CBSE Class 12 Physics Current Electricity Theory and Examples

CBSE Class 12 Physics Current Electricity Theory and Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Electric current is the rate of flow of charge through a conductor. This Class 12 Physics Current Electricity page gives the theory of electric current, current density and drift velocity, Ohm's law and resistivity, with solved examples after each topic, in the order of the chapter.

1. ELECTRIC CURRENT

(a) When a charge flows in a conductor from one place to the other, then the rate of flow of charge is called electric current (i).
(b) The electric current is measured by 'rate of flow of charge'.

(c) Charge flowing per second from any cross section of the conductor is called electric current, Current i = \( \frac{Charge}{Time} \) = \( \frac{dq}{dt} \). If the flow is uniform, i = \( \frac{q}{t} \).

(d) Unit: Ampere (A). 1 ampere = 1 coulomb/second, i.e. if 1 coulomb of charge flows per second then 1 ampere of current is said to be flowing.
(e) Dimension: [M⁰L⁰T⁰A¹]
(f) If n electrons pass through any cross section in every t seconds then i = \( \frac{ne}{t} \), where e = 1.6 × 10^-19 coulomb.
(g) 1 ampere of current means the flow of 6.25 × 10¹⁸ electrons per second through any cross section of the conductor.
(h) Direction of flow of current is taken to be opposite to the direction of flow of electrons.
(i) Value of the current is the same throughout the conductor, irrespective of the cross section of the conductor at different points.
(j) Net charge in a current carrying conductor is zero at any instant of time.

Note: A current carrying conductor cannot be said to be charged, because in a conductor the current is caused by electrons (free electrons). The number of electrons (negative charge) and protons (positive charge) in a conductor is the same. Hence the net charge in a current carrying conductor is zero.

[Figure: A cylindrical conductor of length l with the + end on the left and the - end on the right, an electric field E drawn to the right along its length, and free electrons inside drawn as circled minus signs with arrows to the left showing the force F on each electron]

(k) Electric field outside a current carrying conductor is zero, but it is non zero inside the conductor and is given by E = − \( \frac{V}{l} \).

Note: The electric field inside a charged conductor is zero, but it is non zero inside a current carrying conductor.

(l) Electric current is a scalar quantity. Although in diagrams we represent current in a wire by an arrow, the arrow simply indicates the direction of flow of positive charges in the wire.

Note: Though electric current needs direction for its representation, yet it is a scalar quantity. It is because the current can be added algebraically. Only scalar quantities can be added algebraically, not the vector quantities.

EXAMPLES BASED ON ELECTRIC CURRENT

Ex.1 If a charge of 1.6 × 10^-19 coulomb flows per second through any cross section of any conductor, the current constitute will be-
(A) 2.56 × 10^-19 A
(B) 6.25 × 10^-19 A
(C) 1.6 × 10^-19 A
(D) 3.2 × 10^-19 A

Answer: (C) 1.6 × 10^-19 A

1. From the definition of current, i = \( \frac{q}{t} \).
2. Here q = 1.6 × 10^-19 C and t = 1 s.
3. i = \( \frac{1.6 \times 10^{-19}}{1} \) = 1.6 × 10^-19 A.
4. In general, if a charge dq flows in time dt, i = \( \frac{dq}{dt} \). Hence, the correct option is (C).

Teacher's Note:
a) Current is charge per unit time, so a charge of 1.6 × 10^-19 C per second is the same number in amperes.
b) Do not multiply by the number of electrons again; the charge is already given.

Ex.2 If charge flowing through a conductor is given by q = 1.5 t² + t. The current caused in t = 2 second will be-
(A) 4 ampere
(B) 5 ampere
(C) 6 ampere
(D) 7 ampere

Answer: (D) 7 ampere

1. i = \( \frac{dq}{dt} \) = \( \frac{d}{dt} \)(1.5 t² + t) = 3t + 1.
2. At t = 2 s, i = 3 × 2 + 1 = 7 A.
3. Hence, the correct option is (D).

Teacher's Note:
a) When charge varies with time, the current at an instant is the derivative dq/dt.
b) Substitute t only after differentiating.

Ex.3 The no. of electrons flowing per second through any cross section of wire, if it carries a current of one ampere, will be -
(A) 2.5 × 10¹⁸
(B) 6.25 × 10¹⁸
(C) 12.5 × 10¹⁸
(D) 5 × 10¹⁸

Answer: (B) 6.25 × 10¹⁸

1. i = \( \frac{q}{t} \) = \( \frac{ne}{t} \) because q = ne (quantization of charge).
2. n = \( \frac{i t}{e} \) = \( \frac{1 \times 1}{1.6 \times 10^{-19}} \) = 6.25 × 10¹⁸.
3. Hence, the correct option is (B).

Teacher's Note:
a) One ampere means 6.25 × 10¹⁸ electrons crossing a section every second.
b) This value is the reciprocal of the electronic charge.

Ex.4 The no. of electron passing through a heater wire in one minute, if it carries a current of 8 ampere, will be-
(A) 2 × 10²⁰
(B) 2 × 10²¹
(C) 3 × 10²⁰
(D) 3 × 10²¹

Answer: (D) 3 × 10²¹

1. n = \( \frac{I t}{e} \) with t = 1 minute = 60 s.
2. n = \( \frac{8 \times 60}{1.6 \times 10^{-19}} \) = 3 × 10²¹.
3. Hence, the correct option is (D).

Teacher's Note:
a) Convert one minute to 60 seconds before substituting.
b) Total charge = It, and dividing by e gives the number of electrons.

Ex.5 In hydrogen atom, the electron moves in an orbit of radius 5 × 10^-11 m with a speed of 2.2 × 10⁶ m/sec. the equivalent current will be -
(A) 1.12 mA
(B) 4.32 mA
(C) 3.32 mA
(D) 7.12 mA

Answer: (A) 1.12 mA

1. Time taken by the electron in 1 revolution is T = \( \frac{2π r}{v} \), where r is the radius of the orbit and v is the speed.
2. Number of revolutions in 1 second, n = \( \frac{1}{T} \) = \( \frac{v}{2π r} \) = \( \frac{2.2 \times 10^{6}}{2π \times 5 \times 10^{-11}} \) ≈ 7 × 10¹⁵ per second.
3. In one revolution a charge of 1.6 × 10^-19 C passes any point of the orbit, so charge per second q = 1.6 × 10^-19 × 7.0 × 10¹⁵ = 1.12 × 10^-3 C.
4. Current I = \( \frac{q}{t} \) = 1.12 × 10^-3 A = 1.12 mA, so the correct option is (A).

Teacher's Note:
a) A revolving charge is equivalent to a current I = e × (frequency of revolution).
b) The frequency is the speed divided by the circumference of the orbit.

Ex.6 A conductor of non-uniform cross-sectional area, has cross-sectional area at three points as A₁ = 2cm², A₂ = 4cm², A₃ = 6cm². If a current of 5 ampere is passed through A₁, the current will give values, when passed through A₂ and A₃ respectively as-
(A) 10 ampere, 15 ampere
(B) 20 ampere, 30 ampere
(C) 2.5 ampere, 1.66 ampere
(D) 5 ampere, 5 ampere

Answer: (D) 5 ampere, 5 ampere

1. In a steady flow, the same charge must cross every section of the conductor in each second (no charge piles up).
2. So the current is the same at A₁, A₂ and A₃.
3. Only the current density j = i/A changes: it is largest at the smallest area.
4. Hence, the correct option is (D).

Teacher's Note:
a) Current does not depend on the cross-sectional area; current density does.
b) Current is a scalar and is conserved along a wire in steady state.

Ex.7 A steady current is flowing in a cylindrical conductor. Is there any electric field within the conductor ? [IIT-82]

Answer: Yes, there is an electric field inside the conductor.
1. Current in a conductor flows only when an electric field established in the conductor applies a force on each free electron.
2. The steady drift of the electrons therefore needs a steady electric field along the conductor, E = \( \frac{V}{l} \).

Teacher's Note:
a) A conductor in electrostatic equilibrium has zero field, but a conductor carrying current is not in equilibrium.
b) The field exerts the force that keeps the electrons drifting against collisions.

Ex.8 An electron moves in a circle of radius 10 cm with a constant speed of 4 × 10⁶ m/s find the electric current at a point on the circle.

Answer: I ≈ 1 × 10^-12 A
1. Consider a point A on the circle. The electron crosses this point once in every revolution.
2. Number of revolutions in one second = \( \frac{4 \times 10^{6}}{2π \times 10 \times 10^{-2}} \) = \( \frac{2}{π} \) × 10⁷.
3. Charge crossing A every second = \( \frac{2}{π} \) × 10⁷ × 1.6 × 10^-19 = \( \frac{3.2}{π} \) × 10^-12 C.
4. By the definition of current, this is the current at the point, I ≈ 1 × 10^-12 A.

Teacher's Note:
a) A single revolving charge gives an average current equal to charge times frequency.
b) Convert the radius from cm to m (10 cm = 0.1 m).

2. CURRENT DENSITY

(a) The current density at a point in a conductor is the ratio of the current at that point in the conductor to the area of cross-section of the conductor at that point.
(b) It is denoted by j, i.e. j = \( \frac{i}{A} \), where i = electric current and A = area of cross section.

Note: Area 'A' is normal to current 'I'. If A is not normal to I, but makes an angle θ with the normal to the current, then j = \( \frac{I}{A_{normal}} \) = \( \frac{I}{A \cos θ} \) ⇒ I = j A cos θ = j · A (dot product of the vectors j and A).

[Figure: A cylinder with current I flowing along its axis through a slanted cross-section marked P and A; angle θ is marked between the normal to the slanted area and the direction of the current]

(c) It is a vector quantity. Its direction is the direction of motion of the positive charges at that point.
(d) Units: ampere/meter² (A/m²).
(e) Dimension: [M⁰L^-2T⁰A].
(f) If n = number of free electrons per unit volume of conductor, A = cross sectional area of conductor, v_d = drift velocity, then I = neAv_d and J = nev_d.

(g) Drift velocity: An applied potential difference does not give an accelerated motion to electrons but simply gives them a small constant velocity (≈ 10^-4 m/s) along the length of the wire towards the end at higher potential. This is called the drift velocity of the electrons.

Note: The speed of random motion of electrons is determined by temperature and is given by \( \frac{1}{2} \) mv² = \( \frac{3}{2} \) kT ⇒ v = \( \sqrt{\frac{3kT}{m}} \), where m is the mass of the electron, T is the absolute temperature and k is the Boltzmann constant.

(h) Electrons collide with the ions of the metal while moving. The average time-interval between two successive collisions is called relaxation-time, denoted by τ. The relation between relaxation time (τ) and drift velocity (v_d) is given by v_d = - \( \frac{e E τ}{m} \) (E is the applied electric field).

EXAMPLES BASED ON CURRENT DENSITY

Ex.9 One end of an aluminium wire, whose diameter is 2.5 mm is welded to one end of a copper wire whose diameter is 2.0 mm. The composite wire carries a steady current of 6.25 A. The current densities in Al and Cu respectively will be -
(A) 127 A/cm², 200 A/cm²
(B) 126 A/cm², 180 A/cm²
(C) 125 A/cm², 160 A/cm²
(D) 125 A/cm², 180 A/cm²

Answer: (A) 127 A/cm², 200 A/cm²

1. J = \( \frac{I}{A} \) = \( \frac{I}{\frac{π}{4} d^2} \).
2. Aluminium (d = 2.5 mm): J = \( \frac{6.25}{\frac{π}{4} \times 2.5^{2} \times 10^{-6}} \) = 1.27 × 10⁶ A/m² = 127 A/cm².
3. Copper (d = 2.0 mm): J = \( \frac{6.25}{\frac{π}{4} \times 2^{2} \times 10^{-6}} \) ≈ 2 × 10⁶ A/m² = 200 A/cm².
4. Hence, the correct option is (A).

Teacher's Note:
a) The same current passes through both wires, so the thinner wire has the larger current density.
b) 1 A/cm² = 10⁴ A/m².

Ex.10 (i) A current of 5 amp. is passing through metallic wire of cross-sectional area 4 × 10^-6 / m², the drift speed of electrons will be (in m/sec) (density of electron per unit volume is 5 × 10²⁶) [Roorkee-91]
(A) 1.56 × 10^-2
(B) 1.6 × 10^-19
(C) 1.56 × 10²
(D) 1.6 × 10¹⁹

Answer: (A) 1.56 × 10^-2

1. I = neAv_d, so v_d = \( \frac{I}{neA} \).
2. v_d = \( \frac{5}{5 \times 10^{26} \times 1.6 \times 10^{-19} \times 4 \times 10^{-6}} \).
3. v_d = \( \frac{5}{320} \) = 1.56 × 10^-2 m/s.
4. Hence, the correct option is (A).

Teacher's Note:
a) Take the area as 4 × 10^-6 m².
b) The product neA is the charge per metre of length that can move.

Ex.11 The diameter of a copper wire is 2mm. If a steady current of 6.25 A is caused by 8.5 × 10²⁸ m^-3 electrons flowing through it. The drift velocity of conduction electrons will be -
(A) 0.25 mm/s
(B) 0.35 mm/s
(C) 0.15 mm/s
(D) 0.45 mm/s

Answer: (C) 0.15 mm/s

1. A = π r² = \( \frac{π}{4} \) × (2 × 10^-3)² = 3.14 × 10^-6 m², so J = \( \frac{6.25}{3.14 \times 10^{-6}} \) ≈ 2 × 10⁶ A/m².
2. v_d = \( \frac{J}{ne} \) = \( \frac{2 \times 10^{6}}{8.5 \times 10^{28} \times 1.6 \times 10^{-19}} \) ≈ 1.47 × 10^-4 m/s.
3. v_d ≈ 0.15 mm/s, so the correct option is (C).

Teacher's Note:
a) The printed working shows 8.5 × 10^-25 and 2.5 in the area; the correct values are 8.5 × 10²⁸ and a diameter of 2 mm, and they give the same result.
b) Drift velocity is of the order of 10^-4 m/s, much smaller than random thermal speed.

Ex.12 A silver wire 1mm diameter carries a charge of 90 coulombs in 1 hour and 15 minutes. Silver contains 5.8 × 10²⁸ free electrons per cm³. The current (in amp.) in wire and drift velocity of the electron will be respectively-
(A) 0.02, 2.69 × 10^-7
(B) 0.03, 3.69 × 10^-7
(C) 3.2, 2.69 × 10^-7
(D) 2.3, 3.69 × 10^-7

Answer: (A) 0.02 A and about 2.7 × 10^-6 m/s

1. t = 1 h 15 min = 4500 s, so i = \( \frac{q}{t} \) = \( \frac{90}{4500} \) = 0.02 A.
2. Radius = 0.5 mm, so J = \( \frac{i}{π r^2} \) = \( \frac{0.02}{π (0.5 \times 10^{-3})^2} \) ≈ 2.55 × 10⁴ A/m².
3. With n = 5.8 × 10²⁸ per m³, v_d = \( \frac{J}{ne} \) = \( \frac{2.55 \times 10^{4}}{5.8 \times 10^{28} \times 1.6 \times 10^{-19}} \) ≈ 2.7 × 10^-6 m/s.
4. The current 0.02 A matches option (A); the drift velocity in the options should read 10^-6 instead of 10^-7.

Teacher's Note:
a) The number density of electrons in silver is 5.8 × 10²⁸ per cubic metre, not per cm³.
b) Option (A) is the only one with the correct current of 0.02 A.

Ex.13 The total momentum of electrons in a straight wire of length ℓ = 1000m carrying a current I = 70A, will be - (in N.s)
(A) 0.40 × 10^-6
(B) 0.20 × 10^-6
(C) 0.80 × 10^-6
(D) 0.16 × 10^-6

Answer: (A) 0.40 × 10^-6

1. I = neAv_d, where v_d is the drift velocity and n the number density of electrons.
2. Total number of electrons N = nAℓ, so total momentum P = N m v_d = (nAℓ) m \( \frac{I}{neA} \) = \( \frac{I ℓ m}{e} \).
3. P = \( \frac{70 \times 1000 \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19}} \) ≈ 0.40 × 10^-6 N s.
4. Hence, the correct option is (A).

Teacher's Note:
a) The result depends only on current, length and the electron charge-to-mass ratio, not on the area.
b) Use m = 9.1 × 10^-31 kg for the electron.

3. OHM'S LAW

(a) If there is no change in the physical state of a conductor (such as temperature) then the ratio of the potential difference applied at its ends and the current flowing through it is constant, i.e. V ∝ I or V = IR, where R is a constant. This is called the 'Electrical resistance' of the conductor.

[Figure: Graph of I on the vertical axis against V on the horizontal axis: a straight line through the origin]

(b) This is true for metallic conductors only which have free electrons.
(c) The law is not applicable for ionized gases, transistors, semi-conductors etc.

[Figure: Four I against V graphs. (a) a straight line through the origin; (b) Diode: a curve that stays near zero current and then rises steeply; (c) Semi conductor: a curve rising upwards with increasing slope; (d) Torch bulb: a curve rising steeply at first and then bending towards the V axis]

(d) Units of resistance: ohm (Ω), 1 ohm = 1 volt / 1 ampere.
(e) Dimensions of resistance: [M¹L²T^-3A^-2].
(f) If L = length of conductor, R = resistance of conductor, A = cross sectional area of conductor perpendicular to current, then R ∝ L and R ∝ \( \frac{1}{A} \), so R = ρ \( \frac{L}{A} \). This constant of proportionality ρ is called 'Resistivity' or 'Specific resistance'.

Note: Effect of stretching a wire on its resistance
(i) If the length of wire is changed, then \( \frac{R_1}{R_2} \) = \( \frac{l_1^2}{l_2^2} \).
(ii) If the radius of wire is changed, then \( \frac{R_1}{R_2} \) = \( \frac{r_2^4}{r_1^4} \).

(g) Units of ρ = Ohm-meter. Dimensions of ρ = [M¹L³T^-3A^-2].
(h) If T = temperature in kelvin, R = R₀ [1 + α (T - T₀)], where R₀ = resistance at temperature T₀ and α = thermal coefficient of resistance. So, as T increases, R increases.
(i) Resistivity is also defined as the ratio of the intensity of the electric field E at any point within the conductor and the current density j at that point: ρ = \( \frac{E}{j} \) or j ∝ E.
(j) Resistivity is a characteristic property of the material of the conductor. It does not depend upon the length, area etc. of the conductor, although it depends on temperature. It increases with increase in temperature.
(k) The value of resistivity is least for conductors and most for insulators.
(l) Inverse of resistivity is called conductance of wire, denoted by (σ): σ = \( \frac{1}{ρ} \).
(m) Unit of conductance: mho.

Important points:
(a) If a conductor is stretched to n times its original length, its new resistance will be n² times the older one.
(b) If x% of change is brought in the length of a wire, its resistance will change by 2x%. This is true for x less than 5 only.
(c) If a conductor is stretched such that its radius is reduced to 1/n^th of its original value, then the resistance will increase n⁴ times; similarly the resistance will decrease n⁴ times if the radius is increased n times by contraction.

Effect of temperature on resistance:

[Figure: Graph of R_t on the vertical axis against t°C on the horizontal axis: a straight line rising from a positive intercept R₀ on the vertical axis, with its dotted extension continuing backwards to the left below the horizontal axis]

(a) R_t = R₀ (1 + α Δt), where R_t = resistance at t°C, R₀ = resistance at 0°C, Δt = change in temperature, and α = temperature coefficient of resistance at 0°C.
α is positive for metals, negative for semi conductors and insulators, and zero for alloys.
(b) R₂ = R₁ [1 + α (t₂ - t₁)]. This formula gives an approximate value.
(c) Resistance of a conductor decreases linearly with temperature and becomes zero at a specific temperature. This temperature is called the critical or transition temperature; the conductor becomes a super conductor at this temperature.
(d) There is no loss of energy in a circuit formed by super conductors. Current passed in a loop formed by a superconductor will continue flowing for infinite time if there is no resistance in the loop.

EXAMPLES BASED ON OHM'S LAW

Ex.14 In a wire of length 4m and diameter 6mm, a current of 120 ampere is passed. The potential difference across the wire is found to be 80 volt. The resistance of wire will be-
(A) 0.15 ohm
(B) 0.25 ohm
(C) 6.660 ohm
(D) none

Answer: (D) none of these (R = 0.67 Ω)

1. By the definition of resistance, R = \( \frac{V}{I} \).
2. R = \( \frac{80}{120} \) = 0.67 Ω.
3. This value does not match options (A), (B) or (C), so the correct option is (D).

Teacher's Note:
a) The printed key (A) 0.15 Ω is obtained only if the potential difference is 18 V, since \( \frac{18}{120} \) = 0.15; the printed working "180/120" does not equal 0.15 either.
b) The length and diameter are not needed to find R from V and I.

Ex.15 Is the formula V = iR true for non-ohmic resistance ?

Answer: Yes, this formula defines resistance and not Ohm's law.
1. For any conductor, the ratio V/i at a given point of its I - V curve is the resistance R, so V = iR can always be written.
2. For a non-ohmic conductor R is not constant; it changes with V (or i), so the I - V graph is not a straight line.

Teacher's Note:
a) Ohm's law means V ∝ i with a constant R, which holds only for ohmic conductors.
b) Diodes, thermistors and bulb filaments are non-ohmic examples.

Ex.16 The resistance between two rectangular faces of a block of dimensions 4cm × 4cm × 10cm of maganin (ρ = 48 × 10^-8 ohm. m) will be-
(A) 4.8 μΩ
(B) 3.8 μΩ
(C) 30 μΩ
(D) 3 μΩ

Answer: (A) 4.8 μΩ

1. Take the two rectangular faces (4 cm × 10 cm), separated by a length l = 4 cm, so the area A = 4 × 10 cm² = 4 × 10^-3 m².
2. R = ρ \( \frac{l}{A} \) = \( \frac{4.8 \times 10^{-7} \times 4 \times 10^{-2}}{4 \times 10 \times 10^{-4}} \).
3. R = 4.8 × 10^-6 Ω = 4.8 μΩ, so the correct option is (A).

Teacher's Note:
a) ρ = 48 × 10^-8 Ω m = 4.8 × 10^-7 Ω m; the printed working writes 4.8 × 10^-8, but the final value is right.
b) Resistance depends on the faces chosen: l is the distance between the faces and A is the face area.

Ex.17 If resistance of a wire formed by 1cc of copper be 2.46Ω. The diameter of wire is 0.32mm, then the specific resistance of wire will be-
(A) 1.59 × 10^-6 ohm.cm
(B) 2.32 × 10^-6 ohm.cm
(C) 3.59 × 10^-6 ohm.cm
(D) 1.59 × 10^-8 ohm.cm

Answer: (A) 1.59 × 10^-6 ohm.cm

1. Radius = 0.016 cm, so the area A = π (0.016)² = 8.04 × 10^-4 cm².
2. Length of wire l = \( \frac{Volume}{Area} \) = \( \frac{1}{3.14 \times (0.016)^2} \) cm.
3. ρ = \( \frac{R A}{l} \) = R A² / V = 2.46 × (3.14 × 0.016²)² / 1 = 1.59 × 10^-6 ohm cm.
4. Hence, the correct option is (A).

Teacher's Note:
a) Specific resistance is the same for any shape of wire formed by the metal.
b) Use volume = A × l to find the length.

Ex.18 A given piece of wire of length ℓ, cross sectional area A and resistance R is stretched uniformly to a wire of length 2ℓ. The new resistance will be-
(A) 2R
(B) 4R
(C) R/2
(D) Remains unchanged

Answer: (B) 4R

1. R = \( \frac{ρ ℓ}{A} \) and R' = \( \frac{ρ (2ℓ)}{A'} \).
2. Volume stays the same: A ℓ = A' (2ℓ), so \( \frac{A}{A'} \) = 2.
3. \( \frac{R'}{R} \) = \( \frac{2ℓ}{ℓ} \) × \( \frac{A}{A'} \) = 2 × 2 = 4, so R' = 4R and the correct option is (B).

Teacher's Note:
a) When the length is doubled, the area halves, so the resistance becomes four times.
b) Use the rule R ∝ ℓ² for a fixed volume.

Ex.19 A given piece of wire of length ℓ, radius r and resistance R is stretched uniformly to a wire of radius (r/2). The new resistances will be -
(A) 2R
(B) 4R
(C) 8R
(D) 16R

Answer: (D) 16R

1. Volume remains unchanged: A ℓ = A' ℓ', so \( \frac{A}{A'} \) = \( \frac{ℓ'}{ℓ} \).
2. R = \( \frac{ρ ℓ}{A} \) and R' = \( \frac{ρ ℓ'}{A'} \), so \( \frac{R'}{R} \) = \( \frac{A}{A'} \) × \( \frac{ℓ'}{ℓ} \) = \( \left( \frac{A}{A'} \right)^2 \).
3. With r' = \( \frac{r}{2} \), \( \frac{A}{A'} \) = \( \frac{π r^2}{π r'^2} \) = 4, so \( \frac{R'}{R} \) = 16.
4. R' = 16R, so the correct option is (D).

Teacher's Note:
a) Resistance varies as 1/r⁴ at constant volume; halving the radius gives 2⁴ = 16 times.
b) The length increases to four times when the area becomes one-fourth.

Ex.20 The potential difference across a wire of 10^-3 cm² cross-sectional area and 50cm length is 2 volt, when a current of 0.25amp exists in wire. Calculate-
(i) field strength in the wire (ii) the current density, and (iii) the conductivity of the metal

Answer: (i) 4 V/m (ii) 2.5 × 10⁶ A/m² (iii) 6.25 × 10⁵ mho/m
1. (i) E = \( \frac{V}{l} \) = \( \frac{2 \text{ volt}}{0.5 \text{ metre}} \) = 4 V/m.
2. (ii) J = \( \frac{i}{A} \) = \( \frac{0.25 \text{ amp}}{10^{-3} \times 10^{-4} \text{ m}^2} \) = 2.5 × 10⁶ A/m².
3. (iii) σ = \( \frac{J}{E} \) = \( \frac{2.5 \times 10^{6}}{4} \) = 6.25 × 10⁵ mho/m.

Teacher's Note:
a) Convert cm² to m²: 10^-3 cm² = 10^-7 m².
b) Conductivity is the reciprocal of resistivity, σ = J/E.

Ex.21 A copper wire is stretched to make, 0.1% longer, the percentage change in its resistance will be -
(A) 0.2
(B) 0.4
(C) 0.8
(D) 1.6

Answer: (A) 0.2

1. R = \( \frac{ρ ℓ}{A} \). With density d and mass m of the wire, A ℓ d = m, so A = \( \frac{m}{ℓ d} \).
2. R = \( \frac{ρ d ℓ^2}{m} \), and \( \frac{ρ d}{m} \) is constant for a given wire.
3. Taking logs and differentiating, \( \frac{dR}{R} \) = 2 \( \frac{dℓ}{ℓ} \).
4. \( \frac{dR}{R} \) × 100 = 2 × 0.1 = 0.2%, so the correct option is (A).

Teacher's Note:
a) For a small change, the percentage change in R is twice the percentage change in length.
b) This holds because volume stays constant on stretching.

Ex.22 There are two wires of the same length and of same material and radii r and 2r. The ratio of their specific resistance is-
(A) 1 : 2
(B) 1 : 1
(C) 1 : 4
(D) 4 : 1

Answer: (B) 1 : 1

1. Specific resistance is a characteristic property of the metal and does not depend on the dimensions of the wire used.
2. Both wires are of the same material, so their specific resistances are equal and the ratio is 1 : 1.

Teacher's Note:
a) Resistance changes with radius, but resistivity does not.
b) Do not confuse resistance (1 : 4 here) with resistivity.

Ex.23 The resistance of wire is 50π then the graph between log v and log I is-
(A) straight line passing through origin
(B) parabola
(C) hyperbola
(D) none of the above

Answer: (D) none of the above

1. V = IR, so log V = log I + log R.
2. This is a straight line of slope 1 with the constant log R as its intercept, so it does not pass through the origin.
3. It is none of the options given, so the correct option is (D).

Teacher's Note:
a) The graph is a straight line but with a non-zero intercept log R.
b) The graph of V against I (not logarithms) would pass through the origin.

Ex.24 The current voltage graph for a given metallic wire at two different temperatures T₁ and T₂ are shown in fig. Which is true-
(A) T₁ = T₂
(B) T₁ > T₂
(C) T₁ < T₂
(D) None

[Figure: Graph of I on the vertical axis against V on the horizontal axis with two straight lines from the origin: the steeper one is labelled T₁ and the less steep one is labelled T₂]

Answer: (C) T₁ < T₂

1. The slope of the I - V graph is \( \frac{I}{V} \) = \( \frac{1}{R} \).
2. The slope at temperature T₁ is greater than that at T₂, so the resistance at T₂ is greater than the resistance at T₁.
3. For a metallic wire resistance increases with temperature, hence T₂ is greater than T₁ and the correct option is (C).

Teacher's Note:
a) A steeper I - V line means a smaller resistance.
b) For metals R rises with temperature, so the hotter wire has the flatter line.

Ex.25 The resistance of a tungsten filament at 150º C is 133 ohm. Its resistance at 500º C will be - (The temperature coefficient of resistance of tungsten is 0.0045 per ºC)
(A) 257 Ω
(B) 79 Ω
(C) 50 Ω
(D) none of these

Answer: (A) 257 Ω (approximately)

1. If the resistance at 0°C is R₀ and at t°C is R_t, then R_t = R₀ (1 + αt), so R₀ = \( \frac{R_t}{1 + α t} \).
2. R₀ = \( \frac{133}{1 + (0.0045) \times 150} \) = 79.0 Ω.
3. R₅₀₀ = R₀ (1 + α t₅₀₀) = 79.0 [1 + (0.0045) × 500] = 257 Ω.
4. Hence, the correct option is (A).

Teacher's Note:
a) Without rounding the intermediate step, the result is 258 Ω, which is still closest to option (A).
b) Always bring the resistance to 0°C first, because α is defined with respect to R₀.

Ex.26 The resistance of a conductor at 20º C is 3.15 ohm and at 100º C is 3.75 ohm. Determine the temperature coefficient of resistance of the conductor. The resistance of the conductor at 0º C will be-
(A) 2Ω
(B) 3Ω
(C) 6Ω
(D) 5Ω

Answer: (B) 3 Ω (and α = 0.0025 per °C)

1. R₁ = R₀ (1 + αt₁) and R₂ = R₀ (1 + αt₂).
2. On dividing, \( \frac{R_1}{R_2} \) = \( \frac{1 + α t_1}{1 + α t_2} \), so α = \( \frac{R_2 - R_1}{R_1 t_2 - R_2 t_1} \).
3. α = \( \frac{3.75 - 3.15}{(3.15 \times 100) - (3.75 \times 20)} \) = \( \frac{0.60}{240} \) = 0.0025 per °C.
4. R₀ = \( \frac{R_1}{1 + α t_1} \) = \( \frac{3.15}{1 + (0.0025) \times 20} \) = 3.0 Ω, so the correct option is (B).

Teacher's Note:
a) The printed key marks (C), but the working gives 3.0 Ω, which is option (B).
b) The printed working writes t₂ = 1000°C by mistake; t₂ = 100°C.

4. COMBINATION OF RESISTANCES

(a) Series Combination

[Figure: Three resistors R₁, R₂ and R₃ connected in series in a closed loop with a battery of voltage V; the voltages across the resistors are V₁, V₂ and V₃ and the current I flows in the loop]

(i) The same current passes through each resistance.
(ii) Voltage across each resistance is directly proportional to its value: V₁ = IR₁, V₂ = IR₂.
(iii) The sum of the voltages across the resistances is equal to the voltage applied across the circuit, i.e. V = V₁ + V₂ + V₃ + ......, V = IR₁ + IR₂ + IR₃ + ......, \( \frac{V}{I} \) = R₁ + R₂ + R₃ + ...... = R, where R = equivalent resistance.

Note: If n resistances (each R) are connected in series, the resultant will be nR.

(iv) For a series combination of two resistances:

[Figure: Two resistors R₁ and R₂ in series with a battery of voltage V]

(A) Equivalent resistance R = R₁ + R₂
(B) I = \( \frac{V}{R_1 + R_2} \)
(C) V₁ (voltage across R₁) = IR₁ = \( \frac{R_1 V}{R_1 + R_2} \)
(D) V₂ (voltage across R₂) = IR₂ = \( \frac{R_2 V}{R_1 + R_2} \)

(b) PARALLEL COMBINATION:

[Figure: Three resistors R₁, R₂ and R₃ connected in parallel between points A and B, carrying currents i₁, i₂ and i₃ from a total current i, with a battery V across A and B]

(i) There is the same drop of potential across each resistance.
(ii) Current in each resistance is inversely proportional to the value of the resistance, i.e. i₁ = \( \frac{V}{R_1} \), i₂ = \( \frac{V}{R_2} \), i₃ = \( \frac{V}{R_3} \) etc.
(iii) Current flowing in the circuit is the sum of the currents in the individual resistances, i.e. i = i₁ + i₂ + i₃, i = \( \frac{V}{R_1} \) + \( \frac{V}{R_2} \) + \( \frac{V}{R_3} \) + ........ ⇒ \( \frac{i}{V} \) = \( \frac{1}{R} \) = \( \frac{1}{R_1} \) + \( \frac{1}{R_2} \) + \( \frac{1}{R_3} \) + ...., where R = equivalent resistance.

Note:
(a) You may be asked to find R and not \( \frac{1}{R} \) in the question, so be careful.
(b) The equivalent resistance of a parallel combination is lower than the value of the lowest resistance in the combination.

(c) For a parallel combination of two resistances:

[Figure: Two resistors R₁ and R₂ in parallel, carrying currents i₁ and i₂ from a total current i, connected across a battery V]

(i) i = i₁ + i₂ = \( \frac{V (R_1 + R_2)}{R_1 R_2} \)

Note:
(i) If n resistances (each R) are connected in parallel, their resultant will be R/n.
(ii) If n resistances are connected in series and parallel respectively, the ratio of their resultants will be nR : R/n = n².

EXAMPLES BASED ON COMBINATION OF RESISTANCES

Ex.27 Two wires of the same material having lengths in the ratio of 1 : 2 and diameters in the ratio 2 : 1 are connected with a cell of 6 volt and internal resistance 1Ω. The ratio of the potential difference across the two wires will be-
(A) 1 : 2
(B) 2 : 1
(C) 1 : 8
(D) 8 : 1

Answer: (C) 1 : 8

1. The two wires are in series with the cell, so the same current flows in both.
2. \( \frac{V_1}{V_2} \) = \( \frac{I R_1}{I R_2} \) = \( \frac{R_1}{R_2} \) = \( \frac{ℓ_1}{ℓ_2} \) × \( \frac{π r_2^2}{π r_1^2} \).
3. \( \frac{ℓ_1}{ℓ_2} \) = \( \frac{1}{2} \) and \( \frac{r_2}{r_1} \) = \( \frac{1}{2} \), so \( \frac{V_1}{V_2} \) = \( \frac{1}{2} \) × \( \frac{1}{4} \) = \( \frac{1}{8} \).
4. V₁ : V₂ = 1 : 8, so the correct option is (C).

Teacher's Note:
a) In a series circuit the potential difference is proportional to resistance.
b) Resistance varies directly with length and inversely with the square of the diameter.

Ex.28 Resistance R, (R + 1), (R + 2) ......... (R + n) Ω are connected in series, their resultant resistance will be -
(A) (n + 1) \( \left[ R + \frac{n}{2} \right] \)
(B) (n - 1) \( \left[ R - \frac{n}{2} \right] \)
(C) n (R + n)
(D) n (R - n)

Answer: (A) (n + 1) [R + n/2]

1. There are (n + 1) terms in the series, and the resultant is their sum R' = R + (R + 1) + (R + 2) + ........ + (R + n).
2. This is an arithmetic series with sum S = \( \frac{N}{2} \) [2a + (N - 1)d], where a = R, d = 1 and N = n + 1.
3. R' = \( \frac{n + 1}{2} \) [2R + (n + 1) - 1] = \( \frac{n + 1}{2} \) (2R + n) = (n + 1) \( \left[ R + \frac{n}{2} \right] \).
4. Hence, the correct option is (A).

Teacher's Note:
a) Count the terms carefully: from R to R + n there are n + 1 terms.
b) Series resistances simply add.

Ex.29 The Resistance 4R, 16R, 64R, ............∞ are connected in series, their resultant will be-
(A) 0
(B) ∞
(C) 4/3 R
(D) 3/4 R

Answer: (B) ∞

1. In series, R' = 4R + 16R + 64R + ........ ∞.
2. Each term is larger than the previous one, so the sum keeps growing without limit.
3. R' = ∞, so the correct option is (B).

Teacher's Note:
a) A geometric series with ratio 4 (greater than 1) diverges.
b) Compare with Ex.30, where the ratio 1/2 gives a finite sum.

Ex.30 Resistance R, 2R, 4R, 8R ..................∞ are connected in parallel. Their resultant resistance will be -
(A) R
(B) R/2
(C) 0
(D) ∞

Answer: (B) R/2

1. Let R' be the resultant resistance. Then \( \frac{1}{R'} \) = \( \frac{1}{R} \) + \( \frac{1}{2R} \) + \( \frac{1}{4R} \) + ........ ∞.
2. \( \frac{1}{R'} \) = \( \frac{1}{R} \) [1 + \( \frac{1}{2} \) + \( \frac{1}{4} \) + ........ ∞].
3. The bracket is a geometric series with sum \( \frac{1}{1 - \frac{1}{2}} \) = 2, so \( \frac{1}{R'} \) = \( \frac{2}{R} \).
4. R' = \( \frac{R}{2} \), so the correct option is (B).

Teacher's Note:
a) The sum of an infinite G.P. with first term 1 and ratio 1/2 is 2.
b) The result must be less than the smallest resistor R, as expected for a parallel combination.

Ex.31 The resistance of two conductors in series is 40Ω and their resistance becomes 7.5Ω when connected in parallel. The resistances are-
(A) R₁ = 30Ω, R₂ = 40Ω.
(B) R₁ = 20Ω, R₂ = 30Ω.
(C) R₁ = 30Ω, R₂ = 10Ω.
(D) R₁ = 20Ω, R₂ = 20Ω.

Answer: (C) R₁ = 30Ω, R₂ = 10Ω

1. In series, R₁ + R₂ = 40 Ω ......(A)
2. In parallel, \( \frac{R_1 R_2}{R_1 + R_2} \) = 7.5 Ω, so R₁R₂ = 7.5 × 40 = 300.
3. (R₁ - R₂)² = (R₁ + R₂)² - 4R₁R₂ = 40² - 1200 = 400, so R₁ - R₂ = 20 ......(B)
4. Solving (A) and (B), R₁ = 30 Ω and R₂ = 10 Ω, so the correct option is (C).

Teacher's Note:
a) The sum and product of the two resistances determine them as roots of x² - 40x + 300 = 0.
b) Check: 30 + 10 = 40 and 30 × 10 / 40 = 7.5.

Ex.32 In the following figure the Resistance between A and C will be-
(A) R \( \left( \frac{n^2 + 1}{n} \right) \)
(B) R \( \left( \frac{n + 1}{n} \right) \)
(C) R \( \left( \frac{n^2 - 1}{n} \right) \)
(D) R \( \left( \frac{n - 1}{n} \right) \)

[Figure: Three resistors R in series from A to B (continued up to n resistors), then from B to C a group of resistors R in parallel (up to n resistors)]

Answer: (A) R \( \left( \frac{n^2 + 1}{n} \right) \)

1. Between A and B, n resistances (each R) are in series, so R₁ = nR.
2. Between B and C, n resistances (each R) are in parallel, so R₂ = \( \frac{R}{n} \).
3. The two groups are in series: R' = R₁ + R₂ = nR + \( \frac{R}{n} \) = R \( \left( \frac{n^2 + 1}{n} \right) \).
4. Hence, the correct option is (A).

Teacher's Note:
a) Series gives nR and parallel gives R/n; add the two groups.
b) For n = 1 the result is 2R, which checks with two resistors in series.

Ex.33 In the following fig, the current drawn by the battery of 12V supply (in amp) will be -
(A) 6 (√3 - 1)
(B) 6 (√3 + 1)
(C) 12 (√3 - 1)
(D) 12 (√3 + 1)

[Figure: An infinite ladder network across a 12 V supply with terminals +A at the top and -B at the bottom: each section has a 1 Ω resistor in the top rail, a 1 Ω resistor as a rung joining the rails and a 1 Ω resistor in the bottom rail, repeating to infinity]

Answer: (A) 6 (√3 - 1)

1. Let x be the resultant resistance of the infinite ladder. Adding one more section leaves x unchanged.
2. Each section has 1 Ω in the top rail, 1 Ω in the bottom rail and a 1 Ω rung in parallel with the rest of the ladder, so x = 1 + \( \frac{1 \cdot x}{1 + x} \) + 1 = 2 + \( \frac{x}{1 + x} \).
3. This gives x² - 2x - 2 = 0, so x = 1 + √3 Ω.
4. Current I = \( \frac{V}{R} \) = \( \frac{12}{1 + \sqrt{3}} \) = 6 (√3 - 1) A, so the correct option is (A).

Teacher's Note:
a) An infinite network repeats itself, so the resistance with one extra section equals the original resistance.
b) Reject the negative root of the quadratic.

Ex.34 A wire of resistance 10Ω is bent to form a complete circle. Its resistance between two diametrically opposite points will be (in Ω)-
(A) 3.5
(B) 5
(C) 2.5
(D) 1.5

[Figure: A circular wire with points A and C diametrically opposite; the upper half and the lower half are each marked 5 Ω and drawn in parallel between A and C]

Answer: (C) 2.5

1. The two diametrically opposite points A and C divide the circle into two semicircles, each of resistance 5 Ω.
2. The two halves are in parallel between A and C.
3. R = \( \frac{5 \times 5}{5 + 5} \) = 2.5 Ω, so the correct option is (C).

Teacher's Note:
a) Resistance of a uniform wire is proportional to its length, so half the circle has half the resistance.
b) The parallel combination is always less than either half.

Ex.35 Find the equivalent resistance and hence current supplied by the battery in the diagram shown -

[Figure: A 6 V battery between a and b; the top wire is a - 4 Ω - c - 4 Ω - e - 4 Ω - g; vertical resistors join c to d (3 Ω), e to f (3 Ω) and g to h (2 Ω); the bottom wire joins b, d, f and h]

Answer: R_eq = 6 Ω and the current from the battery = 1 A
1. The 4 Ω resistor between e and g is in series with the 2 Ω resistor, giving 6 Ω across e and f.
2. This 6 Ω is in parallel with the 3 Ω at e: \( \frac{6 \times 3}{6 + 3} \) = 2 Ω. In series with the 4 Ω between c and e this gives 6 Ω.
3. This 6 Ω is in parallel with the 3 Ω at c: \( \frac{6 \times 3}{6 + 3} \) = 2 Ω. In series with the 4 Ω between a and c this gives 6 Ω.
4. R_eq = 6 Ω, so the current supplied by the battery = \( \frac{6 \text{ V}}{6 \text{ Ω}} \) = 1 A.

Teacher's Note:
a) Reduce the network from the far end towards the battery.
b) Series elements add and parallel elements combine by the product-over-sum rule.

Ex36 Find effective resistance between points A and B in figure.

[Figure: Points A and B at the bottom. A is joined to B by a 3 Ω resistor. A is joined to E, D and C (the top vertices) by 6 Ω resistors, and also to E through a left corner by two 3 Ω resistors in series. Along the top, E to D is 3 Ω, D to C is 3 Ω, and C to B is 3 Ω]

Answer: R_AB = 2 Ω
1. The two 3 Ω resistors joined through the left corner are in series, giving 6 Ω between A and E. This is in parallel with the 6 Ω resistor from A to E: 3 Ω.
2. A to D through E: 3 + 3 = 6 Ω, in parallel with the 6 Ω from A to D: 3 Ω.
3. A to C through D: 3 + 3 = 6 Ω, in parallel with the 6 Ω from A to C: 3 Ω.
4. A to B through C: 3 + 3 = 6 Ω, in parallel with the 3 Ω direct from A to B: \( \frac{6 \times 3}{6 + 3} \) = 2 Ω.

Teacher's Note:
a) Reduce the network step by step starting from the point nearest to E.
b) The final answer is less than the smallest direct resistor of 3 Ω, as expected.

5. COMPARATIVE STUDY OF COMBINATION OF RESISTANCES AND CAPACITORS

6. KIRCHHOFF'S LAWS

Kirchhoff in 1842 gave two laws for solving complicated electrical circuits. These laws are as follows -

(a) First law: In an electrical circuit, the algebraic sum of the currents meeting at any junction in the circuit is zero.
OR
Sum of the currents entering the junction is equal to the sum of the currents leaving the junction.
⇒ Σi = 0
i₁ - i₂ - i₃ - i₄ + i₅ = 0 or i₁ + i₅ = i₂ + i₃ + i₄

Note: This law is based on the law of conservation of charge. In other words, when a steady current flows in a circuit then there is neither accumulation of charge at any point in the circuit nor any charge is removed from there.

(b) Second law: In a 'closed' mesh of a circuit the algebraic sum of the products of the current and the resistance in each part of the mesh is equal to the algebraic sum of the e.m.f.'s in that mesh, i.e. ΣiR = ΣE.

Important notes:
(a) In applying this law, when we traverse in the direction of current then the product of the current and the corresponding resistance is taken as positive, and the emf is taken as positive when we traverse from the negative to the positive electrode of the cell through the electrolyte.
(b) This law is based on the 'law of conservation of energy'.

EXAMPLES BASED ON KIRCHHOFF'S LAWS

Ex.36 In the following figure the 5 ohm-resistance develops heat 10.24 cal's due to a current flowing through it. Calculate -
(i) heat developed per second in the 2Ω resistance and
(ii) the potential difference across 6Ω resistance.

[Figure: A current i enters at A and passes through a 2 Ω resistor to point C. From C the circuit divides into an upper branch with 6 Ω and 9 Ω in series (current i₁) and a lower branch with a 5 Ω resistor (current i₂), meeting again at D and leaving at B]

Answer: (i) About 7.3 cal/s (ii) About 5.87 V
1. Let the main current be i. At C it divides into i₁ (through 6 Ω and 9 Ω in series, 15 Ω) and i₂ (through 5 Ω).
2. The p.d. between C and D is the same: i₁ × 15 = i₂ × 5, so \( \frac{i_1}{i_2} \) = \( \frac{5}{15} \) = \( \frac{1}{3} \), i.e. i₂ = 3i₁.
3. As i = i₁ + i₂, i₁ = \( \frac{i}{4} \) and i₂ = \( \frac{3i}{4} \).
4. Heat in 5 Ω per second = \( \frac{i_2^2 R}{4.2} \) = \( \frac{(3i/4)^2 \times 5}{4.2} \) = 10.24 cal/s, so (3i/4)² = 8.60, 3i/4 = 2.93 A and i = 3.91 A.
5. (i) Heat in 2 Ω per second = \( \frac{i^2 R}{4.2} \) = \( \frac{(3.91)^2 \times 2}{4.2} \) = 7.28 cal/s.
6. (ii) Potential difference across 6 Ω = i₁ × 6 = \( \frac{3.91}{4} \) × 6 = 5.87 V.

Teacher's Note:
a) The printed values 3.92 A, 7.31 cal/s and 5.86 V differ slightly because of rounding; the values above follow from the data.
b) The p.d. across 9 Ω is i₁ × 9 = 8.8 V.
c) Divide the heat in joules by 4.2 to express it in calories.

Ex.37 The value of current i in the following circuit is-
(A) 2.7 A
(B) 3.7 A
(C) 3 A
(D) 4 A

[Figure: A wire with junctions A, B and C in a row: 4 A and 3 A flow into A, 2 A flows out of B, 1.3 A flows out of C, and a current i flows out of C to the right]

Answer: (B) 3.7 A

1. Apply Kirchhoff's first law at the junctions A, B and C taken together: current entering = current leaving.
2. Entering: 4 A and 3 A at A. Leaving: 2 A at B, 1.3 A at C and i.
3. 4 + 3 = 2 + 1.3 + i, so i = 7 - 3.3 = 3.7 A.
4. Hence, the correct option is (B).

Teacher's Note:
a) The sum of currents into the whole junction region equals the sum of currents out of it.
b) Mark every arrow as entering or leaving before adding.

Ex.38 Write KCL (first law) and KVL (second law) for the circuit shown below :

[Figure: Two nodes A (left) and B (right) joined by three parallel branches. The top branch has a cell E₁ and a resistor R₁ with current i₁; the middle branch has a cell E₂ and a resistor R₂ with current i₂; the bottom branch has a resistor R₃ with current i₃. Loop 1 (between the top and middle branches) and loop 2 (between the middle and bottom branches) are marked with circular arrows]

Answer: KCL at A: i₁ + i₂ + i₃ = 0. KVL for loop 1: i₁R₁ - i₂R₂ = E₁ - E₂. KVL for loop 2: i₂R₂ - i₃R₃ = E₂.
1. All three branch currents i₁, i₂, i₃ are marked in the same sense towards the junction A, so KCL at A gives i₁ + i₂ + i₃ = 0.
2. Loop 1 (top and middle branches, traversed in the direction shown): i₁R₁ - i₂R₂ = E₁ - E₂.
3. Loop 2 (middle and bottom branches): i₂R₂ - i₃R₃ = E₂.

Teacher's Note:
a) The signs in the loop equations depend on the loop direction and on the polarity of each cell.
b) Three unknown currents need three independent equations, which these provide.

Ex.39 Kirchhoff's first and second laws respectively show the conservation of
(A) Charge and energy
(B) Energy and charge
(C) Mass and charge.
(D) Charge and mass

Answer: (A) Charge and energy

1. The first law (junction rule) says that charge does not accumulate at a junction, which is conservation of charge.
2. The second law (loop rule) says the net change in potential around a closed loop is zero, which is conservation of energy.
3. Hence, the correct option is (A).

Teacher's Note:
a) Junction rule: charge. Loop rule: energy.
b) Remember the order in the question: first law, then second law.

Ex.40 The potential difference between points a and b in the circuit of figure is -
(i₁ = 1.0 A, i₃ = 0.6 A)
(A) 1.4V
(B) 4.6V
(C) 3.6V
(D) 5.6V

[Figure: A circuit with three branches between the top node a and the bottom node b: a left branch with cell E₁ and resistor R carrying i₁ downwards, a middle branch with R₂ = 3.5 Ω and cell E₂ = 6 V carrying i₂ downwards, and a right branch with cell E₂ and resistor R₁ carrying i₃ towards a along the top wire]

Answer: (B) 4.6 V

1. At junction a, i₃ = i₁ + i₂, so i₂ = 0.6 - 1.0 = -0.4 A (it actually flows in the direction opposite to that marked).
2. In going from a (potential V_a) to b (potential V_b) along the middle branch: V_a - i₂R₂ - E₂ = V_b.
3. V_a - V_b = E₂ + i₂R₂ = 6 + (-0.4)(3.5) = 6 - 1.4 = 4.6 V.
4. Hence, the correct option is (B).

Teacher's Note:
a) A negative current means the actual direction is opposite to the assumed one.
b) Keep the sign of i₂ when substituting in the loop equation.

Ex.41 The steady state current in a 2Ω resistor shown in fig will be - (The internal resistance of the battery is negligible and the capacitance of the condenser C is 0.2μF)
(A) 1.5 A
(B) 0.9 A
(C) 1.2 A
(D) 1.3 A

[Figure: A 6 V battery with a 2.8 Ω resistor in series, feeding a parallel group: a 2 Ω resistor and a 3 Ω resistor (points p and q at their ends, currents i₁ and i₂) and a branch with a capacitor C and a 4 Ω resistor]

Answer: (B) 0.9 A

1. In steady state the branch containing the capacitor acts as an open circuit, since a capacitor offers infinite resistance to d.c. It simply collects charge, so no current flows in the 4 Ω resistor.
2. The effective resistance of the 2 Ω and 3 Ω resistors in parallel is R' = \( \frac{2 \times 3}{2 + 3} \) = \( \frac{6}{5} \) = 1.2 Ω.
3. Total resistance = 1.2 + 2.8 = 4 Ω, so the current drawn from the cell i = \( \frac{E}{R} \) = \( \frac{6}{4} \) = 1.5 A.
4. Potential difference across pq = 1.5 × 1.2 = 1.8 V, so the current in the 2 Ω resistor i₁ = \( \frac{1.8}{2} \) = 0.9 A.
5. Hence, the correct option is (B).

Teacher's Note:
a) A capacitor blocks steady (d.c.) current once it is fully charged.
b) The 2 Ω and 3 Ω resistors share the same p.d. because they are in parallel.

Ex.42 From the fig. determine
(i) potential at A,
(ii) potential at C, and
(iii) reading of the voltmeter connected across the 10V battery -

[Figure: A rectangular circuit ABCDFG: A at the top left, B at the top middle, C at the top right, D at the bottom right and F at the bottom left, with G on the bottom wire connected to earth. A 10 V cell with 1 Ω internal resistance lies between A and B, a 16 V cell with 0.5 Ω internal resistance lies between B and C, a 0.5 Ω resistor lies between C and D, and a 4 Ω resistor lies between A and F. The current i flows from A down through the 4 Ω resistor]

Answer: (i) V_A = 4 V (ii) V_C = -0.5 V (iii) 11 V
1. Consider the loop CBAFGDC. The 16 V cell drives the current and the 10 V cell opposes it: I = \( \frac{E_2 - E_1}{r_1 + r_2 + R_1 + R_2} \) = \( \frac{16 - 10}{1 + 0.5 + 4 + 0.5} \) = \( \frac{6}{6} \) = 1 A.
2. (i) V_A - V_F = IR = 4 V. Since F is grounded, V_F = 0, so V_A = 4 V.
3. (ii) V_D - V_C = 1 × 0.5 = 0.5 V. Since D is grounded, V_D = 0, so V_C = -0.5 V.
4. (iii) The 10 V battery is being charged, so the voltmeter across it reads V = E + Ir = 10 + 1 × 1 = 11 V.

Teacher's Note:
a) The cell in which the current enters the positive terminal is being charged, so its terminal voltage is greater than its emf.
b) The earth connection fixes the zero of potential; the wire F to D is the same potential as G.

7. CELLS

(a) Electro Motive Force (EMF): The potential difference across the terminals of a cell when it is not giving any current is called EMF of the cell.
or
The energy given by the cell in the flow of unit charge in the whole circuit (including the cell) is called the EMF of the cell. E = \( \frac{W}{Q} \).

(b) Terminal voltage:
(i) The resistance offered by the electrolyte of the cell to the flow of current through it is called internal resistance of the cell.
(ii) When current is drawn through the cell or current is supplied to the cell then the potential difference across its terminals is called terminal voltage.
(iii) When current i is drawn from the cell, then the terminal voltage is less than its emf E: V = E - i r.

[Figure: A cell of emf E in series with its internal resistance r]

Where V = terminal voltage, r = internal resistance of the battery.
(iv) When current is supplied to the cell, the terminal voltage is greater than the emf E, i.e. V = E + i r.
(v) Units of both emf and terminal voltage are volt.

Combinations of cells:

[Figure: Three cells E₁ (internal resistance r₁), E₂ (r₂) and E₃ (r₃) connected in series with an external resistance R, current i]

(a) Series Combination:
(i) Equivalent emf E = E₁ + E₂ + E₃ ......... Note: Direction of emf is taken into consideration.
(ii) Equivalent internal resistance r is given by r = r₁ + r₂ + r₃ .............
(iii) Current, i = \( \frac{E}{r + R} \) = \( \frac{ΣE_i}{Σr_i + R} \).

Imp:
(iv) For maximum power in the load, R = Σr, i.e. the load resistance must be equal to the equivalent internal resistance.
(v) If all emf are equal (E), then for a series combination of n such cells, I = \( \frac{nE}{R + nr} \).

Cases:
(a) If nr >> R, I = \( \frac{E}{r} \).
(b) If nr << R, I = \( \frac{nE}{R} \).
(c) Cells are employed in series only when the internal resistance is less than the load resistance.

(b) Parallel Combination:
(i) Equivalent internal resistance r is given by \( \frac{1}{r} \) = \( \frac{1}{r_1} \) + \( \frac{1}{r_2} \) + \( \frac{1}{r_3} \) + ........
(ii) Equivalent emf: E = \( \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2} + \frac{E_3}{r_3} + \ldots}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \ldots} \) = \( \frac{Σ \frac{E_i}{r_i}}{Σ \frac{1}{r_i}} \).

[Figure: Cells E₁ (r₁), E₂ (r₂), E₃ (r₃) connected in parallel across an external resistance R, with a main current i]

(iii) Current, i = \( \frac{E}{R + r} \).

[Figure: A cell E with internal resistance r in series with an external resistance R in a single loop]

(iv) When all 'n' cells with emf E and internal resistance r each are connected in parallel, then the equivalent emf = E and the equivalent internal resistance = \( \frac{r}{n} \).
(v) In this case I = \( \frac{E}{R + \frac{r}{n}} \) = \( \frac{nE}{nR + r} \).

Cases:
(a) If r << nR, I = \( \frac{E}{R} \).
(b) If r >> nR, I = \( \frac{nE}{r} \).
(c) This combination is used only when the load resistance is lower than the internal resistance.

(c) Mixed combination: i = \( \frac{mnE}{mR + nr} \), where m is the number of parallel rows and n is the number of cells in series in each row. For maximum current, internal resistance = external resistance, i.e. R = \( \frac{nr}{m} \).

[Figure: A mixed combination: m identical rows connected in parallel, each row consisting of n cells in series (each of emf E and internal resistance r), connected to an external resistance R with a main current i]

Ex.43 A battery of emf 2 volts and internal resistance 0.1Ω is being charged with a current of 5A. The potential difference between terminal of the battery is?
(A) 1.5V
(B) 2.5V
(C) 3.5V
(D) 4.5V

[Figure: A 5 A current flowing from left to right through a 2 V cell and its 0.1 Ω internal resistance]

Answer: (B) 2.5 V

1. The battery is being charged, so the current enters its positive terminal and the terminal voltage is greater than the emf.
2. V = E + Ir = 2 + 5 × 0.1 = 2 + 0.5 = 2.5 V.
3. Hence, the correct option is (B).

Teacher's Note:
a) The potential drop across the internal resistance is 0.1 × 5 = 0.5 V.
b) For a discharging cell V = E - Ir.

8. WHEAT STONE BRIDGE

[Figure: A Wheatstone bridge: a quadrilateral ABCD with resistors P (A to B), Q (B to C), R (A to D) and S (D to C), a galvanometer G joining B and D, and a cell between A and C at E. The current from the cell is i, splitting into i₁ (towards B) and i₂ (towards D); the galvanometer current is i_g, and the currents through Q and S are i₁ - i_g and i₂ + i_g]

(a) The configuration in the adjacent figure is called the Wheat stone bridge.
(b) If i_g = 0, i.e. the current in the galvanometer is zero, then the bridge is said to be balanced.
(c) For i_g = 0: (i) V_D = V_B (ii) \( \frac{P}{Q} \) = \( \frac{R}{S} \).
(d) Equivalent resistance in the balanced condition = \( \frac{(P + Q)(R + S)}{P + Q + R + S} \).
(e) If \( \frac{P}{Q} \) < \( \frac{R}{S} \) then V_B > V_D and current will flow from B to D.
(f) If \( \frac{P}{Q} \) > \( \frac{R}{S} \), then V_B < V_D and current will flow from D to B.
(g) Meter bridge and post office box work on this principle.

EXAMPLES BASED ON WHEAT STONE BRIDGE

Ex.44 In the following figures, the resistance A and B will respectively be-
(A) R/2, R
(B) 2R, R
(C) R, R
(D) 3R, R

[Figure: Figure A: a diamond between A (left) and B (right) with a resistor R in each of the four sides and a resistor between the top junction C and the bottom junction D. Figure B: from A the wire goes to a, then a resistor R to b, a resistor R from b to d, a resistor R from d to c and then to B; another resistor R joins a to d over the top, and another resistor R joins b to c underneath]

Answer: (C) R, R

1. Figure A: the diamond with four resistors R and a middle resistor R between C and D is a balanced bridge, since \( \frac{P}{Q} \) = \( \frac{R}{R} \) = \( \frac{R}{S} \).
2. In a balanced bridge no current flows in the middle resistor between the junctions, so it can be ignored.
3. The equivalent resistance is \( \frac{(P + Q)(R + S)}{P + Q + R + S} \) = \( \frac{2R \times 2R}{4R} \) = R.
4. Figure B: with a, b, d and c as the junctions, the resistors a to b, a to d, b to c and d to c are all R and the resistor R between b and d is again a balanced bridge arm, so it can be ignored and the resistance between A and B is also R.
5. Hence, the correct option is (C).

Teacher's Note:
a) In a balanced Wheatstone bridge the middle arm carries no current.
b) Both circuits reduce to two R in series in parallel with two R in series, which is R.

Ex.45 Figure shows a cube made of 12 resistances, each of resistance R. The equivalent resistance across a and b will be -
(A) (7/12) R
(B) (12/7) R
(C) (9/5) R
(D) (5/9) R

[Figure: A cube drawn with 12 resistors R on its edges, with corners numbered 1 to 8; the terminal a is joined to corner 1 and the terminal b to corner 2, where 1 and 2 are the ends of one vertical edge]

Answer: (A) (7/12) R

1. The terminals a and b are connected to the adjacent corners 1 and 2 of the cube.
2. By symmetry alone, points 4 and 5 must be at the same potential, and so must points 3 and 6.
3. The circuit can then be redrawn with points 4 and 5 joined and points 6 and 3 joined, which reduces step by step to the combination R/2, R/2, R, R/2 and 2R between 1 and 2.
4. The equivalent resistance between a and b is \( \frac{7}{12} \) R, so the correct option is (A).

Teacher's Note:
a) Points of equal potential can be joined without changing the circuit.
b) The resistance across an edge of a cube is 7R/12 and across a body diagonal is 5R/6.

Ex.46 The equivalent resistance between points A and B in the following circuit will be-
(A) 7R/5
(B) 5R/7
(C) 6R/4
(D) 4R/6

[Figure: A bridge circuit between A (bottom left) and B (top right). The left side has two resistors R in series with the junction C between them, and the right side has two resistors R in series with the junction D between them. A resistor R joins C and D across the middle. A resistor R along the top wire joins the top of the left side to the top of the right side near B, and a resistor R along the bottom wire joins the bottom of the left side at A to the bottom of the right side]

Answer: (A) 7R/5

1. The circuit is a Wheatstone bridge but it is unbalanced, since the ratio of the arms on the two sides is not equal. Imagine a cell of emf E connected between A and B; let the current from the cell be i.
2. Let i₁ flow in R from A to C, and i₂ in the resistor R from C to D. The current in the 2R from A to D is i - i₁, in the 2R from C to B is i₁ - i₂, and in the R from D to B is i - i₁ + i₂.
3. For the loop ACDA: 2i = 3i₁ + i₂ ......(A). For the loop BDCB: i = 3i₁ - 4i₂ ......(B).
4. Solving, i₁ = \( \frac{3}{5} \) i and i₂ = \( \frac{1}{5} \) i.
5. E = (i - i₁) 2R + (i - i₁ + i₂) R = \( \frac{4R}{5} \) i + \( \frac{3R}{5} \) i = \( \frac{7R}{5} \) i.
6. R_eq = \( \frac{E}{i} \) = \( \frac{7R}{5} \), so the correct option is (A).

Teacher's Note:
a) When a bridge is unbalanced, use Kirchhoff's laws with an imaginary cell across the terminals.
b) The printed working writes (i - i₂) in the energy equation; the correct term is (i - i₁).

9. ELECTRIC ENERGY AND POWER

Electric energy:
When a potential difference is applied across a wire, current starts flowing in it. The free electrons collide with the positive ions of the metal and lose energy. Thus energy taken from the battery is dissipated. The battery constantly provides energy to continue the motion of electrons and hence the electric current in the circuit. This energy is given to ions of the metal during collision and thus the temperature of the wire rises. Thus, energy taken from the battery gets transferred into heat. This energy is called electrical energy. This effect is also called 'Heating Effect of Current'.

If R = resistance of wire, I = current in wire, V = potential difference across wire, flow of charge in 'dt' time = Idt.
Energy dissipated dW = Vdq = VIdt.
∵ V = IR, ∴ dW = VIdt = I²Rdt = \( \frac{V^2}{R} \) dt = Vdq.
This energy is equal to the work done by the battery or the heat produced in the wire.
If energy is to be written in calorie, then dW = \( \frac{dW}{4.2} \) cal = 0.24 dW cal.

Electrical power:
The rate of loss of energy in an electrical circuit is called electrical power. It is denoted by 'P'.
P = \( \frac{dW}{dt} \) = I²R = IV = \( \frac{V^2}{R} \).
Units of power = joule/sec, watt, horse power.
1 watt = 1 joule/sec, 1 HP = 746 watt.
Unit of electrical energy = watt second, kilowatt hour.
1 kilowatt hour (kwh) = 36 × 10⁵ joule.

Combination of electrical instruments -
(a) If 220V and 40W is written on an electrical instrument then this is called its standard Ratings. It means that if 220V is applied across this instrument then 40W of power will be generated. Thus the resistance will be given by R = \( \frac{V^2}{P} \) = \( \frac{(220)^2}{40} \) ohm.

(b) Series combination

[Figure: Three instruments with ratings V₁, P₁; V₂, P₂; V₃, P₃ connected in series with a source E]

(i) If the total power dissipated is P, then \( \frac{1}{P} \) = \( \frac{1}{P_1} \) + \( \frac{1}{P_2} \) + \( \frac{1}{P_3} \), where P₁, P₂, P₃ are the standard powers of the instruments.
(ii) In this combination, the bulb with least power will glow most and the bulb with highest power will glow least; we can say that the bulb with highest R will glow brightest and the bulb with least R will glow least.

(c) Parallel combination

[Figure: Three instruments with ratings V₁, P₁; V₂, P₂; V₃, P₃ connected in parallel across a source E]

(i) Net power dissipation P = P₁ + P₂ + P₃.
(ii) The bulb with least power will glow least, or the bulb in which maximum current is flowing will glow brightest, and vice-versa.

Note:
(a) These formulae are applicable only if the voltage ratings of all the instruments are equal along with the power source. If voltage ratings are different then the circuit is solved by considering equivalent resistances of the instruments as follows.
(b) Replace the instrument by its equivalent resistance. If the standard rating is (V/P) then the resistance is R = V²/P.
(c) Find the currents and voltages in different branches using Kirchhoff's first and second laws.
(d) If the rating of a bulb is changed from V₁/P₁ to V₂/P₂ then \( \frac{V_1^2}{P_1} \) = \( \frac{V_2^2}{P_2} \) = R, or P₂ = \( \frac{V_2^2}{V_1^2} \) P₁.

EXAMPLES BASED ON ELECTRIC ENERGY AND POWER

Ex.47 A 60-watt lamp is operated at 220-volt power line. What will be resistance when the lamp is lighting ? How much current will flow in it? Estimate, how much current will flow in the lamp when it is connected to a 2-volt battery?

Answer: R = 806.7 Ω; current at 220 V = 0.27 A; estimated current with the 2 V battery = 0.0025 A
1. If the resistance of the lamp is R ohm and it lights at V volt, the electric power consumed (in watt) is given by P = \( \frac{V^2}{R} \).
2. Here P = 60 watt and V = 220 volt, so R = \( \frac{V^2}{P} \) = \( \frac{(220)^2}{60} \) = 806.7 ohm.
3. According to Ohm's law, the current flowing in the lamp is i = \( \frac{V}{R} \) = \( \frac{220}{806.7} \) = 0.27 ampere.
4. On connecting the lamp with a 2-volt battery, the estimated current is i = \( \frac{V}{R} \) = \( \frac{2}{806.7} \) = 0.0025 ampere.

Teacher's Note:
a) The last value is only an estimation, not the actual current. The actual current will be slightly larger, because the resistance was determined at the very high temperature of the glowing filament.
b) With a 2 V battery the filament stays at ordinary temperature, so its resistance is less than 806.7 Ω.

Ex.48 The resistance of each of the three wires joined as in the fig 2 ohm, and each one can have maximum power of 18 watt (otherwise it will melt). The maximum power the whole circuit can take is -
(A) 27 W
(B) 9 W
(C) 81 W
(D) 18 W

[Figure: Two 2 Ω wires in parallel joined in series with a third 2 Ω wire]

Answer: (A) 27 W

1. Electric power expended in a wire of resistance R is P = i²R, so the maximum current in any wire is i = √(\( \frac{P}{R} \)) = √(\( \frac{18 \text{ watt}}{2 \text{ ohm}} \)) = 3 ampere.
2. The circuit has two 2 Ω wires in parallel and a third 2 Ω wire in series with this parallel combination.
3. Equivalent resistance R' = \( \frac{2 \times 2}{2 + 2} \) + 2 = 3 Ω.
4. The series wire carries the whole current, so the maximum current of the circuit is 3 A and P_max = i²R' = (3)² × 3 = 27 W, which is option (A).

Teacher's Note:
a) The series wire limits the total current to 3 A; each parallel wire then carries only 1.5 A, which is safe.
b) Check each wire against its own limit before choosing the maximum current.

Ex.49 A heater is designed to operate with a power of 1000 watt in a 100-volt line. It is connected to two resistance of 10 ohm and R ohm as shown in the fig. If the heater is now giving a power of 62.5 watt. The value of the resistance R, will be -
(A) 5 Ω
(B) 10 Ω
(C) 2.5 Ω
(D) 1.25 Ω

[Figure: A 100 volt supply with a 10 Ω resistor in series with a parallel combination of the heater (1000 watt) and the resistance R]

Answer: (A) 5 Ω

1. Let the resistance of the heater be r. Then P = \( \frac{V^2}{r} \), i.e. 1000 = \( \frac{(100)^2}{r} \), so r = 10 ohm.
2. When connected as shown, let the potential difference across the heater be V'. Then 62.5 = \( \frac{V'^2}{r} \), so V' = 25 volt.
3. Potential difference across the 10 Ω = 100 - 25 = 75 volt, and the current in the 10 Ω = \( \frac{75 \text{ volt}}{10 \text{ ohm}} \) = 7.5 ampere.
4. Current in the heater i = \( \frac{P}{V'} \) = \( \frac{62.5}{25} \) = 2.5 ampere.
5. Current in R = 7.5 - 2.5 = 5.0 ampere, so R = \( \frac{25 \text{ volt}}{5.0 \text{ ampere}} \) = 5 Ω, which is option (A).

Teacher's Note:
a) The heater and R are in parallel, so they have the same potential difference V'.
b) First find the heater resistance from its rated power and voltage.

Ex.50 In a house there are 12 bulbs each of resistance 440 ohm, one motor of 1/10 horse-power, and 4 fans each of 100 watt. These are operated 5 hours per day. The expenditure of electricity in a month of 30 days will be - (1 H.P. = 746 watt, cost of electricity is 40 paisa per unit and voltage of power is 220 volt.
(A) Rs. 269.19
(B) Rs. 107.68
(C) Rs. 1794.6
(D) Rs. 179.46

Answer: (B) Rs. 107.68

1. Electric power expended in each bulb = \( \frac{V^2}{R} \) = \( \frac{(220)^2}{440} \) = 110 watt, so 12 bulbs use 12 × 110 = 1320 watt.
2. Electric power expended in 4 fans = 4 × 100 = 400 watt.
3. Electric power expended in the motor = \( \frac{1}{10} \) H.P. = \( \frac{1}{10} \) × 746 = 74.6 watt.
4. Total power expended = 1320 + 400 + 74.6 = 1794.6 watt.
5. Units consumed in 30 days = \( \frac{\text{watt} \times \text{hour}}{1000} \) = \( \frac{1794.6 \times 5 \times 30}{1000} \) = 269.19 units.
6. Cost of electricity = 269.19 × 0.40 = Rs. 107.68, so the correct option is (B).

Teacher's Note:
a) Rs. 269.19 is the number of units, not the cost; the cost is 40 paisa per unit.
b) The printed working writes (200)²/440 for the bulb power, but 220 V is the supply voltage and the value 110 W is right.

Your stage: Now you may solve the questions from Q. No. 43 to 54 of Ex.# 1

POINTS TO REMEMBER

1. In liquids, the charge carriers are positive and negative ions.
2. In gases, the charge carriers are positive ions and free electrons.
3. In semiconductors, the charge carriers are holes and free electrons. The conventional direction of flow of current is opposite to the direction of flow of electrons.
4. Current is a scalar quantity. Current density is a vector quantity. The direction of j is the same as that of E.
5. The reciprocal of the slope of the V - I graph is equal to the resistance.
6. The resistance of a conductor depends on the material of the conductor, the shape and size of the conductor as well as on the physical state (temperature) of the conductor.
7. The reciprocal of resistance is called conductance, G = 1/R.
8. Fuse Wire: A fuse wire is used in a circuit to control the maximum current flowing in the circuit. It is a thin wire having high resistance and is made up of a material with low melting point.
9. While calculating the resistance of a wire by the formula R = ρℓ/A, it should be remembered that 'A' is the area normal to the direction of current flow and 'ℓ' is the length in the direction of current flow.
10. Free electron density in a metal is given by n = \( \frac{N x d}{A} \), where N is the Avogadro number, x is the number of free electrons per atom, d is the density of the metal and A is the atomic weight of the metal; n is of the order of 10²³ per cm³.
11. For carbon, India rubber, mica, electrolytes and insulators the value of α is negative. It means that their resistance decreases with increase in temperature.
12. The resistance of a conductor increases with decrease in density or when it is subjected to mechanical stress.
13. The direction of current density is the direction of motion of positive charge at that point.
14. House wiring circuits are in parallel, therefore the voltage across each bulb is constant. The power of the bulb is given by the formula P = V²/R. For constant voltage P ∝ (1/R); therefore, the greater the resistance, the smaller is the power. Hence, if we take two bulbs of 60W and 100W, then the resistance of the 60W bulb will be more than the resistance of the 100 watt bulb.
15. The filament of the 60W bulb is thinner than the filament of the 100 watt bulb.
16. If the two bulbs, one of 60W and the other of 100 W, are connected in parallel, then the 100 W bulb will glow more.
17. If two bulbs, one of 60W and the other of 100W, are connected in series, then the 60 W bulb will glow more, because in a series combination the current through both the bulbs is the same and the power consumed P = I²R or P ∝ R. Now, since the resistance of the 60W bulb is more than the resistance of the 100 watt bulb, the 60W bulb will glow more.
18. Specific resistance is minimum for conductors and maximum for insulators.
19. The specific resistance ρ = E/J. If j = 1 amp/m², then ρ = E. Hence, specific resistance is the electric field per unit current density. The greater the current density, the greater is the field.
20. If two bulbs of power P₁ and P₂ are connected in parallel and the rated voltage is applied, then the total power consumed is P = P₁ + P₂.
21. If two bulbs of power P₁ and P₂ (of the same rated voltage) are connected in series across a supply of that rated voltage, then the total power consumed is \( \frac{1}{P} \) = \( \frac{1}{P_1} \) + \( \frac{1}{P_2} \).

22. In a series resistance circuit, it should be remembered that:
(i) The current is the same in every part of the series circuit.
(ii) The total resistance in the circuit is the sum of the individual resistances including the internal resistance of the cell (if any).
(iii) Voltage across any part of a circuit is proportional to the resistance of that part.
(iv) Current in the circuit is independent of the relative positions of the various resistances in the series.
(v) The effective resistance of n equal resistances each of resistance r in series is R = nr.
(vi) If σ is the effective conductance then \( \frac{1}{σ} \) = \( \frac{1}{σ_1} \) + \( \frac{1}{σ_2} \) + \( \frac{1}{σ_3} \) + ....................

23. In a parallel combination of resistances, it should be remembered that:
(i) The total current through the combination is the sum of the individual currents through the various branches.
(ii) The potential difference across all the resistances is the same.
(iii) The current through each branch is inversely proportional to the resistance of that branch.
(iv) The reciprocal of the total resistance of the combination is equal to the sum of the reciprocals of the individual resistances.
(v) If two resistances R₁ and R₂ are in parallel, then the currents I₁ and I₂ in them will be distributed as I₁ = \( \frac{R_2 I}{R_1 + R_2} \), I₂ = \( \frac{R_1 I}{R_1 + R_2} \), where I is the net current that flows through the circuit.
(vi) The total resistance of n equal resistances each of resistance r, in parallel is r/n.
(vii) For n equal resistances, \( \frac{R_{series}}{R_{parallel}} \) = n².
(viii) If σ is the effective conductance then σ = σ₁ + σ₂ + σ₃.

24. Using 'n' conductors of equal resistances, the number of combinations one can have, using all at a time, is 2^n-1.
25. If the resistances of the 'n' conductors are all different, then the number of possible combinations is larger (for example, 8 for three different resistances).
26. The resistivity of a conductor changes with impurity. The impurity increases the resistivity; the resistivity of an alloy is greater than the resistivity of its constituents.
27. Resistivity increases with temperature.
28. Mechanical stress increases the resistivity.
29. Resistivity of most of the metals increases on heating.
30. Resistivity of antimony, bismuth and semi-conductors decreases with increase of temperature.
31. When the charge is stationary, the energy stored is calculated by the formula \( \frac{VQ}{2} \), while if the charge is flowing then the energy dissipated is calculated by VQ.
32. When a wire is drawn to n times its length, its new resistance becomes n²R, where R is the original resistance.
33. When wires are drawn from the same volume but with different areas of cross-section, then resistance ∝ \( \frac{1}{(\text{Area of cross-section})^2} \) or resistance ∝ \( \frac{1}{(\text{radius})^4} \).
34. When a wire is folded n times on its own length to 1/n^th length then the new resistance becomes R/n² (R is the initial resistance of the wire).
35. Temperature coefficient of materials is given by α = \( \frac{R_t - R_0}{R_0 t} \). Its units are per °C.
36. If α is positive the resistance increases with increase in temperature.
37. If α is negative the resistance decreases with increase in temperature.
38. For some materials α is nearly zero. There is no change in resistance with temperature.
39. The temperature coefficient of thermistors is negative and it is high.
40. The direction of flow of current in the cell is from the negative electrode to the positive electrode, while outside the cell it is from the positive electrode to the negative electrode.
41. Open circuit means that no current is being drawn from the cell.
42. Closed circuit means that current is being drawn from the cell.
43. Energy is consumed inside the cell due to its internal resistance.
44. Due to the presence of internal resistance in the cell, there is a potential drop inside the cell.
45. E.m.f. (E) is the characteristic of each cell and its value remains constant for the cell, while the potential difference (V) goes on decreasing on taking more and more current from the cell.
46. While charging a cell the positive terminal of the charger should be connected to the positive terminal of the cell and the negative terminal of the charger should be connected to the negative terminal of the cell.
47. While charging a cell, we send a current in the cell by some other electric source (such as a battery), called the charger; then the direction of current inside the cell will be from the positive plate to the negative plate. In this case the potential difference V between the plates of the cell will be greater than the emf of the cell. V = E + ir.
48. The greater the length of the potentiometer wire, the smaller is the potential gradient (k = V/l) and the more is the balancing length, hence the more is the accuracy of observations.
49. Potential difference across the terminals of the cell when current is drawn from it is V = E - Ir.
50. Potential difference across the terminals of the cell when the cell is charged is V = E + Ir.
51. Internal resistance of the cell r = \( \left( \frac{E - V}{V} \right) \) R, where E = emf of the cell, V = potential difference across the terminals of the cell, R = external resistance.
52. Energy supplied by the cell in t seconds = Eit (if r = 0) = Vit (if r is finite).
53. Energy dissipated in the internal circuit of resistance r is i²rt.
54. Energy dissipated in the external circuit of resistance R is i²Rt.
55. If the diameter of the potentiometer wire is not uniform, then the value of the potential gradient will not be uniform throughout the wire.
56. The current flowing in the circuit if a cell of emf E and internal resistance r is connected to an external resistance R is I = \( \frac{E}{r + R} \).
57. If two cells of emf E₁ and E₂ and internal resistances r₁ and r₂ are connected so as to support each other, then the current in a resistor R is I = \( \frac{E_1 + E_2}{r_1 + r_2 + R} \).
58. If the two cells oppose each other, then I = \( \frac{E_1 \sim E_2}{r_1 + r_2 + R} \) (where E₁ ~ E₂ means the difference of the two emfs).
59. In a battery of N cells each of emf E, if n cells are wrongly connected, then the net emf of the battery is E' = NE - 2nE, as one wrongly connected cell destroys one more cell.
60. Charging current = \( \frac{\text{emf of charger} - \text{emf of cell}}{\text{total resistance of the circuit}} \).
61. When cells of emf E₁ and E₂ are connected in parallel:
(a) net emf = E₁ - E₂ if E₁ > E₂.
(b) Total internal resistance is (r₁ + r₂).
(c) Current i = \( \frac{E_1 - E_2}{r_1 + r_2} \).
(d) As E₁ > E₂, E₁ is discharging while E₂ is charging.
(e) The common terminal voltage V = \( \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \).
62. When an external resistance R is connected to a battery of emf E and internal resistance r, the power transferred from the cell to the resistance is maximum when r = R.
63. Nichrome is used as an electrical heater element because of its high specific resistance, high melting point and low temperature coefficient of resistance.
(i) Nichrome is an alloy of nickel, chromium and iron.
(ii) Manganin is an alloy of copper, manganese and nickel.
(iii) Constantan is an alloy of copper and nickel.
(iv) Fuse wire is an alloy of lead and tin.
64. Heater wire has high resistance and high melting point.
65. Heater wire is always connected in parallel to the mains.
66. Fuse wire is always connected in series in the circuit.
67. In resistance boxes manganin wires are used as their temperature coefficient is negligibly small.
68. Wheatstone bridge is the arrangement of resistances used to measure the resistance of a conductor.
69. The bridge is balanced if \( \frac{P}{Q} \) = \( \frac{R}{S} \).
70. At balance, the potential difference across the galvanometer arm is zero and the current through the galvanometer is also zero (i_g = 0).
71. The bridge is most sensitive when the current in all the four branches of the bridge is of the same order.
72. The balance is not affected on interchanging the position of battery and galvanometer.
73. Post office box works on the principle of Wheatstone bridge and it is used to measure the resistance of a wire, but its sensitivity is small.
74. Metre bridge is an instrument to measure the resistance of a wire accurately. It works on the principle of Wheatstone's bridge.
75. Meter bridge is more sensitive than a post office box.
76. The drawback of a meter bridge is the appearance of end resistances.
77. The effect of end resistances is reduced by interchanging the gaps.
78. Complete elimination of end resistances is possible in the Carey Foster Bridge, which has four gaps instead of two.
79. Potentiometer is an instrument used to measure accurate potential differences.
80. The resistance of a potentiometer can be considered as infinity while measuring the emf.
81. The potentiometer and vacuum tube voltmeter (VTVM) are ideal voltmeters.
82. The potentiometer or VTVM does not draw any current from the points across which the potential difference is measured.
83. If l₁ and l₂ are the balancing lengths for two cells of emf E₁ and E₂, then \( \frac{E_1}{E_2} \) = \( \frac{l_1}{l_2} \).
84. If l₁ and l₂ are the lengths in open and closed circuit across a cell and R is the external resistance, then the internal resistance of the cell is r = R \( \left( \frac{l_1}{l_2} - 1 \right) \).