NCERT Solutions Class 12 Physics Chapter 3 Current Electricity

Why batteries have internal resistance. Inside any real cell the charge carriers move through the electrolyte, the electrodes, and the connecting tabs - none of which are perfect conductors. The total of these losses is bundled into a single symbolic resistor r drawn in series with the ideal emf source. A car battery's r is small ∼ 0.01–0.5 Ω so it can deliver the huge cranking current to start the engine.

Why you should NEVER short a car battery. A 30 A short through battery terminals dissipates P = I² r = 30²× 0.4 = 360 W inside the battery and a similar amount in any small wire. Tools melt, hydrogen gas can ignite, and the battery can explode. The maximum-current calculation is a theoretical limit, not a recommendation.

Sanity comparison - torch cell. A 1.5 V AA cell has r ≈ 0.5 Ω, so its short-circuit current is ∼ 3 A - ten times less than the car battery despite a comparable r, because the emf is smaller. Cranking current capability scales linearly with emf and inversely with r.

Common mistake. Forgetting that I_max is for R=0. If you accidentally include the external load say a starter motor of 0.05 Ω, the answer drops to 12/(0.05+0.4)=26.7 A. Read the question carefully - "maximum" means worst-case short circuit.

EMF vs terminal voltage. The emf is intrinsic to the battery - it doesn't depend on what is connected outside. The terminal voltage, however, depends on the current being drawn. Two limits help:

• Open circuit I=0: V = ε. A voltmeter across an idle cell reads the emf directly.
• Short circuit R=0: V = 0 and I = ε/r (the maximum).

Everywhere in between, V drops linearly with I. Plotting V vs I and extrapolating to I=0 is how one measures emf without disturbing the circuit.

Why r=3 Ω is "high." Old, partially discharged, or chemistry-unfriendly cells (e.g., zinc–carbon torch cells near end-of-life) acquire r of a few ohms. A fresh AA alkaline has r ∼ 0.2 Ω. When r approaches R, the battery wastes as much power inside itself as it delivers to the load - that's why old batteries make torches dim even though they still register an open-circuit emf close to 1.5 V.

Power balance. Total power from emf: P_tot = ε I = 10× 0.5 = 5 W. Power to the resistor: P_R = I² R = 0.25× 17 = 4.25 W. Power lost in the battery: P_r = I² r = 0.25× 3 = 0.75 W. They add to 5 W exactly - energy conservation holds.

Why heaters are made of high-resistance alloys. A heating element (toaster wire, kettle coil, room heater) needs to dissipate a lot of power as heat. Power dissipated is P = V²/R, so for a fixed mains voltage you'd think low R is better - but then the current I = V/R would blow the fuse. The right balance is moderate R made of an alloy (nichrome, kanthal) that withstands red-hot temperatures without oxidising.

Limits of the linear formula. The relation R(T) = R₀1+-T₀ is a Taylor expansion to first order. It works well for ∼ few hundred degrees range. At very high T, higher-order terms matter and R grows slightly slower than linear. For semiconductors and insulators, α is negative - resistance falls with temperature, because more charge carriers are thermally promoted into the conduction band.

Real-world sanity check. Nichrome heating elements regularly run at 800–1100 the orange-red glow of a toaster coil is around 900 . Our answer 1027 °C is right in the operating range.

Common mistake. Plugging T₁ and T₂ in kelvins gives the same answer because Δ T is the same in both scales - but ONLY for differences. Never mix scales T₁ in Kelvin, T₂ in Celsius, or vice versa.

Resistivity vs conductivity. The reciprocal of resistivity is the conductivity σ: σ = 1ρ. For our wire, σ = 1/2× 10^-7 = 5× 10⁶ S/m (siemens per metre). Copper is ∼ 6× 10⁷ S/m, so this material conducts about 12 times less than copper - consistent with an alloy used for precision resistors where stability is more important than raw conductance.

Why "negligibly small current" matters. If a large current flowed, I² R heating would warm the wire, raising its resistance (α > 0 for metals) and giving a falsely high ρ. Standard practice in resistance measurement is to use a small "sense" current and a 4-wire (Kelvin) probe to eliminate lead-resistance errors.

Microscopic picture (Drude model). Resistivity can be expressed as ρ = m_en e² τ, where n is the free-electron density, τ the mean time between collisions, m_e and e the electron mass and charge. Alloys have lower τ than pure metals because impurity atoms scatter electrons more - explaining their higher ρ.

Common mistake. Mixing units. Make sure A is in m², not mm². Here 6.010^-7 m² = 0.6 mm² - a wire about 0.87 mm in diameter, ordinary jeweller-wire thickness.

Why α is positive for metals. Heating a metal lattice makes the ions vibrate more vigorously. Free electrons collide with these vibrating ions more frequently, shortening their mean free time τ. Since ρ ∝ 1/τ, ρ rises with T - hence α > 0.

Why α is negative for semiconductors. In Si or Ge, the dominant temperature effect is exponential creation of new charge carriers (electron-hole pairs) as temperature rises. This effect overwhelms the lattice-scattering increase, so resistivity DROPS with temperature - useful for thermistors that detect heat by their falling resistance.

Subtle point: α of resistivity vs α of resistance. The question asks for the temperature coefficient of resistivity, but we computed it from resistance. These are equal because the wire's length and area expand only slightly with temperature ∼ 10^-5/ - negligible compared to α ∼ 410^-3, so Δ R/R ≈ Δρ/ρ.

Choice of reference temperature. Textbooks tabulate α at 0 or 20 . Our calculation effectively used 27.5 as the reference. Since α is itself temperature-dependent (slightly), values at different references can differ by ∼ 10%. For high-precision metrology this matters; for engineering, the value we got is fine.

Why the initial current "settles." Three time scales are at play. (i) Electrical transient ∼ microseconds for the current to establish along the wire. (ii) Thermal transient ∼ seconds for the wire to reach steady temperature. (iii) Radiative / convective equilibrium ∼ tens of seconds for the surroundings. The "few seconds" mentioned refers to (ii) - the wire reaches a temperature where the electrical power input equals the heat lost to surroundings.

Steady-state energy balance. P_electrical = V I₂ = 230× 2.8 = 644 W. This is also the power radiated + conducted + convected away to the room at T₂ ≈ 867 . Practically all of it is radiated (Stefan–Boltzmann): P_rad = ε σ A T⁴ - which is why glowing nichrome looks yellow-orange.

Why nichrome (and not copper). Copper has the same α sign but oxidises catastrophically above ∼ 200 . Nichrome (Ni-Cr alloy) forms a protective Cr₂O₃ layer that prevents further oxidation up to 1200 . It also has high ρ, which means a manageable wire length gives the desired resistance.

Common mistake. Many students assume the wire is at supply voltage's "rated" temperature. The supply voltage is fixed 230 V; the wire temperature is whatever the energy balance dictates. If you change the wire's length or thickness, both R and T change.

The Wheatstone-bridge symmetry trick. When a network has the topology R₁/R₃ = R₂/R₄, the bridge resistor carries zero current and can be ignored. Here the ratios 10/5 ≠ 5/10, so the bridge is unbalanced - and indeed I₃ ≠ 0.

Why Kirchhoff and not series-parallel? Series-parallel reduction works only when the network can be unwound into nested combinations. The presence of the bridge resistor makes this impossible, so Kirchhoff's two laws (KCL + KVL) are the systematic way out. For n independent loops you get n equations; for m nodes you get m-1 independent KCL equations. The total exactly matches the number of unknown branch currents.

Sign conventions. Always assign an arrow to each branch BEFORE writing equations. If the algebra gives a negative current, the physical current simply flows opposite to your guessed arrow - no need to redo the algebra.

Total resistance of the network. R_eq = εI_total = 1010/17 = 17 Ω. Useful check: if the bridge were balanced, you would simply have two series paths in parallel - verify your numerical answer against the cleaner geometry to build intuition.

Why terminal voltage in charging EXCEEDS emf. During discharge, V = ε - I r because the internal resistance drops the voltage seen by an external load. During charging, the current is reversed and the internal drop I r ADDS to the emf - you must push at least ε + I r volts against the battery to drive current into it. This is why a charger nominally for a 12 V car battery actually outputs ∼ 14.4 V.

Power flow during charging. Of the 112 V net driving voltage, the battery's internal resistance dissipates I² r = 49× 0.5 = 24.5 W as heat, the series resistor dissipates I² R = 49× 15.5 = 759.5 W as heat, and ε I = 8× 7 = 56 W is stored as chemical energy in the battery. Efficiency = 56/(56+24.5+759.5) ≈ 6.7% - very wasteful! Real chargers use a switched-mode regulator instead of a fixed series resistor.

Why DC, not AC. A battery stores energy by reversible chemical reduction–oxidation. Alternating current would oxidise and reduce the electrodes back and forth without net charging. DC ensures a steady forward push.

Common mistake. Writing V = ε - Ir for the charging case. The sign of I flips, so the formula physically becomes V = ε + Ir. Always set up the Kirchhoff loop equation carefully rather than memorising signed formulas.

The drift-velocity paradox. Electrons drift incredibly slowly ∼ 0.1 mm/s, yet a bulb lights instantly when you flip the switch. How? Because the electric signal - the wave of electric field propagating along the wire - travels at nearly the speed of light. The signal tells every electron along the length to start drifting almost simultaneously. So the bulb has electrons already in its filament ready to start moving the instant you switch on.

Compare drift to thermal speed. Random thermal motion of conduction electrons at room temperature has speed v_th ∼ 10⁶ m/s - about 10¹⁰ times faster than v_d! The drift is a tiny systematic bias on top of this enormous random motion. It's like a stadium where everyone wanders randomly but, on average, drifts very slowly toward the exit.

Where the slowness comes from. The mean free path of an electron in copper is ∼ 40 nm, and the mean free time between collisions ∼ 2.5× 10^-14 s. The drift speed is v_d = eEτ/m_e; a typical field of 0.01 V/m gives v_d ∼ 10^-4 m/s - matching our calculation.

What if you doubled the current? v_d doubles (linear in I) - but the wire would also heat more P ∝ I². For ten times the current, drift becomes ∼ 10^-3 m/s - still pedestrian.

Why the balance condition is so robust. Notice the balance condition involves only the RATIO of resistances and the RATIO of wire lengths - not their absolute values. This makes the metre bridge a brilliant null method: temperature drift in the bridge wire, slight changes in the supply voltage, or aging of the cell all cancel because they affect both sides equally. Null measurements are the gold standard of precision physics.

Best position of the balance point. The bridge is most sensitive - galvanometer deflection per unit imbalance is maximum - when ≈ 50 cm (i.e., the standard resistor Y is chosen so the balance falls near the middle). The data = 39.5 cm is acceptable but slightly off-centre. If fell at 5 cm or 95 cm, the experimenter should pick a different Y and re-measure.

End corrections. Real metre-bridge wires have small unaccounted bits of resistance at the soldered ends called "end corrections" α, β. Precision work calibrates these by swapping X and Y (exactly what part (b) of the question hints at) and averaging the two readings to eliminate the systematic error.

Why the galvanometer–cell swap is invariant. Mathematically, the balance condition follows from setting the determinant of the loop equations to zero. The galvanometer arm and the cell arm play symmetric roles in this determinant - swapping them is a relabelling, not a physical change.

Why V ≈ ε here. Since r ≪ R, only a tiny fraction of the emf is lost inside the cell. In general, the terminal voltage approaches the emf when the load resistance dominates. This is the regime where the cell acts almost like an ideal voltage source.

Maximum power transfer. A classical result: the external load draws maximum power when R = r. At that point P_R,max = ε²/(4r). For this cell, P_max = 4/0.4 = 10 W - forty times more than the 0.25 W we computed. But at matched impedance, the cell loses an equal amount internally, so efficiency is only 50%. Real applications choose R ≫ r to maximise efficiency at the cost of less raw power.

Battery vs solar cell. A modern Li-ion cell has r ∼ 0.05 Ω; a solar cell has r of a few Ω (much "softer" source). The same external load draws very different fractions of available power. This is why solar-charge controllers do MPPT (maximum power point tracking) - they dynamically adjust R to keep it near r.

Heat dissipated inside the battery. P_r = I² r = (0.128)²× 0.1 ≈ 1.6 mW. Tiny - that's why this AA-sized cell stays cool. A 100-A short would dissipate 1 kW inside it and explode.

Why a potentiometer beats a voltmeter. A voltmeter draws a small but non-zero current from the cell under test, slightly reducing its terminal voltage below the true emf. A potentiometer, at balance, draws zero current - making it an "ideal" voltmeter, infinite input impedance. Standard cells (Weston cell, Cd-amalgam) are calibrated this way.

Internal resistance measurement. If you first measure the open-circuit balance length of a test cell, then close a known resistor R across it and measure the new balance length ', the internal resistance follows: r = R - ''. This is a beautifully indirect, error-free way to find r.

What can go wrong. (i) The driver-cell current must remain constant during the experiment - use a rheostat to lock it. (ii) The wire must be uniform - checked at manufacture. (iii) The test cell's emf must be LESS than the total potential drop across the wire, else no balance exists. (iv) Connections at the wire ends contribute "end corrections" - calibrate by swapping the test cell to the other end.

Sanity check. ₂ = 2.25 V corresponds to a lead-acid cell open-circuit emf ∼ 2.1–2.2 V per cell - a plausible match to a real-world standard.

The two speeds living in the wire. Each conduction electron has:

• A huge random thermal speed ∼ 10⁵–10⁶ m/s. This is the same with or without current flow - it's the equilibrium thermal motion.
• A tiny drift velocity ∼ 10^-4 m/s - the net systematic motion in the direction opposite to the electric field (electrons being negative).

Why your light still turns on instantly. Repeating from Q3.9: the electric field itself propagates at near light speed through the wire. Every electron along the length starts drifting almost in unison. So although each electron only travels millimetres per second, the bulb has electrons already inside it that begin drifting the moment the switch closes.

Quantum sidenote. Strictly speaking, at room temperature copper electrons follow a Fermi–Dirac distribution. Their typical speed at the Fermi surface is v_F ∼ 1.57× 10⁶ m/s - even bigger than the classical thermal speed. The drift is still negligible compared to this. So the picture is: relativistic-grade Fermi motion + a barely perceptible drift = the metallic current.

Common mistake. Reporting drift velocity in cm/s without converting. A bigger conceptual error is to assume the drift speed equals the speed at which the bulb lights up - they are entirely different physical quantities.