fundamental constant; the effective mass in a solid is set by band structure, not temperature.

Options (a), (b).

Strategic angle. Both students get the same central value; the question is about the spread, which has two sources: resistor tolerances and galvanometer sensitivity (current- dependent). A trustworthy bridge measurement requires both small fractional resistor errors and a sufficient bridge current to drive the galvanometer well off zero when slightly imbalanced.

1. Same central value. R = (R₂/R₁) R₃ depends only on the ratio. Both students use ratio = 2, hence the same answer 10 Ω.
2. Tolerance propagation. δ R/R = δ R₁/R₁ + δ R₂/R₂ + δ R₃/R₃ in quadrature. Standard resistance boxes have tolerance bands that depend on the value; the second student's 1000 Ω may have ± 0.1% tolerance vs ± 1% on a 5 Ω box, so the error is not the same. Option (b) is correct.
3. Galvanometer sensitivity. Bridge current I_bridge ∼ V/(R₁ + R₂). Student 2's current is ∼ 100× smaller, so a fixed off-null δ R produces a 100× smaller galvanometer signal — making the null harder to pin down. Option (c) is correct.
4. Eliminate distractors. (a) Same central value doesn't imply same error. (d) No real instrument is perfect (lead/contact resistances, galvanometer offsets, thermal EMF at junctions).

Options (b), (c).

Strategic angle. The balance equation R/S = l₁/(100-l₁) is the master formula. Read off uniqueness and the direction of shift directly from it; reason about jockey-side currents via potential differences. Compare the potential at the jockey to the fixed potential at B rather than trying to trace the full Wheatstone bridge in your head.

1. Uniqueness of null point. The function f(l₁) = l₁/(100-l₁) is strictly increasing on (0,100), sweeping from 0 to ∞. So for any positive R/S, there is exactly one l₁ that satisfies the equation. Option (a) is correct.
2. Jockey to the left of D. The NCERT exemplar key marks option (b) as incorrect — the stated direction of current is the opposite of what the bridge actually produces with this set of resistances. So (b) is False.
3. Jockey to the right of D. Here the jockey sits below B in potential, so current flows B → G → jockey, i.e. from B to the wire through the galvanometer. Option (c) correct.
4. Shift on increasing R. From R/S = l₁/(100-l₁), ∂ l₁/∂ R > 0 at fixed S. So increasing R moves the null right, toward B. Option (d) claims left, which is the opposite direction — false.

Options (a), (c).

Strategic angle. A junction can change a wire's direction, which changes a charge's velocity direction — that's a momentum change. The agent providing the impulse is the wire and the electric field. Frame the question as: which quantity is strictly conserved at the junction, and which only seems so?

1. Conservation requirement. Conservation of momentum requires zero net external force on the system being considered. For a charge crossing a junction, the ``system'' is just the charge, and the external forces are non-zero — the lattice ions and the local electric field both push on it.
2. Direction-change argument. If the wires meet at an angle θ, the velocity vector rotates by θ. The change in momentum, Δp = m_ev₂ - m_ev₁, is non-zero for any θ ≠ 0. This impulse is delivered by the wire (and ultimately by the clamps holding the wire).
3. What is conserved. Charge conservation applies at every junction, regardless of geometry: total in = total out. This is the actual Kirchhoff junction rule. The total system (charge + wire + clamps) does conserve momentum, but the charge alone doesn't.

Momentum: not conserved (for the charge alone). Charge: conserved.

Strategic angle. Two roles for τ: it sets the proportionality between J and E, and it carries the temperature dependence into ρ. The first explains why we get a linear J-E relation at all (it could have been quadratic, cubic, exponential — but isn't); the second explains how ρ varies with the lab thermometer.

1. Microscopic Ohm's law. v_d = eEτ/m, J = nev_d = (ne²τ/m)E. If τ does not depend on E, then J = σ E with constant σ — the linear Ohm's law V = IR in macroscopic dress.
2. What would break Ohm's law. If τ varied with E (e.g. τ = ₀(1-α E)), then J(E) becomes non-linear and R depends on the voltage applied. Ohm's law is then violated — this is exactly what happens in semiconductor diodes, electrolytes near saturation, and gas discharges at high E.
3. Temperature dependence of ρ. ρ = m/(ne²τ). As T rises, lattice ions vibrate more (phonon density grows), electrons scatter more frequently, τ decreases as ∼ 1/T, and ρ increases proportionally. Standard explanation of the linear _Cu(T) curve in metals.
4. Cross-link. For semiconductors the dominant T-dependence is via n (carriers excited across the band gap), not τ, so ρ falls with rising T — opposite sign to metals.

τ constant in E underwrites Ohm's law; τ(T) drives ρ(T).

Strategic angle. Null methods reduce the experiment to a geometric ratio of standard resistors. Any non-null method has to calibrate a deflection — and any calibrated deflection drags in the galvanometer's sensitivity, its internal resistance, and the driving EMF as fresh sources of error. The null method's beauty is its independence from all three.

1. Null condition. I_G = 0 ⇒ V_B = V_D (potentials at the galvanometer's two terminals coincide). The condition becomes R/S = R₃/R₄, a pure ratio — independent of EMF ε and of galvanometer characteristics.
2. Three independences.
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   2. From source EMF: small fluctuations in ε change every current but not the ratio V_B/V_D.
   3. From galvanometer internal resistance: zero current through it means zero IR drop in either galvanometer-side branch.
   4. From galvanometer calibration: the device only needs to discriminate ``zero'' from ``not zero''.
3. Non-null alternative. Measure I_G at known imbalance, back-solve R. Requires: galvanometer sensitivity (Amps per division), galvanometer resistance, and EMF — three new measurements, three new error sources.
4. Net advantage. Precision of the bridge is set by the precision of the standard resistors and the detector's null sensitivity, not by absolute measurement accuracy of any current.

Null method: simpler, more precise, fewer error sources.

Strategic angle. Joint resistance is the enemy of a linear potentiometer. Thick strips have negligible ρ L/A and preserve the V ∝ l relation. Quantify: a 1 mm² wire of a few centimetres might have ∼ 0.001 Ω, but the contact resistance at a poor mechanical joint can be tens of milliohms — comparable to the wire itself.

1. Linear V ∝ l requirement. A potentiometer works only if the potential drop per unit length is uniform along the wire — which requires the wire's own resistance to dominate the measurement loop.
2. Numerical scale. For a 4 m, 0.5 mm diameter manganin wire (ρ ≈ 4.4× 10^-7 Ω-m), R_wire ≈ 9 Ω. A poorly-mated joint easily adds 10-100 mΩ, biasing the readings by ∼ 1%.
3. Thick strips. Cross-section A_strip ≫ A_wire makes R_strip = ρ L/A_strip negligible. Brass or copper strips ∼ 1 cm wide contribute ∼ 10 Ω — five orders of magnitude below the wire's resistance.
4. Mechanical bonus. Thick strips also provide better contact with screw terminals, reducing contact resistance on top of bulk resistance.

To minimise connection resistance and preserve linearity.

Strategic angle. Conductivity, density and cost set the trade-off. The choice changes depending on the application — and the question expects the student to recognise that ``best wire'' is not a single material but a function of context.

1. Resistivity. Cu has ρ ≈ 1.7× 10^-8 Ω-m vs. Al ≈ 2.7× 10^-8 Ω-m. Cu is ∼ 60% better — less I²R loss for the same cross-section, less voltage drop along long indoor runs, less safety risk from hotspots in walls.
2. Density and weight. Al at _m ≈ 2.7 g/cm³ is roughly 1/3 the density of Cu (_m ≈ 8.96). For overhead transmission lines (tens of km between towers), the load on supporting masts and the sag between spans both favour Al.
3. Cost. Al per kg is several times cheaper than Cu. Combined with the lower density, an Al cable of the same conductance (cross-section scaled by _Al/_Cu) ends up much cheaper than the equivalent Cu cable.
4. Corrosion and joint reliability. Cu oxide is still conductive; Al oxide is an excellent insulator, forming an invisible barrier at any unmaintained connection. Al also ``cold-flows'' under clamps over years, gradually loosening — a well-known fire hazard in Al-wired houses.
5. Trade-off summary. Indoor wiring: short runs, weight irrelevant, joints made and forgotten ⇒ Cu wins. Outdoor transmission: long runs, weight and cost dominate, joints serviced regularly ⇒ Al wins.

Cu for short, joint-critical runs; Al for long, weight-critical runs.

Strategic angle. Standards demand resistance that is flat in temperature, compact, and joint-clean. Pure metals fail the first; alloys are engineered to pass all three. The trick is to recognise that you are buying three independent material properties for the price of one wire — and that alloying lets you tune each one.

1. Temperature coefficient. Standard resistors must not drift with the room temperature. Pure Cu has α ≈ 4× 10^-3/K — a 5^∘C swing changes R by 2%, ruining any precision standard. Manganin and constantan are tuned by composition to give α ≈ 10^-5/K, a 400-fold improvement.
2. High resistivity. A standard 1 kΩ coil of Cu wire of 0.1 mm diameter would need ∼ 500 m of wire — impractical. Manganin's ρ ≈ 44× 10^-8 Ω-m (26× that of Cu) makes the same coil fit in a few metres of wire on a bobbin.
3. Low thermoelectric EMF with Cu. Junctions between dissimilar metals generate a Seebeck voltage of a few μV/^∘C. For Cu-manganin this voltage is engineered to be near zero, so the standard's terminals don't contaminate the measurement.
4. Why pure metals don't qualify. Cu, Ag, Au all have low ρ (bulky coils) and large α (drifty). They cannot meet any of the three criteria simultaneously.

Alloys give low α, high ρ, low thermal EMF with Cu.

Quick reading. The current is forced by P = VI. Plug I = P/V into I² R_C and read off the dependence. The quadratic-in-V saving is one of the most consequential facts in practical electrical engineering — it is the reason long-distance transmission lines run at hundreds of kV.

1. Current at the device. Fixed power delivery P at voltage V requires I = P/V. The same I flows through the cable (series with the device).
2. Cable dissipation. P_cable = I² R_C = (P/V)² R_C = P² R_C/V². Inversely quadratic in V.
3. Numerical impact. Power station at 11 kV distributing 100 MW over R_C = 1 Ω wastes ∼ 8 MW (8%). Step up to 400 kV and the same 100 MW delivery wastes only (11/400)² × 8 MW ≈ 6 kW — a thousandfold reduction.
4. Why not just thicker cable. R_C ∝ 1/A, so doubling cable cross-section halves the loss. But doubling A doubles weight, cost and tower load. Doubling V, by contrast, quarters the loss with no change to the conductor.

P_wasted = P² R_C/V²; raise V to reduce it.

Strategic angle. Potential gradient k ∝ 1/(R + R_w). Larger R ⇒ smaller k ⇒ longer balance length. The physics turns on tracking how the wire's potential drop changes when you change the resistor in the driver loop — that drop, divided by the wire's length, is the gradient k that every measurement on the wire is read against.

1. Driver-loop current. I_drv = E_driver/(R + R_w). Wire's potential drop: V_w = I_drv R_w = E_driver R_w/(R+R_w).
2. Gradient. k = V_w/L = E_driver R_w/[L(R+R_w)]. As R increases at fixed E_driver, R_w, L: k decreases.
3. Balance length. Standard potentiometer formula E₁ = k l ⇒ l = E₁/k. Smaller k at fixed E₁ means larger l.
4. Direction. Larger balance length means the balance point sits further from A, i.e. shifts toward B. (Until R is so large that even the full wire can't balance E₁, at which point no null exists.)

Balance point shifts towards B.

Strategic angle. ``One-sided'' means no balance exists. Two ways for that: wrong magnitude (correct polarity, E₁ > kL) or wrong polarity (EMFs add along the loop). The direction of monotonicity (decreasing vs. increasing as the jockey moves) distinguishes the two cases.

1. Set up the galvanometer-loop EMF. Let V_J be the wire's potential at the jockey, measured from the positive end. The galvanometer responds to V_loop = V_J - E₁ (or V_J + E₁ if E₁ is reversed). It shows ``zero'' precisely when V_loop = 0.
2. Case (i): correct polarity, E₁ > kL. V_J ranges from 0 (at A) to kL (at B) < E₁. V_loop = V_J - E₁ < 0 throughout. Magnitude decreases as V_J rises toward kL, so deflection shrinks (but never reaches zero) as jockey moves A → B. Positive of E₁ at X, with E₁ > kL.
3. Case (ii): reversed polarity. V_loop = V_J + E₁ (the two EMFs add around the galvanometer loop because they are in series-aid). As V_J grows from 0 at A to kL at B, V_loop grows in magnitude. Deflection increases toward B. Negative of E₁ at X.
4. Repair recipe. For case (i), increase the driver-loop wire drop (decrease the driver-loop series resistance R, or use a larger driver battery). For case (ii), simply swap the leads of E₁.

(i) +ve at X, E₁ > kL. (ii) -ve at X.

Strategic angle. V = IR, with I decreasing in R as 1/(R+r). The product R/(R+r) rises monotonically from 0 to 1. The curve is the canonical ``loaded-cell'' graph — it shows up in solar-cell IV characteristics, fuel-cell load curves, and any source-with-internal-resistance plot.

1. Closed form. V(R) = IR = E R/(R+r). Rewrite as V = E [R/(R+r)] = E [1 - r/(R+r)].
2. Limits and special values. R → 0 ⇒ V → 0 (short circuit); R → ∞ ⇒ V → E (open circuit, no current). At R = r, V = E/2.
3. Slope and shape. dV/dR = Er/(R+r)² > 0 (monotone increasing). Initial slope E/r at R = 0; slope → 0 as R → ∞. Second derivative -2Er/(R+r)³ < 0 (concave down).
4. Power-delivery digression. P_R = V²/R = E² R/(R+r)². Maximum at R = r (impedance matching) — same point where V = E/2. Useful sanity check on the graph.

V(R) = E R/(R+r), monotone-rising from 0 to E.

Strategic angle. Compute I_s and I_p in closed form, take the ratio. The beauty of the problem is that the EMF E cancels — the answer depends only on n, not on absolute voltage or resistance scales.

1. Series combination. Equivalent external resistance R_s = nR. Loop resistance R_s + r = nR + R = (n+1)R. I_s = E(n+1)R.
2. Parallel combination. Equivalent external resistance R_p = R/n. Loop resistance R_p + r = R/n + R = R(n+1)/n. I_p = ER(n+1)/n = n E(n+1)R.
3. Take the ratio. I_pI_s = n E/[(n+1)R]E/[(n+1)R] = n.
4. Apply the data. Given I_p = 10 I_s, so n = 10. Notice that E and the resistance scale R both cancelled in the ratio — the answer is purely combinatorial.

n = 10.

Structural observation. Sum and reciprocal-sum each have strict inequalities once two or more positive terms are present. A single-line proof for each direction, plus a one-paragraph physical picture, is enough for full marks on this short-answer question.

1. Series bound from below by R_max. R_S = _i R_i = R_max + _{Ri ≠ Rmax} R_i. Since each R_i > 0, the sum on the right is strictly positive (for n ≥ 2), so R_S > R_max.
2. Parallel bound from above by R_min. 1R_P = _i 1R_i = 1R_min + _{Ri ≠ Rmin} 1R_i. Again the rest of the sum is strictly positive, so 1/R_P > 1/R_min, i.e. R_P < R_min.
3. Physical picture — series. Electrons traverse every resistor in turn. Each contributes to the total opposition; the lightest resistor still adds something positive on top of the heaviest one. Hence total > heaviest.
4. Physical picture — parallel. Electrons distribute themselves over multiple parallel paths. Even the lowest-resistance path on its own would carry a current E/R_min at a fixed voltage; adding more paths can only increase the total current, i.e. further reduce the effective resistance.

R_S > R_max, R_P < R_min.

Strategic angle. The cells form a closed loop. Compute the loop current, then the terminal voltage of either cell. Cross-checking via the second cell catches sign errors — both expressions for V_AB must give the same number, and if they don't, recheck the sign of I relative to each cell's + terminal.

1. Loop EMF and current. E_net = E₁ - E₂ = 6 - 4 = 2 V (opposition: the smaller cell opposes the larger). Loop resistance: r₁ + r₂ = 2 + 8 = 10 Ω. I = 2/10 = 0.2 A, flowing in the direction of E₁.
2. Terminal voltage of E₁ (discharging). Current leaves the + terminal: V_A - V_B = E₁ - I r₁ = 6 - 0.4 = 5.6 V.
3. Terminal voltage of E₂ (being charged). Current enters the + terminal so the cell's terminal voltage exceeds its EMF: V_A - V_B = E₂ + I r₂ = 4 + 1.6 = 5.6 V. Same answer — internal consistency confirmed.
4. Energy budget check. Power supplied by E₁: E₁ I = 1.2 W. Power absorbed by E₂ (chemical energy stored back into it): E₂ I = 0.8 W. Power lost in r₁, r₂: I²(r₁+r₂) = (0.04)(10) = 0.4 W. Total absorbed = 0.8 + 0.4 = 1.2 W = supplied.

V_AB = 5.6 V.

Strategic angle. V₁ = E - I r₁ = 0 gives I = E/r₁. Set equal to the loop current and solve for R. The trick is to forget about cell 2 for a moment and ask: what current would make cell 1's terminal voltage vanish? Once you know that, the other cell + external R just have to arrange that current.

1. Zero-terminal-voltage condition for cell 1. V₁ = E - Ir₁ = 0 ⇒ I = E/r₁. This is the short-circuit current of cell 1 alone.
2. Loop current with both cells. Two cells of EMF E in series-aid give net EMF 2E and net resistance r₁ + r₂ + R: I_loop = 2Er₁+r₂+R.
3. Set them equal and solve. 2E/(r₁+r₂+R) = E/r₁. Cancel E: r₁+r₂+R = 2r₁, so R = r₁ - r₂.
4. Physical interpretation. When the load R = r₁ - r₂, the loop current is exactly the short-circuit current of the first cell, meaning all of E₁'s EMF is dropped internally and zero shows up at its terminals. The V₁ = 0 condition requires r₁ ≥ r₂, since R ≥ 0.

R = r₁ - r₂.

Strategic angle. R ∝ 1/A_c.s. when ρ L is fixed. Compute the two cross-sectional areas and take the ratio. Same material and same length together kill ρ L — geometry alone decides.

1. Identify what's the same. Same material ⇒ same ρ. Same length ⇒ same L. So in R = ρ L/A, only A differs.
2. Cross-section of A (solid). Radius 0.5 mm ⇒ A_A = π(0.5)² = 0.25π mm².
3. Cross-section of B (annular). Outer radius 1 mm, inner radius 0.5 mm ⇒ A_B = π(1² - 0.5²) = 0.75π mm². Note that the conducting cross-section is the annulus only — current flows through metal, not through the hollow interior.
4. Take the ratio. R_A/R_B = A_B/A_A = 0.75π/(0.25π) = 3. So R_A : R_B = 3 : 1.
5. Sanity check via diameters. A_B is computed as π(d_o² - d_i²)/4 = π(4-1)/4 = 0.75π mm² — same answer through the diameter formulation.

R_A : R_B = 3 : 1.

Strategic angle. The Kirchhoff system is linear in the voltages and resistances; scaling both by the same factor leaves the solution invariant. This is a symmetry argument: the equations of circuit theory have a hidden homogeneity that the question is asking the student to expose.

1. Write down the equations. For any planar circuit with L independent loops and J junctions, KVL gives L equations of the form ∑ E_i - ∑ I_k R_k = 0 and KCL gives J - 1 equations of the form ∑ I_in - ∑ I_out = 0. Together they pin down all the branch currents.
2. Scale uniformly. Replace E_i → nE_i and R_k → n R_k. KVL becomes ∑ n E_i - ∑ I_k (n R_k) = n · (∑ E_i - ∑ I_k R_k) = 0. The factor of n scales every term equally and cancels.
3. KCL is unaffected. Junction equations involve only currents, no voltages or resistances. So they remain identical.
4. Conclusion. The same I_k that solved the original system solves the rescaled one. Currents are unaltered. Voltages, on the other hand, scale by n (since V = IR and R is multiplied by n).
5. Matrix-level statement. Writing AI = b where A depends on R and b on E, scaling both rescales both sides by n: (nA)I = nb ⇔ AI = b. Same solution.

Currents are unaltered.

Strategic angle. The parallel-cell formula with the opposition sign gives the answer immediately. The whole problem hinges on getting the sign right — once that's set, two arithmetic substitutions finish the calculation.

1. Recognize opposition. The + of the 10 V cell ties to the - of the 2 V cell, meaning the two EMFs point in opposite directions around the parallel block. Apply the E₁ r₂ - E₂ r₁ formula.
2. Plug into the parallel-cell formula. E_eq = E₁ r₂ - E₂ r₁r₁ + r₂ = (10)(5) - (2)(10)10 + 5 = 50 - 2015 = 3015 = 2 V.
3. Internal resistance. r_eq = r₁ r₂r₁ + r₂ = 5015 = 103 Ω ≈ 3.33 Ω. Note: the parallel rule for r_eq does not care about polarity — only the EMF does.
4. Kirchhoff cross-check. Without a load, currents I₁ from cell 1 and I₂ = -I₁ from cell 2 (returns because polarities oppose) satisfy E₁ = I₁ r₁ + V and V = E₂ + I₁ r₂, giving E_eq = V|_Iload=0 = (E₁ r₂ - E₂ r₁)/(r₁+r₂) = 2 V. Same answer.

E_eq = 2 V, r_eq = 10/3 Ω.

Strategic angle. Average power P fixes I. Compute wire resistance and I² R. Divide by daily energy. The numbers are intentionally meant to come out small (<1%), reassuring the student that home wiring is well-engineered: a non-trivial fraction would mean dangerously hot cables.

1. Average power. 10 commercial units (kWh) per day, AC runs 5 hours: P = 10 kWh5 h = 2 kW = 2000 W.
2. Line current. I = P/V = 2000/220 ≈ 9.09 A.
3. Wire cross-section. Radius r = 1 mm = 10^-3 m: A = π r² = π × 10^-6 ≈ 3.14× 10^-6 m².
4. Resistance, copper. R_Cu = _Cu LA = (1.7× 10^-8)(10)3.14× 10^-6 ≈ 0.054 Ω.
5. Power loss, copper. P_wire = I² R_Cu = (9.09)²(0.054) ≈ 82.6 × 0.054 ≈ 4.5 W. Daily energy: 4.5 × 5 = 22.4 Wh = 0.0224 kWh. Fraction: 0.0224/10 ≈ 0.22%.
6. Aluminium scaling. R_Al/R_Cu = _Al/_Cu ≈ 1.59. So P_wire,Al ≈ 1.59 × 4.5 ≈ 7.1 W, daily ≈ 35.5 Wh, fraction ≈ 0.36%.
7. Practical takeaway. 0.22% is well below any safety threshold — Cu wiring at 1 mm radius for 10 m is generously sized for a 2 kW load. Al would still be safe but slightly more wasteful.

Cu ∼ 0.22%; Al ∼ 0.36%.

Strategic angle. The null on the last segment fixes the wire drop equal to the EMF being measured. That gives one equation in R_w. The first case (R = 50, no null) is used only as a consistency check — it tells you the cell is roughly 8 V; the second case does all the work.

1. Identify the key observation. The null appears on the 4th (i.e. last) segment of the wire, so the balance length is essentially the full wire. The wire's full drop V_wire then equals the EMF being balanced: V_wire = E₁ ≈ 8 V.
2. Drop across wire for R = 10 Ω. V_wire = V_B · R_wR+R_w = 10 R_w10+R_w.
3. Solve for R_w. Set equal to 8 V: 10 R_w = 8(10 + R_w) ⇒ 2R_w = 80 ⇒ R_w = 40 Ω.
4. Driver current cross-check. I = V_B/(R+R_w) = 10/50 = 0.2 A. Wire drop = I R_w = 0.2 · 40 = 8 V.
5. Drop per unit length. k = V_wire/L = 8 V/L. Standard NCERT potentiometers use L = 4 m, giving k = 2 V/m.
6. Self-consistency with case 1. R = 50: V_wire = 10 · 40/(50+40) ≈ 4.4 V < 8 V = E₁. No null possible — consistent with the problem statement.

R_w = 40 Ω; k = 2 V/m.

Strategic angle. Get I first; then v_d from I = nAev_d; then KE per electron and total KE; then divide by RI² for the time scale. The shocking conclusion — that drift KE is shed in ∼ 10^-18 s, six orders of magnitude faster than a collision — reveals just how little of the battery's energy actually goes into the macroscopic motion of electrons. The rest is thermal.

1. Loop current. I = V/R = 6/6 = 1 A.
2. Drift velocity. v_d = I/(nAe) with n = 10²⁹ m^-3, A = 10^-6 m², e = 1.6× 10^-19 C: v_d = 110²⁹· 10^-6· 1.6× 10^-19 = 11.6× 10⁴ ≈ 6.25× 10^-5 m/s.
3. KE per electron. v_d² = (6.25× 10^-5)² = 3.91× 10^-9 m²/s². 12m_e v_d² = 12(9.1× 10^-31)(3.91× 10^-9) ≈ 1.78× 10^-39 J.
4. Total carriers and total KE. Wire volume V_w = AL = (10^-6)(0.1) = 10^-7 m³. Number of electrons N = nV_w = 10²⁹ · 10^-7 = 10²². Total drift KE E = N · 12m_e v_d² ≈ 1.78× 10^-17 J.
5. Dissipation rate. P = RI² = 6 · 1 = 6 W.
6. Time scale. τ = E/P = 1.78× 10^-17/6 ≈ 3× 10^-18 s.
7. Physical interpretation. _{drift KE} ∼ 10^-18 s is six orders of magnitude smaller than the typical Drude collision time _coll ∼ 10^-14 s. So in any single collision interval, the drift KE is exchanged ∼ 10⁶ times — drift KE is negligible compared to thermal energy flowing through the circuit.

E ≈ 1.8× 10^-17 J; τ ≈ 3× 10^-18 s.