NCERT Solutions for Class 12 Physics Chapter 3: Current Electricity
All 9 exercise questions of the latest NCERT textbook (2026–27 reprint) solved like a topper’s answer-sheet — emf and internal resistance, Ohm’s law and resistivity, temperature dependence of resistance, drift velocity, charging a battery and Kirchhoff’s rules for a bridge network — with diagrams linked to the exact NCERT pages, comparison tables, complete revision notes and exam tips.
Chapter at a glance — complete revision notes
3.1 Introduction
- In Chapter 1 all charges, free or bound, were considered at rest. Charges in motion constitute an electric current. Currents occur in nature — e.g. lightning, in which charge flows from clouds to the earth through the atmosphere (not steady, with disastrous results) — and in devices (torch, heater, electric motors). Most of our daily technology relies on steady currents driven by cells and batteries. This chapter studies steady currents in conductors and the basic laws governing them.
3.2 Electric current
- Definition: current through a given area = net charge passing per unit time: I = ΔQ/Δt (steady current) or I = dQ/dt. If equal charge crosses in equal times the current is steady. SI unit: ampere (A) = 1 C s⁻¹.
- By convention the direction of current is the direction of flow of positive charge (opposite to electron flow). Current is a scalar (it does not obey the vector law of addition) though drawn with an arrow. To maintain a steady current a closed circuit is required, with an external agency (cell) that moves charge from lower to higher potential energy.
- In a wire, positive and negative charges may both carry current (in electrolytes both ions move); in metals the carriers are electrons, the positive ions being fixed.
3.3 Electric currents in conductors
- An electric charge experiences a force in an electric field, so it accelerates; in a conductor placed in an electric field the free charges move. Imagine charges +Q and −Q placed at the ends of a metallic cylinder (Fig. 3.1): the field they create drives the electrons toward the positive end to neutralise the charges — a momentary current. A steady current needs a continuous supply (a source such as a battery).
- For a conductor the net force from the field is balanced by frequent collisions with ions, so electrons acquire an average drift speed, not unlimited acceleration.
3.4 Ohm’s law
- Ohm’s law (Georg Simon Ohm, 1826): the current I through a conductor is proportional to the voltage across its ends: V ∝ I or V = RI, where R is the resistance (SI unit ohm, Ω = V A⁻¹). It holds at constant temperature for metallic conductors.
- Resistance depends on the geometry: R = ρ l/A (Fig. 3.2: slab of length l and area A); the constant of proportionality ρ is the resistivity — a property of the material (not of shape), unit Ω m. Conductivity σ = 1/ρ (unit S m⁻¹, siemens per metre). Resistances in series add and those in parallel add as reciprocals (Class XII/XI).
- Current density j: the amount of charge flowing per second per unit area normal to the flow: j = I/A; vector j = nq v_d for charge carriers of charge q and number density n. Vector form of Ohm’s law: E = ρ j (or j = σE) — ρ independent of E for an ohmic material. Using E = V/l, j = I/A, V = IR gives R = ρl/A.
3.5 Drift of electrons and the origin of resistivity
- Free electrons in a metal move randomly with high thermal speeds, colliding with ions (Fig. 3.3). With no field the average velocity is zero; the paths are straight lines between collisions. In a field E each electron gets an acceleration a = −eE/m. Between collisions its velocity changes by a τ, where τ is the average time between collisions (relaxation time), so the drift velocity is v_d = −eEτ/m — proportional to E, not growing without limit.
- Current and drift velocity (Fig. 3.4): I = n e A v_d, j = n e v_d (magnitude). Comparing with Ohm’s law gives ρ = m/(n e² τ) and σ = ne²τ/m: resistivity is lower for larger n and τ.
- Example 3.1: copper wire A = 1.0 × 10⁻⁷ m², I = 1.5 A; one conduction electron per atom, density 9.0 × 10³ kg m⁻³, atomic mass 63.5 u ⇒ n ≈ 8.5 × 10²⁸ m⁻³ and v_d ≈ 1.1 mm s⁻¹; compare thermal speeds ~10² m s⁻¹ and the speed of propagation of the field (≈ c).
- Example 3.2 (conceptual): (a) current is established almost instantly because the field is set up throughout the circuit at nearly the speed of light, causing local drift everywhere; (b) electrons accelerate between collisions but lose the drift speed at each collision, so only an average drift speed results; (c) large currents arise because n is enormous (~10²⁹ m⁻³); (d) not all free electrons move in the same direction — the drift is superposed on large random velocities; (e) paths are straight between collisions without a field and, in general, curved with a field.
3.5.1 Mobility
μ = |v_d|/E = qτ/m — drift speed per unit field (SI unit m² V⁻¹ s⁻¹). It is a positive quantity, same for both signs of charge; the current density in terms of mobility: σ = n e μ (for electrons), more generally sum over all carrier types (electrons and holes in semiconductors).
3.6 Limitations of Ohm’s law
- Ohm’s law is not a fundamental law of nature; many materials and devices do not obey it. V = IR is the definition of resistance and applies to every device; Ohm’s law is the statement that R is independent of V (linear I–V). Deviations: (a) V ceases to be proportional to I (Fig. 3.5: V vs I curve for a good conductor at high I — non-linear); (b) the relation between V and I depends on the sign of V for the same absolute value (diode, Fig. 3.6 — conducts in one direction); (c) the V–I relation is non-unique — more than one voltage for the same current (e.g. GaAs, Fig. 3.7). Rectifiers combine (a) and (b).
3.7 Resistivity of various materials
- Materials are classified by resistivity: conductors (metals, ρ ≈ 10⁻⁸ to 10⁻⁶ Ω m), semiconductors (Si, Ge — intermediate), insulators (glass, rubber — 10²² to 10²⁴ times that of metals). The resistivity of metals increases with temperature; that of semiconductors decreases. Carriers are electrons in most substances but ions in ionic crystals and electrolytes. Alloys like manganin, constantan and nichrome have high resistivity and a small temperature coefficient.
3.8 Temperature dependence of resistivity
- In the range where ρ increases linearly with T: ρ_T = ρ₀ [1 + α (T − T₀)], where α is the temperature coefficient of resistivity (SI unit °C⁻¹ or K⁻¹) — fractional increase per degree. Likewise R_T = R₀[1 + α(T − T₀)].
- Copper (Fig. 3.8): ρ increases linearly with T at higher temperatures (not at very low temperatures). Nichrome (Fig. 3.9): exhibits a very weak dependence on temperature. Semiconductor (Fig. 3.10): ρ decreases with rising T (more carriers freed), so α is negative.
- Example 3.3: toaster element — 75.3 Ω at 27 °C; steady current 2.68 A at 230 V ⇒ R = 85.8 Ω; with α = 1.70 × 10⁻⁴ °C⁻¹ the steady temperature is about 847 °C. Example 3.4: platinum resistance thermometer: R₀ = 5 Ω (ice point), R₁₀₀ = 5.23 Ω (steam point), R_t = 5.795 Ω ⇒ t = (R_t − R₀)/(R₁₀₀ − R₀) × 100 ≈ 345.65 °C.
3.9 Electrical energy, power
- A conductor of resistance R carrying current I for time t: work done by the source W = VIt; the power dissipated is P = VI = I²R = V²/R (watt). The energy is dissipated as heat (Joule heating), because electrons gain kinetic energy from the field and lose it to the lattice in collisions. SI unit of energy: joule; commercial unit: kilowatt-hour (1 kWh = 3.6 × 10⁶ J).
- Power transmission: for a fixed power P = VI delivered, the Joule loss in the lines = I²R_line = P²R_line/V²; hence power is transmitted at high voltage (low current) and stepped down near the consumer — a key use of transformers (Chapter 7).
3.10 Cells, emf, internal resistance
- A cell (electrolytic cell with positive terminal P and negative terminal N; Fig. 3.12) maintains a steady potential difference between its terminals by chemical action. In open circuit the terminal potential difference is the emf ε (electromotive force) — not actually a force: the work done by the source per unit charge in taking charge from the lower to the higher potential energy (i.e. from one terminal to the other inside the cell).
- The electrolyte offers internal resistance r to current. When connected to an external resistance R (Fig. 3.11): I = ε/(R + r); terminal voltage V = ε − I r (V_ext = IR = ε R/(R + r)); V < ε while current flows. Maximum current I = ε/r on short-circuit (R = 0). During charging (the cell absorbs current from a higher-voltage supply), V = ε + I r.
3.11 Cells in series and in parallel
- Series: ε_eq = ε₁ + ε₂ (with proper signs; if the negative terminal of one is connected to the positive of the next), r_eq = r₁ + r₂. Parallel (Fig. 3.14): ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂); 1/r_eq = 1/r₁ + 1/r₂ — in general ε_eq/r_eq = Σ εᵢ/rᵢ.
3.12 Kirchhoff’s rules
- Not every circuit can be reduced by series–parallel combinations; Kirchhoff’s rules (1845) handle any network. Junction rule (KCL): at any junction the sum of currents entering equals the sum leaving (conservation of charge; Fig. 3.15). Loop rule (KVL): the algebraic sum of changes in potential around any closed loop is zero (conservation of energy; electrostatic field is conservative).
- Sign conventions: going through a resistor in the direction of current gives −IR (potential drop); against the current +IR; going from − to + terminal of a cell, +ε; from + to −, −ε. Example 3.5: a cube of twelve 1 Ω resistors across a body diagonal: R_eq = 5R/6 = 5/6 Ω, I = 12 A for 10 V, with 4 A, 2 A and 4 A in the successive sets of edges. Example 3.6 solves a bridge-like network with both rules.
3.13 Wheatstone bridge
- Four resistors R₁, R₂, R₃, R₄ arranged in a quadrilateral with a battery across one diagonal and a galvanometer across the other (Fig. 3.18). Applying Kirchhoff’s rules, at the balance (null) point where the galvanometer current is zero: R₁/R₂ = R₃/R₄, which enables the measurement of an unknown resistance.
- Example 3.7: arms AB = 100 Ω, BC = 10 Ω, CD = 5 Ω, DA = 60 Ω, galvanometer 15 Ω across BD, 10 V across AC: the galvanometer current is ≈ 4.87 mA.
Summary & “Points to ponder” (NCERT)
- I = dQ/dt; emf is the potential difference across the terminals in open circuit; Ohm’s law V = RI; R = ρl/A; j = n q v_d, ρ = m/(ne²τ); α = fractional change in resistivity per unit temperature; V = ε − Ir; Kirchhoff’s rules; Wheatstone bridge R₁/R₂ = R₃/R₄.
- Current is a scalar; V = IR defines resistance (applies to every conducting device), Ohm’s law claims R is independent of V; E = ρj gives another statement of Ohm’s law (ρ independent of E); the drift velocity arises only from the applied field (random motion averages to zero); in a neutral conducting wire the net charge density is zero (ρ₊ + ρ₋ = 0), so j = ρv must be applied to each type of carrier separately; Kirchhoff’s junction rule is based on conservation of charge and does not depend on the shape/orientation of the wires.
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?
Concept and context. A cell or battery is a source that maintains a steady potential difference between its terminals by converting chemical energy into electrical energy. Its emf ε (electromotive force) is the potential difference between the terminals in open circuit (no current drawn); it is not a force but the work done per unit charge by the source in moving charge from the lower to the higher potential energy. The electrolyte and electrodes inside the cell offer a resistance to the flow of charge, called the internal resistance r. When a cell is connected to an external resistance R the current is I = ε/(R + r) and the terminal voltage across the cell is V = ε − Ir. The current is therefore largest when the external resistance is as small as possible — R = 0 (short circuit) — and then it is limited only by the internal resistance of the battery.
Given: emf ε = 12 V; internal resistance r = 0.4 Ω.
- Write the circuit equation: for an external resistance R, I = ε/(R + r).
- Condition for maximum current: I is maximum when the denominator is minimum, i.e. when the external resistance R → 0 (the terminals are directly connected by a thick conductor).
- Substitute: I_max = ε/r = 12 V / 0.4 Ω = 30 A.
Imax = ε / r = 12 / 0.4 = 30 A
- Physical significance: in this limit the terminal voltage V = ε − Ir = 12 − 30 × 0.4 = 0 — the whole emf is dropped across the internal resistance and all the chemical energy is dissipated as heat inside the battery (power ε²/r = 360 W), which can overheat or damage it. That is why one must never short-circuit a battery.
- Practical context: a car’s starter motor draws a current of the order of tens to hundreds of amperes for a short time, which a storage (lead–acid) battery can supply only because its internal resistance is very small (a fraction of an ohm). A battery with small r delivers large currents; the heavy cables to the starter motor keep the circuit resistance low.
- The “maximum current” is the value at the extreme of zero external resistance — in practice the value actually drawn depends on the load.
| External resistance R | Current I = ε/(R + r) | Terminal voltage V = ε − Ir |
|---|---|---|
| ∞ (open circuit) | 0 | 12 V (= emf) |
| 2.6 Ω | 4 A | 10.4 V |
| 0.4 Ω | 15 A | 6 V |
| 0 (short circuit) | 30 A (maximum) | 0 V |
Conclusion: the maximum current that can be drawn from the battery is ε/r = 30 A, obtained when the external resistance is zero (terminals short-circuited), at which the terminal voltage falls to zero.
A battery of emf 10 V and internal resistance 3 Ω 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?
Concept and context. For a cell of emf ε and internal resistance r connected across an external resistor R, the same current I flows through R and through r (they are in series), so ε = I(R + r). The potential difference between the terminals of the cell when current flows (the terminal voltage) is less than the emf by the potential drop across the internal resistance: V = ε − Ir = IR. Only in open circuit (I = 0) does the terminal voltage equal the emf.
Given: ε = 10 V; r = 3 Ω; I = 0.5 A.
(i) Resistance of the resistor
- Use I = ε/(R + r) ⇒ R + r = ε/I = 10/0.5 = 20 Ω.
- Therefore R = 20 − r = 20 − 3 = 17 Ω.
(ii) Terminal voltage when the circuit is closed
- V = ε − Ir = 10 − (0.5 × 3) = 10 − 1.5 = 8.5 V.
- Check: V must equal the p.d. across the external resistor: IR = 0.5 × 17 = 8.5 V ✓.
R = 17 Ω ; Vterminal = 8.5 V
| Part of circuit | Resistance | Voltage drop (I = 0.5 A) |
|---|---|---|
| Internal resistance r | 3 Ω | 1.5 V (lost inside the cell) |
| External resistor R | 17 Ω | 8.5 V (= terminal voltage) |
| Total | 20 Ω | 10 V = emf ✓ |
- Why is the terminal voltage less than the emf? Part of the emf (1.5 V here) is used up in driving the current through the internal resistance of the cell; the rest (8.5 V) appears across the external circuit. The larger the current drawn, the lower the terminal voltage — which is why car headlights dim when the starter motor draws a heavy current.
- Energy aspect: power supplied by the cell εI = 5 W; power delivered to R = I²R = 4.25 W; power wasted as heat in the cell I²r = 0.75 W (5 = 4.25 + 0.75 ✓).
Conclusion: the external resistance is 17 Ω and the terminal voltage of the battery with the circuit closed is 8.5 V (the remaining 1.5 V of the 10 V emf being dropped across its internal resistance).
At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10⁻⁴ °C⁻¹.
Concept and context. The resistivity (and hence the resistance) of a metallic conductor increases with temperature because the ions vibrate more strongly and the electrons collide more often (relaxation time τ falls, ρ = m/ne²τ). Over a limited temperature range in which the variation is nearly linear, ρ_T = ρ₀[1 + α(T − T₀)], where α is the temperature coefficient of resistivity, the fractional increase in resistivity per unit rise in temperature (SI unit °C⁻¹ or K⁻¹). Since the dimensions of the wire change negligibly, the same relation holds for resistance: R_T = R₀[1 + α(T − T₀)]. This is how a resistance measurement can be used to find temperature (resistance thermometry).
Given: R₀ = 100 Ω at T₀ = 27.0 °C; R_T = 117 Ω; α = 1.70 × 10⁻⁴ °C⁻¹.
- Write the relation: R_T = R₀[1 + α(T − T₀)].
- Rearrange for the temperature rise: T − T₀ = (R_T − R₀)/(R₀ α) = (117 − 100)/(100 × 1.70 × 10⁻⁴).
- Calculate: numerator 17 Ω; denominator 100 × 1.70 × 10⁻⁴ = 1.70 × 10⁻² Ω °C⁻¹; so T − T₀ = 17/(1.70 × 10⁻²) = 1000 °C.
- Final temperature: T = T₀ + 1000 = 27.0 + 1000 = 1027 °C.
T = 1027 °C
| Quantity | Value |
|---|---|
| R₀ (at 27.0 °C) | 100 Ω |
| R_T | 117 Ω |
| Fractional change ΔR/R₀ | 0.17 (17 %) |
| α | 1.70 × 10⁻⁴ °C⁻¹ |
| Temperature rise ΔT | 1000 °C |
| Final temperature | 1027 °C |
- Remark: a rise of 1000 °C in the temperature increases the resistance by only 17 % because α for such alloys (e.g. nichrome, α ≈ 1.7 × 10⁻⁴ °C⁻¹) is small — one reason these alloys are used in heaters: the heating power does not change much as the element warms up. For pure metals such as copper α ≈ 4 × 10⁻³ °C⁻¹ (about 25 times larger).
- Values like this assume α is constant over the 27 °C – 1027 °C range, which is the approximation (an “averaged” coefficient) stated in the question.
Conclusion: the heating element is at a temperature of 1027 °C when its resistance has risen from 100 Ω to 117 Ω, a rise of 1000 °C above the room temperature of 27.0 °C.
A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10⁻⁷ m², and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Concept and context. The resistance of a conductor of uniform cross-section depends on its dimensions as R = ρ l/A: it is directly proportional to the length l and inversely proportional to the cross-sectional area A. The constant of proportionality ρ is the resistivity — a property of the material (and of temperature) that does not depend on the shape or size of the sample (SI unit ohm-metre, Ω m). Passing only a “negligibly small current” ensures that Joule heating does not raise the temperature, so the measured resistance corresponds to the temperature of the surroundings (the temperature at which ρ is wanted).
Given: l = 15 m; A = 6.0 × 10⁻⁷ m²; R = 5.0 Ω.
- Use R = ρl/A and solve for ρ: ρ = RA/l.
- Substitute: ρ = (5.0 Ω)(6.0 × 10⁻⁷ m²)/(15 m) = (30 × 10⁻⁷)/15 Ω m.
- Simplify: 30/15 = 2, so ρ = 2.0 × 10⁻⁷ Ω m.
ρ = 2.0 × 10⁻⁷ Ω m
| Material class | Typical resistivity (Ω m) | This wire |
|---|---|---|
| Good conductors (Cu, Ag) | 10⁻⁸ to 10⁻⁶ | 2.0 × 10⁻⁷ Ω m — in the metallic/alloy range |
| Semiconductors (Ge, Si) | ~10⁻⁵ to 10⁶ | — |
| Insulators (glass, rubber) | ~10¹⁴ – 10¹⁸ (10²² – 10²⁴ times that of metals) | — |
- Interpretation: the value 2.0 × 10⁻⁷ Ω m lies in the range of metals (10⁻⁸–10⁻⁶ Ω m); it is higher than copper (1.7 × 10⁻⁸ Ω m) and closer to an alloy such as constantan (~5 × 10⁻⁷ Ω m) or iron-like metals (~10⁻⁷ Ω m). The corresponding conductivity is σ = 1/ρ = 5 × 10⁶ S m⁻¹.
- Unit check: Ω × m² / m = Ω m ✓.
Conclusion: the resistivity of the material of the wire at the temperature of the experiment is 2.0 × 10⁻⁷ Ω m, a value typical of metals/alloys.
A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Concept and context. The temperature coefficient of resistivity α is defined as the fractional increase in resistivity per unit rise in temperature in the range where resistivity varies linearly with temperature: α = (ρ_T − ρ₀)/[ρ₀(T − T₀)]. Because the length and cross-section of a metallic wire change only negligibly with temperature, the resistance of the wire varies in the same proportion as resistivity, so the same definition may be written with resistance: α = (R_T − R₀)/[R₀ (T − T₀)]. For metals α is positive (resistance rises with temperature) because the lattice ions vibrate more at higher temperature and scatter the conduction electrons more often, shortening the relaxation time τ.
Given: R₁ = 2.1 Ω at T₁ = 27.5 °C (reference); R₂ = 2.7 Ω at T₂ = 100 °C.
- Linear relation: R₂ = R₁[1 + α(T₂ − T₁)].
- Solve for α: α = (R₂ − R₁)/[R₁(T₂ − T₁)].
- Substitute: R₂ − R₁ = 2.7 − 2.1 = 0.6 Ω; T₂ − T₁ = 100 − 27.5 = 72.5 °C; R₁(T₂ − T₁) = 2.1 × 72.5 = 152.25 Ω °C.
- Calculate: α = 0.6/152.25 = 3.94 × 10⁻³ °C⁻¹ ≈ 3.9 × 10⁻³ °C⁻¹.
α(silver) ≈ 3.9 × 10⁻³ °C⁻¹
| Quantity | Value |
|---|---|
| R at 27.5 °C | 2.1 Ω |
| R at 100 °C | 2.7 Ω |
| Fractional increase ΔR/R₁ | 0.6/2.1 = 0.286 (28.6 %) |
| Temperature interval | 72.5 °C |
| α = (ΔR/R₁)/ΔT | 3.9 × 10⁻³ °C⁻¹ |
- Comparison: this agrees with the tabulated value for silver (≈ 4 × 10⁻³ °C⁻¹) and shows that pure metals have a large positive α, whereas alloys like nichrome (1.7 × 10⁻⁴ °C⁻¹) have a very small one and semiconductors have a negative one.
- Choice of reference: the value of α depends on the reference temperature chosen (here 27.5 °C) because the coefficient is defined per unit resistance at the reference temperature.
Conclusion: the temperature coefficient of resistivity of silver is α ≈ 3.9 × 10⁻³ °C⁻¹.
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 the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10⁻⁴ °C⁻¹.
Concept and context. When a heating element at room temperature is first switched on, its resistance is low, so the initial current is large. As the current flows, Joule heating (power I²R) raises the temperature of the element; since the resistance of the metallic alloy increases with temperature, the current falls. A steady state is reached when the heat produced equals the heat lost to the surroundings — the temperature, resistance and current then become constant. The initial current corresponds to the room temperature (the element has not yet warmed up), while the steady current gives the resistance at the steady temperature. Using Ohm’s law and R_T = R₀[1 + α(T − T₀)] we find the steady temperature (same method as NCERT Example 3.3).
Given: V = 230 V; initial current I₁ = 3.2 A at T₁ = 27.0 °C (room temperature); steady current I₂ = 2.8 A at T₂ (unknown); α = 1.70 × 10⁻⁴ °C⁻¹.
- Resistance at room temperature: R₁ = V/I₁ = 230/3.2 = 71.875 Ω ≈ 71.9 Ω.
- Resistance at the steady temperature: R₂ = V/I₂ = 230/2.8 = 82.143 Ω ≈ 82.1 Ω.
- Use R₂ = R₁[1 + α(T₂ − T₁)]: T₂ − T₁ = (R₂ − R₁)/(R₁α) = (82.143 − 71.875)/(71.875 × 1.70 × 10⁻⁴).
- Calculate: numerator = 10.268 Ω; denominator = 71.875 × 1.70 × 10⁻⁴ = 1.2219 × 10⁻² Ω °C⁻¹; so T₂ − T₁ = 840.3 °C. (Equivalently R₂/R₁ = 3.2/2.8 = 1.1429 so ΔT = 0.1429/1.7 × 10⁻⁴ = 840.3 °C.)
- Steady temperature: T₂ = 27.0 + 840.3 = 867.3 °C ≈ 867 °C.
Tsteady ≈ 867 °C
| Stage | Current | Resistance R = V/I | Temperature |
|---|---|---|---|
| Initial (just switched on) | 3.2 A | 71.9 Ω | 27.0 °C |
| Steady state | 2.8 A | 82.1 Ω | ≈ 867 °C |
- Power check: initial power = 230 × 3.2 = 736 W; steady power = 230 × 2.8 = 644 W — the heater delivers slightly less power once hot.
- Why nichrome? It has a high resistivity (so a short wire gives sufficient resistance), a small temperature coefficient (so the current and power change by only ~12 % over a rise of 840 °C), and a high melting point/oxidation resistance, making it ideal for heaters (NCERT Fig. 3.9: weak temperature dependence).
Conclusion: the steady temperature of the nichrome heating element is about 867 °C, at which its resistance has risen from 71.9 Ω to 82.1 Ω and the current has settled at 2.8 A.
Determine the current in each branch of the network shown in Fig. 3.20.
Concept and context. A network like this (a bridge circuit with a battery) cannot be reduced by simple series–parallel combinations because the arm BD joins the two parallel branches; it is solved with Kirchhoff’s rules: (i) the junction rule — the algebraic sum of currents at any junction is zero (conservation of charge); (ii) the loop rule — the algebraic sum of potential changes around any closed loop is zero (conservation of energy). The network of NCERT Fig. 3.20 has a bridge ABCD with AB = 10 Ω, BC = 5 Ω, AD = 5 Ω, DC = 10 Ω, BD = 5 Ω, and a 10 V battery (negligible internal resistance) connected between A and C through a series resistor of 10 Ω, the positive terminal being on the side of A.
Method 1 — Equivalent resistance, then branch currents
- Potentials: let V_A = V and V_C = 0 (the bridge sits between A and C). Let V_B = x and V_D = y be the unknown potentials of B and D.
- Junction rule at B: current in from A = current out to C + current out to D: (V − x)/10 = x/5 + (x − y)/5 ⇒ V − x = 2x + 2(x − y) ⇒ V = 5x − 2y …(1).
- Junction rule at D: current in from A + current in from B = current out to C: (V − y)/5 + (x − y)/5 = y/10 ⇒ 2(V − y) + 2(x − y) = y ⇒ 2V + 2x = 5y …(2).
- Solve (1) and (2): from (2) y = (2V + 2x)/5. Substituting in (1): V = 5x − (4V + 4x)/5 ⇒ 5V = 25x − 4V − 4x ⇒ 9V = 21x ⇒ x = 3V/7, then y = (2V + 6V/7)/5 = 4V/7.
- Branch currents in terms of V: I_AB = (V − 3V/7)/10 = 2V/35; I_BC = (3V/7)/5 = 3V/35; I_AD = (V − 4V/7)/5 = 3V/35; I_DC = (4V/7)/10 = 2V/35; I_BD = (x − y)/5 = (−V/7)/5 = −V/35, i.e. V/35 flows from D to B.
- Total bridge current and equivalent resistance: I = I_AB + I_AD = 5V/35 = V/7 ⇒ R_bridge = V/I = 7 Ω.
Method 2 — Include the battery and the 10 Ω series resistor
- The 10 Ω resistor is in series with the bridge, so the total resistance of the circuit = 10 + 7 = 17 Ω.
- Total current from the battery: I = 10 V/17 Ω = 10/17 A (≈ 0.59 A).
- Voltage across the bridge: V_AC = I × 7 = 70/17 V. Putting V = 70/17 into the branch-current expressions: I_AB = 2V/35 = 4/17 A; I_BC = 3V/35 = 6/17 A; I_AD = 3V/35 = 6/17 A; I_DC = 2V/35 = 4/17 A; I_BD = V/35 = 2/17 A from D to B (opposite to the B → D direction assumed in the figure, i.e. −2/17 A along B → D).
| Branch | Resistance | Current (A) | Direction |
|---|---|---|---|
| AB | 10 Ω | 4/17 ≈ 0.235 | A → B |
| BC | 5 Ω | 6/17 ≈ 0.353 | B → C |
| AD | 5 Ω | 6/17 ≈ 0.353 | A → D |
| DC | 10 Ω | 4/17 ≈ 0.235 | D → C |
| BD | 5 Ω | 2/17 ≈ 0.118 | D → B (against the sketched B → D arrow) |
| Total (battery, 10 Ω resistor) | 10 + 7 = 17 Ω | 10/17 ≈ 0.588 | out of + terminal into A |
- Verification with Kirchhoff’s rules: junction B: 4/17 (from A) + 2/17 (from D) = 6/17 (to C) ✓; junction D: 6/17 (from A) = 4/17 (to C) + 2/17 (to B) ✓; junction A: 10/17 = 4/17 + 6/17 ✓; loop ABDA (going A → B → D → A, with the current in BD taken as −2/17 A along B → D): −10(4/17) − 5(−2/17) + 5(6/17) = (−40 + 10 + 30)/17 = 0 ✓.
- Why the bridge is unbalanced: AB/BC = 2 whereas AD/DC = 0.5, so potentials of B and D differ (V_B = 3V/7 ≠ V_D = 4V/7) and the “galvanometer” arm BD carries a current. The current direction D → B is as expected because D is at the higher potential.
Conclusion: the currents are I_AB = 4/17 A, I_BC = 6/17 A, I_AD = 6/17 A, I_DC = 4/17 A, I_BD = 2/17 A (from D to B), and the total current drawn from the battery is 10/17 A.
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Concept and context. A storage (secondary) battery can be recharged by passing current through it in the direction opposite to the one in which it delivers current: the external dc source (higher voltage) drives current into the battery’s positive terminal. During charging the battery behaves like a source of emf ε connected against the supply, so the net driving voltage is (V_supply − ε), and the terminal voltage of the battery is higher than its emf: V = ε + Ir (the supply must overcome both the emf and the voltage drop in the internal resistance). While discharging we have the opposite relation, V = ε − Ir.
Given: battery emf ε = 8.0 V; internal resistance r = 0.5 Ω; supply voltage V_s = 120 V; series resistor R = 15.5 Ω.
(i) Terminal voltage of the battery during charging
- Apply the loop rule going around the circuit (supply → resistor → battery): V_s − IR − Ir − ε = 0, because the battery’s emf opposes the supply (the positive terminal of the battery is connected to the positive terminal of the supply).
- Charging current: I = (V_s − ε)/(R + r) = (120 − 8.0)/(15.5 + 0.5) = 112/16 = 7.0 A.
- Terminal voltage: V = ε + Ir = 8.0 + (7.0 × 0.5) = 8.0 + 3.5 = 11.5 V (greater than the emf, since the battery is absorbing current).
- Check: voltage drop in the series resistor = 7 × 15.5 = 108.5 V; 108.5 + 11.5 = 120 V ✓.
(ii) Purpose of the series resistor
- The supply voltage (120 V) is far greater than the emf of the battery (8.0 V) and the internal resistance of the battery (0.5 Ω) is very small. Without the resistor the charging current would be I = (120 − 8)/0.5 = 224 A — an enormous current that would overheat (I²r = 25 kW!) and ruin the battery and the supply.
- The series resistor limits the charging current to a safe, moderate value (7 A) so that the battery is charged steadily and not damaged; most of the surplus voltage (108.5 V) is dropped across the resistor (it dissipates 7² × 15.5 ≈ 760 W as heat — hence a robust rheostat/power resistor is required).
- It also allows the current to be controlled: as the battery charges and its emf rises, the resistor keeps the current from rising uncontrollably; reducing R raises the charging rate, increasing R reduces it.
| Quantity | With 15.5 Ω series resistor | Without the resistor |
|---|---|---|
| Charging current | 7 A | 224 A |
| Terminal voltage ε + Ir | 11.5 V | 120 V (supply) |
| Heat in battery I²r | 24.5 W | ≈ 25 kW |
| Result | safe charging | battery damaged |
Conclusion: the terminal voltage of the battery during charging is 11.5 V (charging current 7 A), and the series resistor is used to limit the charging current to a safe value, preventing the large current (≈ 224 A) that would otherwise flow because the battery’s internal resistance is only 0.5 Ω.
The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10²⁸ 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 × 10⁻⁶ m² and it is carrying a current of 3.0 A.
Concept and context. In a metallic conductor the free electrons move randomly with very large thermal speeds (~10⁵–10⁶ m s⁻¹) which average to zero; when an electric field is applied, the electrons acquire an additional, tiny, average velocity opposite to the field — the drift velocity v_d (the net result of acceleration between collisions with lattice ions, v_d = −eEτ/m). The current is related to drift velocity by I = n e A v_d, where n is the number density of free electrons (~10²⁸–10²⁹ m⁻³ in metals), e the electronic charge and A the area of cross-section. Because n is enormous, a large current needs only a very small drift speed — typically fractions of a millimetre per second.
Given: n = 8.5 × 10²⁸ m⁻³; L = 3.0 m; A = 2.0 × 10⁻⁶ m²; I = 3.0 A; e = 1.6 × 10⁻¹⁹ C.
- Drift speed: from I = neAv_d, v_d = I/(neA).
- Denominator: neA = (8.5 × 10²⁸)(1.6 × 10⁻¹⁹)(2.0 × 10⁻⁶) = 8.5 × 1.6 × 2.0 × 10³ = 27.2 × 10³ = 2.72 × 10⁴ C m⁻¹.
- v_d = 3.0/(2.72 × 10⁴) = 1.1 × 10⁻⁴ m s⁻¹ (0.11 mm s⁻¹).
- Time to drift 3.0 m: t = L/v_d = 3.0/(1.1 × 10⁻⁴) = 2.7 × 10⁴ s.
- Convert: 2.7 × 10⁴ s ÷ 3600 = 7.5 hours (about 7½ hours).
t = L/v_d ≈ 2.7 × 10⁴ s ≈ 7.5 hours
- Surprise! An electron takes about 7½ hours to drift just 3 metres, yet a lamp lights almost instantly when the switch is pressed. NCERT Example 3.2 explains this: the electric field is established throughout the circuit at nearly the speed of light, so electrons everywhere start drifting simultaneously; the current is set up at once without any electron needing to travel from the switch to the lamp.
- Why so slow, yet a large current? Because the number density of free electrons is enormous (~10²⁹ m⁻³): even the minuscule drift of each electron carries a large charge per second across any cross-section.
- Compare speeds: drift ≈ 10⁻⁴ m s⁻¹; thermal speed of electrons ≈ 10⁵ m s⁻¹; speed of propagation of the field ≈ 3 × 10⁸ m s⁻¹.
| Quantity | Value |
|---|---|
| n e A | 2.72 × 10⁴ C m⁻¹ |
| Drift speed v_d = I/neA | 1.1 × 10⁻⁴ m s⁻¹ |
| Time for 3.0 m | 2.7 × 10⁴ s |
| In hours | ≈ 7.5 h |
Conclusion: the drift speed of the conduction electrons is only about 1.1 × 10⁻⁴ m s⁻¹, so an electron takes about 2.7 × 10⁴ s (≈ 7.5 hours) to drift from one end of the 3.0 m wire to the other.