Page No 129: - Chapter 3 Current Electricity Additional Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Question 3.15:

(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Answer:

(a) Number of secondary cells, n = 6

Emf of each secondary cell, E = 2.0 V

Internal resistance of each cell, r = 0.015 Ω

series resistor is connected to the combination of cells.

Resistance of the resistor, R = 8.5 Ω

Current drawn from the supply = I, which is given by the relation,

Terminal voltage, V = IR = 1.39 × 8.5 = 11.87 A

Therefore, the current drawn from the supply is 1.39 A and terminal voltage is

11.87 A.

(b) After a long use, emf of the secondary cell, E = 1.9 V

Internal resistance of the cell, r = 380 Ω

Hence, maximum current

Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.

Question 3.16:

Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρ_Al = 2.63 × 10⁻⁸ Ω m, ρ_Cu = 1.72 × 10⁻⁸ Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

Answer:

Resistivity of aluminium, ρ_Al = 2.63 × 10⁻⁸ Ω m

Relative density of aluminium, d₁ = 2.7

Let l₁ be the length of aluminium wire and m₁ be its mass.

Resistance of the aluminium wire = R₁

Area of cross-section of the aluminium wire = A₁

Resistivity of copper, ρ_Cu = 1.72 × 10⁻⁸ Ω m

Relative density of copper, d₂ = 8.9

Let l₂ be the length of copper wire and m₂ be its mass.

Resistance of the copper wire = R₂

Area of cross-section of the copper wire = A₂

The two relations can be written as

It is given that,

And,

Mass of the aluminium wire,

m₁ = Volume × Density

= A₁l₁ × d₁ = A₁ l₁d₁ … (3)

Mass of the copper wire,

m₂ = Volume × Density

= A₂l₂ × d₂ = A₂ l₂d₂ … (4)

Dividing equation (3) by equation (4), we obtain

It can be inferred from this ratio that m₁ is less than m₂. Hence, aluminium is lighter than copper.

Since aluminium is lighter, it is preferred for overhead power cables over copper.

Question 3.17:

What conclusion can you draw from the following observations on a resistor made of alloy manganin?

It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7 Ω.Answer:

Question 3.18:

Answer the following Questions:

(a) A steady current flows in a metallic conductor of non-uniform cross- section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?

(b) Is Ohm’s law universally applicable for all conducting elements?

If not, give examples of elements which do not obey Ohm’s law.

(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?

(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

Answer:

(a) When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant.

(b) No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.

(c) According to Ohm’s law, the relation for the potential is V = IR

Voltage (V) is directly proportional to current (I).

R is the internal resistance of the source.

If V is low, then R must be very low, so that high current can be drawn from the source.

(d) In order to prohibit the current from exceeding the safety limit, a high tension supply must have a very large internal resistance. If the internal resistance is not large, then the current drawn can exceed the safety limits in case of a short circuit.

Question 3.19:

Choose the correct alternative:

(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.

(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.

(c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.

(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10²²/10³).

Answer:

(a) Alloys of metals usually have greater resistivity than that of their constituent metals.

(b) Alloys usually have lower temperature coefficients of resistance than pure metals.

(c) The resistivity of the alloy, manganin, is nearly independent of increase of temperature.

(d) The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 10²².

Question 3.20:

(a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?

(c) Determine the equivalent resistance of networks shown in Fig. 3.31.

Answer:

(a) Total number of resistors = n

Resistance of each resistor = R

(i) When n resistors are connected in series, effective resistance R₁is the maximum, given by the product nR.

Hence, maximum resistance of the combination, R₁ = nR

(ii) When n resistors are connected in parallel, the effective resistance (R₂) is the minimum, given by the ratio.

Hence, minimum resistance of the combination, R₂ = 

(iii) The ratio of the maximum to the minimum resistance is,

(b) The resistance of the given resistors is,

R₁ = 1 Ω, R₂ = 2 Ω, R₃ = 3 Ω2

1. Equivalent resistance, 

Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,

2. Equivalent resistance, 

Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,

(iii) Equivalent resistance, R^’ = 6 Ω

Consider the series combination of the resistors, as shown in the given circuit.

Equivalent resistance of the circuit is given by the sum,

R^’ = 1 + 2 + 3 = 6 Ω

(iv) Equivalent resistance, 

Consider the series combination of the resistors, as shown in the given circuit.

Equivalent resistance of the circuit is given by,

(c) (a) It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in series.

Hence, their equivalent resistance = (1+1) = 2 Ω

It can also be observed that two resistors of resistance 2 Ω each are connected in series.

Hence, their equivalent resistance = (2 + 2) = 4 Ω.

Therefore, the circuit can be redrawn as

It can be observed that 2 Ω and 4 Ω resistors are connected in parallel in all the four loops. Hence, equivalent resistance (R^’) of each loop is given by,

The circuit reduces to,

All the four resistors are connected in series.

Hence, equivalent resistance of the given circuit is 

(b) It can be observed from the given circuit that five resistors of resistance R each are connected in series.

Hence, equivalent resistance of the circuit = R + R + R + R + R

= 5 R

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