DC Pandey Solutions: Current Electricity | Physics Class 12 - NEET PDF Download

Introductory Exercise 20.1

Q.1. In the Bohr model of the hydrogen atom, the electron is pictured to rotate in a circular orbit of radius 5 × 10^-11 m, at a speed of 2.2 × 10⁶ m/s. What is the current associated with electron motion?
Ans.

Current associated with a charge in circular motion is the charge passing a point per unit time. For one electron,

Ans: I = e/T = e·v/(2πr)

Substitute e = 1.6 × 10^-19 C, v = 2.2 × 10⁶ m/s and r = 5 × 10^-11 m:

I = (1.6 × 10^-19 × 2.2 × 10⁶)/(2π × 5 × 10^-11)

Numerator = 3.52 × 10^-13, Denominator = 3.1416 × 10^-10

I ≃ 1.12 × 10^-3 A = 1.12 mA

Q.2. Copper has one conduction electron per atom. Its density is 8.89 g/cm³ and its atomic mass is 63.54 g/mol. If a copper wire of diameter 1.0 mm carries a current of 2.0 A. What is the drift speed of the electrons in the wire?
Ans.

ρ = 8.89 g/cm³ = 8.89 × 10³ kg/m³

Number of moles per unit volume = density / molar mass

= (8.89 × 10³ kg m^-3)/(63.54 × 10^-3 kg mol^-1)

= 1.399 × 10⁵ mol m^-3

Number density of atoms (and free electrons, one per atom):

n = 1.399 × 10⁵ × 6.02 × 10²³ = 8.42 × 10²⁸ m^-3

Area of wire, diameter = 1.0 mm ⇒ radius = 0.5 × 10^-3 m

A = πr² = π(0.5 × 10^-3)² = 7.85 × 10^-7 m²

Current relation: i = n e A v_d ⇒ v_d = i/(n e A)

Substitute i = 2.0 A, e = 1.6 × 10^-19 C, n and A above:

n e A = 8.42 × 10²⁸ × 1.6 × 10^-19 × 7.85 × 10^-7 ≃ 1.06 × 10⁴ A s m^-1

v_d = 2.0 / (1.06 × 10⁴) ≃ 1.9 × 10^-4 m/s

Thus the drift speed is ≃ 1.9 × 10^-4 m/s.

Q.3. When a steady current passes through a cylindrical conductor, is there an electric field inside the conductor?

Ans: Yes. Under electrostatic equilibrium (no current) the electric field inside a conductor is zero. When a steady current flows, an electric field exists along the conductor to drive the charges. The field and current density are related by J = σE (or E = ρJ), where ρ is resistivity and σ is conductivity.

Q.4. Electrons in a conductor have no motion in the absence of a potential difference across it. Is this statement true or false?

Ans: False. Free electrons undergo random thermal motion even without any applied potential difference. These random motions result in no net charge transport (net current is zero) unless an external electric field produces an organised drift velocity.

Q.5. In an electrolyte, the positive ions move from left to right and the negative ions from right to left. Is there a net current? If yes, in what direction?

Ans: Yes. Conventional current is in the direction of positive charge flow. Positive ions moving left→right give a current left→right. Negative ions moving right→left also produce a conventional current left→right (because negative charge moving left is equivalent to positive charge moving right). Therefore the net current is from left to right; the contributions add.

Q.6. The current through a wire depends on time as, i = (10 + 4t)
Here, i is in ampere and t in seconds. Find the charge crossed through a section in time interval between t = 0 to t = 10 s.

Ans: Charge q = ∫ i dt from 0 to 10 s.

q = ∫₀¹⁰ (10 + 4t) dt = [10t + 2t²]₀¹⁰

q = 10×10 + 2×10² = 100 + 200 = 300 C

Introductory Exercise

Q.1. In household wiring, copper wire 2.05 mm in diameter is often used. Find the resistance of a 35.0 m long wire. Specific resistance of copper is 1.72 ×10^-8 Ω-m.

Ans:

Diameter d = 2.05 mm ⇒ radius r = 1.025 × 10^-3 m

Area A = πr² = π(1.025 × 10^-3)² = 3.30 × 10^-6 m²

Resistance R = ρL/A = (1.72 × 10^-8 × 35.0)/ (3.30 × 10^-6)

R ≃ 0.183 Ω ≃ 0.18 Ω

Q.2. An aluminium wire carrying a current has diameter 0.84 mm. The electric field in the wire is 0.49 V/m. What is,
(a) the current carried by the wire?
(b) the potential difference between two points in the wire 12.0 m apart?
(c) the resistance of a 12.0 m length of this wire? Specific resistance of aluminium is 2.75 × 10^-8 Ω-m.

Ans:

Diameter = 0.84 mm ⇒ radius = 0.42 × 10^-3 m

Area A = π(0.42 × 10^-3)² = 5.54 × 10^-7 m²

Resistivity ρ = 2.75 × 10^-8 Ω·m ⇒ resistance per metre R_1m = ρ/A = (2.75 × 10^-8)/(5.54 × 10^-7) = 4.96 × 10^-2 Ω

(a) Electric field E = 0.49 V/m ⇒ potential difference across 1 m is 0.49 V. Using V = iR,

i = V/R_1m = 0.49/(4.96 × 10^-2) ≃ 9.9 A

(b) Potential difference across 12.0 m = E × 12.0 = 0.49 × 12.0 = 5.88 V

(c) Resistance of 12.0 m = 12 × R_1m = 12 × 4.96 × 10^-2 ≃ 0.596 Ω

Q.3. A conductor of length l has a non-uniform cross-section. The radius of the cross-section varies linearly from a to b. The resistivity of the material is ρ. Find the resistance of the conductor across its ends.

Ans.

Let the length be l and x measured from end P. Radius at x is r(x) = a + (b - a)x/l.

Cross-sectional area A(x) = π[r(x)]² = π[a + (b - a)x/l]²

Elemental resistance dR = ρ dx / A(x) = (ρ dx)/(π[a + (b - a)x/l]²)

Integrate from x = 0 to l:

R = (ρ/π) ∫₀^l dx / [a + (b - a)x/l]²

Substitute u = a + (b - a)x/l ⇒ du = (b - a)/l dx ⇒ dx = (l/ (b - a)) du

Limits: x = 0 ⇒ u = a, x = l ⇒ u = b

R = (ρ l)/(π (b - a)) ∫_a^b du / u² = (ρ l)/(π (b - a)) [ -1/u ]_a^b

R = (ρ l)/(π (b - a)) (1/a - 1/b) = (ρ l)/(π a b)

Thus R = (ρ l)/(π a b).

Q.4. The product of resistivity and conductivity of a conductor is constant. Is this statement true or false?

Ans: True. Conductivity σ is the reciprocal of resistivity ρ, so ρ σ = 1, which is a constant (dimensionless).

Q.5. The resistance of a copper wire and an iron wire at 20°C are 4.1 Ω. and 3.9 Ω respectively. Neglecting any thermal expansion, find the temperature at which resistances of both are equal.
α_Cu = 4.0 × 10^-3 K^-1 and α_Fe = 5.0 × 10^-3 K^-1.

Ans:

Resistance at temperature θ: R(θ) = R₀[1 + α(θ - 20)]

Set equal: 4.1[1 + 4.0 × 10^-3(θ - 20)] = 3.9[1 + 5.0 × 10^-3(θ - 20)]

Expand: 4.1 + 4.1×4.0×10^-3(θ - 20) = 3.9 + 3.9×5.0×10^-3(θ - 20)

Solve for θ: (4.1 - 3.9) = (3.9×5.0×10^-3 - 4.1×4.0×10^-3)(θ - 20)

0.2 = (0.0195 - 0.0164)(θ - 20) = 0.0031(θ - 20)

θ - 20 = 0.2 / 0.0031 ≃ 65 ⇒ θ ≃ 85°C

Introductory Exercise

Ans: If the potential difference across each resistor is 10 V (as indicated),

Current through 2 Ω: i = V/R = 10/2 = 5 A

Current through 4 Ω: i = 10/4 = 2.5 A

Q.2. In the circuit shown in figure, find the potentials of A, B, C and D and the current through 1Ω and 2Ω resistance.

Ans:

From the given circuit (refer to the figure):

V_A = 0 V (earthed)

V_C - V_A = 5 V ⇒ V_C = 5 V

V_B - V_A = 2 V ⇒ V_B = 2 V

V_D - V_C = 10 V ⇒ V_D = 10 + V_C = 15 V

Current through 1 Ω (between C and B): (V_C - V_B)/1 Ω = (5 - 2)/1 = 3 A, from C to B.

Current through 2 Ω (between D and A): (V_D - V_A)/2 Ω = (15 - 0)/2 = 7.5 A, from D to A.

Q.3. For what value of E the potential of A is equal to the potential of B?

Ans:

For V_A = V_B, the potential difference V_AB = 0.

From the branch equation, V_AB = E - i r = 0 ⇒ E = i r.

Solving the circuit equations (see diagram) gives E = 5 V.

Q.4. Ten cells each of emf 1V and internal resistance 1Ω are connected in series. In this arrangement polarity of two cells is reversed and the system is connected to an external resistance of 2Ω . Find the current in the circuit.

Ans:

Net emf = (n - 2m)E where n = 10 cells, m = 2 reversed cells, E = 1 V

Net emf = (10 - 4) × 1 = 6 V

Total internal resistance = 10 × 1 Ω = 10 Ω

Total resistance = internal + external = 10 + 2 = 12 Ω

Current I = net emf / total resistance = 6 / 12 = 0.5 A

Q.5. In the circuit shown in figure, R₁ = R₂ = R₃ = 10Ω. Find the currents through R₁ and R₂

Ans.

Q.6. Draw:
(a) current versus load, and
(b) current versus potential difference graph for a cell.

Ans:

(a)

The i versus R graph shows that current decreases as load resistance increases (i = E/(R + r)).

(b) V = E - i r. The V versus i graph is a straight line with slope -r; intercept on V-axis is E.

Introductory Exercise 20.4

Ques 1: In the circuit shown in figure, a 12 V power supply with unknown internal resistance r is connected to a battery with unknown emf E and internal resistance 1Ω and to a resistance of 3Ω carrying a current of 2 A. The current through the rechargeable battery is 1 A in the direction shown. Find the unknown current i, internal resistance r and the emf E.

Ans: Applying loop law equation in upper loop we have,
E + 12 - ir - 1 = 0 ...(i)
Applying loop law equation in lower loop we have where i = 1 + 2 = 3A
E + 6 - 1 = 0 ...(ii)
Solving these two equations we get, E = -5Vand r = 2 Ω

Ques 2: In the above example, find the power delivered by the 12 V power supply and the power dissipated in 3 W resistor.
Ans: 

From the circuit analysis i = 3 A (total current supplied by the 12 V source).

Power delivered by the 12 V supply = E·i = 12 × 3 = 36 W

Current through the 3 Ω resistor is 2 A (given), so power dissipated in the 3 Ω resistor = i²R = (2)² × 3 = 12 W

Introductory Exercise 20.5

Ques 1: Find the equivalent emf and internal resistance of the arrangement shown in Fig.

Ans: 

r = 0.5 W

Ques 2: If a battery of emf E and internal resistance r is connected across a load of resistance R. Show that the rate at which energy is dissipated in R is maximum when R = r and this maximum power is P = E²/4r.

Ans: Current i = E/(R + r)
Power delivered to R: P = i² R = [E² R]/(R + r)²
To find maximum, differentiate P with respect to R and set dP/dR = 0:
dP/dR = E²[(R + r)² - 2R(R + r)]/(R + r)⁴ = 0
⇒ (R + r) - 2R = 0 ⇒ r - R = 0 ⇒ R = r
Maximum power P_max = E² R/(R + r)² with R = r ⇒ P_max = E² r/(4 r²) = E²/(4 r)

Ques 3: Two identical batteries each of emf E = 2 volt and internal resistance r =1 ohm are available to produce heat in an external resistance by passing a current through it. What is the maximum power that can be developed across an external resistance R using these batteries.

Ans:

When used together in series, net emf E_net = 2 + 2 = 4 V, net internal resistance r_net = 1 + 1 = 2 Ω.

Maximum power on external R occurs when R = r_net = 2 Ω.

P_max = E_net²/(4 r_net) = (4)²/(4 × 2) = 16/8 = 2 W

Ques 4: The full scale deflection current of a galvanometer of resistance 1Ω is 5 mA. How will you convert it into a voltmeter of range 5 V ?

Ans:

To convert to a voltmeter of range V = 5 V, a series resistance R is required so that at full-scale current i_g = 5 mA the voltage across the combination is 5 V.

V = i_g(G + R) ⇒ R = V/i_g - G = 5/0.005 - 1 = 1000 - 1 = 999 Ω

Ques 5: A micrometer has a resistance of 100Ω and full scale deflection current of 50μA. How can it be made to work as an ammeter of range 5 mA ?
Ans:

Ans:

Here the meter (galvanometer) resistance G = 100 Ω and full-scale current i_g = 50 μA = 50 × 10^-6 A. Desired range I = 5 mA = 5 × 10^-3 A.

Current through shunt = I - i_g = 5 × 10^-3 - 50 × 10^-6 = 4.95 × 10^-3 A.

Voltage across galvanometer at full-scale: V = G i_g = 100 × 50 × 10^-6 = 5 × 10^-3 V.

Shunt resistance R_s = V / (I - i_g) = (5 × 10^-3)/(4.95 × 10^-3) ≃ 1.01 Ω

Thus a shunt of ≃ 1.01 Ω in parallel with the meter will give a range of 5 mA.

Ques 6: A voltmeter has a resistance G and range V. Calculate the resistance to be used in series with it to extend its range to nV.

Ans:

For original range V, current through the meter at full-scale i_g = V/G.

For new range nV, the total series resistance R must satisfy nV = i_g(G + R).

So R = (nV/i_g) - G = (nV)/(V/G) - G = nG - G = (n - 1) G.

Ques 7: The potentiometer wire AB is 600 cm long.
(a) At what distance from A should the jockey J touch the wire to get zero deflection in the galvanometer.
(b) If the jockey touches the wire at a distance 560 cm from A, what will be the current through the galvanometer.

Ans: 

(a)

Using the balance condition of potentiometer, the length corresponding to the emf equality is found by equating potential drops along the wire to the emf of the lower battery (see figure). Solving the balance equation (refer to the diagram) gives l = 320 cm.

(b)

Resistance of 560 cm portion = (560/600) × total resistance = 14 r (as per diagram notation). Using loop equations in the upper and lower loops (see the circuit diagram) and solving the two simultaneous equations for the branch currents i₁ and i₂ gives the galvanometer current (expressions and final numerical result are obtained from the displayed solution in the figure).

Introductory Exercise 20.6

Ques 1: For the given carbon resistor, let the first strip be yellow, second strip be red, third strip be orange and fourth be gold. What is its resistance?

Ans:

Colour code values: Yellow → 4, Red → 2, Orange → multiplier 10³, Gold → tolerance 5%.

Resistance = 4 2 × 10³ Ω = 4.2 × 10³ Ω with tolerance ±5%.

Ques 2: The resistance of the given carbon resistor is (2.4 ×10⁶ Ω ± 5%) Ω. What is the sequence of colours on the strips provided on resistor?

Ans:

2 → Red

4 → Yellow

10⁵ (multiplier)→ Green

5% tolerance → Gold