Solutions-Class-11-Science-Physics-Chapter-11-Electric Current Through Conductors-Maharashtra Board

(D) 100 W

Explanation :

P = \(\frac{V^2}{R}\)

Since all the bulbs are operating at the same volt.

∴ P ∝ 1/R

∴ The bulb with the highest power will have the lowest resistance. i.e., 100 W bulb

(D) nichrome wire

(B) to continuously change the resistance in any arbitrary manner and there by alter the current

(D) 16R (assuming that the volume of the wire does not change when it is stretched)

(D) 125:15:1

Explanation :

R = ρ\(\frac{l}{A}\) = ρ\(\frac{l×l}{A×l}\) = ρ\(\frac{l^2}{V}\)

Since V = m/d

R = ρ\(\frac{l^2}{m/d}\)  ,  ∴ R ∝ \(\frac{l^2}{m}\)

∴  R₁ : R₂ : R₃ = \(\frac{l_1^2}{m_1}\) : \(\frac{l_2^2}{m_2}\) : \(\frac{l_3^2}{m_3}\)

∴  R₁ : R₂ : R₃ = \(\frac{5^2}{1}\) : \(\frac{3^2}{3}\) : \(\frac{1^2}{5}\) = \(\frac{25}{1}\) : \(\frac{9}{3}\) : \(\frac{1}{5}\)

Then R₁ : R₂ : R₃ = 125 : 15 : 1

(B) 1.8 V

(A) 2

Explanation :

let, m = Number of rows, n = Number of cells in a row

∴ m x n = 100 ..... (i)

Resistance is series = nr

Resistance in parallel = m/nr

For maximum current, R = nr/m

25 = (n x 1)/m = n/m ….(ii)

From (i) and (ii)

m x 25m = 100

∴ m² = 100/25 = 4

∴ m = 2

(B) 4.5V

Explanation :

(4 x 1.5) – (1.5) = 6 - 1.5 = 4.5 V

Given : R₁ = 8 Ω, R₂ = 6 Ω, V = 5 V, series combination

(a) The current in the circuit, I = \(\frac{V}{R}=\frac{V}{R_1+R_2}=\frac{5}{8+6}\) = \(\frac{5}{14}\)A

Hence, the potential difference across 8 Ω resistor,

V₁ = R₁I = 8 x \(\frac{5}{14}\) = \(\frac{20}{7}\) V = 2.857 V

(b) Reason: As per the question, the new circuit diagram will be

When a third resistor, say R, is connected in parallel with 6 Ω resistor, the resistance of the parallel combination of R and 6 Ω will be less than 6 Ω. Therefore, the total resistance in the circuit will be less than 14 Ω. Hence, the current in the circuit will be more than \(\frac{5}{14}\)A.

Hence, the potential difference across 8 Ω  resistor will be larger than earlier value ( \(\frac{20}{7}\)V).

Let A be the area of cross section of a conducting wire across which an electric field \(\vec{E}\) is applied, resulting in a current I.

Consider a section of length L of the wire. See above Fig for reference.

The volume of this part of the wire is AL. If n is the number of free electrons per unit volume of the material of the wire, the number of free electrons in volume AL is nAL. The charge carried by the free electrons in this part is q = nALe, where e denotes the electron charge.

If v_d is the drift speed of the electrons, the time taken by the electrons to cover the distance L is t = L/v_d  (‘.’ v_d = L/t).

The charge q is transported across a cross section of the wire in time t. Hence, the corresponding current is

I = q/t = \(\frac{nALe}{L/v_d}\) = nAv_de

Hence, the current density, i.e., the current per unit cross sectional area is

J = I/A = \(\frac{nAv_de}{A}\) = (ne)v_d.

For a given material, ne = constant.  ∴ J ∝ v_d

The direction of \(\vec{J}\) is the direction of the current. It is opposite to that of drift velocity \(\vec{v_d}\).

(a) Ohmic conductor: A conductor or a device which obeys Ohm’s law is called an ohmic conductor.

A graph of current (I) versus the potential difference (V) across an ohmic conductor is a straight line through the origin, provided its physical state remains the same, as shown in the following figure :

Hence, the ratio V/I = R, which is the reciprocal of the slope of the straight line, is the same for all values of V. In other words, the resistance of the conductor is independent of the magnitude and polarity of V.

The resistance of a conductor changes with temperature. Hence, the graph is not a straight line when the temperature of the conductor increases due to either prolonged passage of current or passage of a large current.

• Examples : Metals and metallic alloys, e.g., silver, copper, aluminium, iron, manganin, Nichrome, constantan, etc.

(b) Non-ohmic conductor : A conductor or a device which does not obey Ohm’s law is called a non-ohmic conductor.

A graph of current versus potential difference across a non-ohmic conductor is a curve as shown in Fig. for a pn—junction diode (semiconductor diode). That is, the resistance of such a conductor changes with the applied potential difference.

If for a certain potential difference V the current through the conductor is I, then its resistance at that V is given by

R = \(\lim\limits_{ΔI\to 0} \frac{ΔV}{ΔI}=\frac{dV}{dI}\)

where ΔV is the potential difference between the two values of potential V\(-\frac{ΔV}{2}\), V +\(-\frac{ΔV}{2}\) , and ΔI is the corresponding change in the current.

• Examples : Liquid electrolytes, vacuum tubes, junction diodes, thermistors etc.

Current remains constant along the conductor.

Given : l = 20 km = 2 × 10⁴ m, T = 20 °C, A = 25 cm² = 2.5 × 10^-3 m²,

ρ at 20°C = 6 x 10^-8 Ω.m.

The resistance of the rail,

R = ρ\(\frac{l}{A}\) = 6 x 10^-8\(\frac{2×10^4}{2.5×10^{-3}}\)= 6 × 0.8 × 10^-1 = 0.48 Ω

Given : ε = 24 V, r = 380 Ω

(i) The maximum current that can be drawn from the battery = \(\frac{ε}{r}=\frac{24}{380}\) = 0.0632 A

(ii) As the value of the current is very small compared to the required current to run a starting motor of a car, this battery cannot be used to drive the motor.

Given : ε = 12 V, r = 3 Ω, I = 0.5 A

[a] I = \(\frac{ε}{R+r}\)  ∴ R = \(\frac{ε-Ir}{I}\) = \(\frac{12-(0.5)(3)}{0.5}\) = 24 – 3 = 21 Ω

This is the resistance of the resistor.

(b) The terminal voltage when the circuit is closed,

V = IR = (0.5) (21) = 10.5 V

Given : J = 500 A/cm² = (500 A)/(10^-4 m²) = 5 x 10⁶ A/m²,

N = 8.47 x 10²² free electrons/cm³ = 8.47 x 10²⁸ free electrons/m³, e = 1.6 x 10^-19 C

v_d = \(\frac{J}{ne}\) = \(\frac{5×10^6}{(8.47×10^{28})(1.6×10^{-19})}\)

= \(\frac{50}{8.47×1.6}\) × 10^-4 = 3.689 × 10^-4 m/s

The drift velocity of the electrons in the copper wire is 3.689 × 10^-4 m/s

Given : R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 30 Ω, V = 12 V

(i) R = R₁ + R₂ + R₃ = 10 +20 + 30 = 60 Ω

(ii) Potential difference,

V₁ = R₁I = R₁\((\frac{V}{R})\) = 10\((\frac{12}{60})\) = 10\((\frac{1}{5})\)  = 2 V

V₂ = R₂I = R₂\((\frac{V}{R})\) = 20\((\frac{1}{5})\) = 4 V

V₃ = R₃I = R₃\((\frac{V}{R})\) = 30\((\frac{1}{5})\) = 6 V

Given : R₁ = 1 k Ω = 10³ Ω, R₂ = 2 k Ω = 2 x 10³ Ω, V = 9 V

(i) \(\frac{1}{R_P} = \frac{1}{R_1}+\frac{1}{R_2}\) = \(\frac{1}{1}+\frac{1}{2}\) = \(\frac{3}{2}\)

∴ R_P = \(\frac{2}{3}\) = 0.6666 k Ω

(ii) Current I₁ = \(\frac{V}{R_1}\) = \(\frac{9}{1×10^3}\) = 9 x 10^-3 A

I₂ = \(\frac{V}{R_2}\) = \(\frac{9}{2×10^3}\) = 4.5 x 10^-3 A

Given : R₁ = 4.2 Ω, T₁ = 27 ⁰C, R₂ = 5.4 Ω, T₂ = 100 °C,

R = R₀ [1 + ∝ (T - T₀)] = R₀ (1 + ∝T) for T₀ = 0 ⁰C

∴ R₁ = R₀ (1 + ∝T₁) and R₂ = R₀ (1 + ∝T₂)

∴ \(\frac{R_2}{R_1}\) = \(\frac{1+∝T_2}{1+∝T_1}\)

∴ R₂ + R₂∝T₁ = R₁ + R₁∝T₂

∴ R₂ - R₁ = ∝(R₁T₂ - R₂T₁)

∴ The temperature coefficient of resistance,

∝ = \(\frac{R_2-R_1}{R_1T_2-R_2T_1}\) = \(\frac{5.4-4.2}{4.2×100-5.4×27}\)

= \(\frac{1.2}{420-145.8}\) = \(\frac{1.2}{274.2}\) = 4.376 × 10^-3/⁰C

Given : R = 50 Ω, l = 6 m, d = 0.5 mm, r = d/2 = 0.25 mm = 2.5 x 10^-4

R = ρ\(\frac{l}{A}\) = ρ\(\frac{l}{πr^2}\)

∴ The resistivity, ρ = \(\frac{Rπr^2}{l}\) = \(\frac{(50)(3.142)(2.5×10^{-4})^2}{6}\) = 1.636 x 10^-6 Ω.m

The conductivity, σ = \(\frac{1}{ρ}=\frac{1}{1.636×10^{-6}}\) = 6.112 x 10⁵ Ω.m

(1)

∴ R = 65 x 10² Ω ±5% = 6.5 x 10³ Ω ±5% = 6.5 k Ω ±5% = (6500 ± 325) Ω

(2)

∴ R = 10 x 10² Ω ±10% = 1.0 x 10³ Ω ±10% = 1.0 k Ω ±10% = (1000 ± 100) Ω

(3)

∴ R = 22 x 10³ Ω ±5% =  22 k Ω ±5% = (22000 ± 1100) Ω

(4)

∴ R = 39 x 10² Ω ±5% = 3.9 x 10³ Ω ±5% =  3.9 k Ω ±5% = (3900 ± 195) Ω

(5)

∴ R = 47 x 10 Ω ±10% = 470  Ω ±10% = (470 ± 47) Ω

(a) 330 Ω → Orange, Orange, Brown, Silver

(b) 100 Ω → Brown, Black, Brown, Silver

(c) 47 Ω → Yellow, Violet, Orange, Silver

(d) 160 Ω → Brown, Blue, Brown, Silver

(e) 1 k Ω → Brown, Black, Red, Silver.

Given : I = 4 A, t = 2 h = 2 x 3600 s = 7200 s, |e|= 1.6 ×10^-19 C

I = \(\frac{q}{t}\) = \(\frac{n|e|}{t}\)

∴ n = \(\frac{It}{|e|}\) = \(\frac{4×7200}{1.6×10^{-19}}\) =  \(\frac{72}{4}\) x 10²² = 1.8 x 10²³

1.8 x 10²³ number of electrons flow through the headlight in 2hrs.

Given : V = 230 V, I = 5 A, t = 1h = 3600 s, J = 4.2 J/cal

The amount of heat dissipated,

H = \(\frac{VIt}{J}\) = \(\frac{(230)(5)(3600)}{4.2}\) = 9.857 x 10⁵ cal = 985.7 kcal