Resistivity Formula Important Questions

Resistivity Formula can be denoted by, ρ = RA/l, where: R = Resistance of conductor, A = Cross-Sectional area, and l = Length of the conductor. Resistivity can be expressed as the resistance of a conductor when its length and surface area are in unity. The SI unit of resistivity is ohm-meter. It is usually represented by the greek letter ‘ρ’.

Resistivity

Discover about the Chapter video:

Current Electricity Detailed Video Explanation:

Read Also: NCERT Solutions for Class 12 Physics Current Electricity

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Very Short Answer Questions [1 mark Questions]

Ques. What is the resistivity of a conductor?

Ans. The resistivity of a material can be expressed as, R = ρ. Thus, it can be said that the resistivity of a material is equivalent to the resistivity of a conductor that has a unit length and unit area of cross-section.

Ques. What is the net charge in a current carrying conductor?

Ans. The net charge in a current carrying conductor is Zero.

Ques. What is the current which is assumed to flow in a circuit from the positive terminal to the negative one called?

Ans. The current assumed to flow in a circuit from a positive terminal to a negative terminal is known as Conventional current.

Ques. What is Electric Current?

Ans. Electric current can be defined as the directed rate of electric charge’s flow via any cross-section of a conductor.

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Short Answer Questions [2 marks Questions]

Ques. A carbon resistor is found to contain three strips of red colour, alongside a gold strip. Find the value of the resistor and its tolerance.

Ans. As per the questions, RRR Gold = (22 \(\times\) 10²)\(\pm 5 \%\)

Thus, the value of resistor = 2200 \(\Omega\), and the tolerance it has = \(\pm 5 \%\)

Ques. Considering that the temperature of a good conductor decreases, how does the relaxation time of the electrons vary in the conductor?

Ans. \(\rho = \frac{m}{ne_2 \tau}\). Thus, if the temperature of a good conductor decreases, the collision decreases as well, causing the relaxation time of electrons to decrease in return. This finally decreases the resistivity.

Ques. What is the variation of resistivity of Silicon (Si) with temperature in graph. [Delhi 2014]

Ans. The graph can be represented as:

Variation of resistivity of Silicon

Ques. Considering an equation AB = C, where A equals current density, and C equals electric field. Then what is B?

Ans. B is resistivity.

Now, as per the question, the following can be derived:

If, J = σE

⇒ Jρ = E

(\(\because\) J = current density, E = electric field)

Thus, B = ρ = Resistivity.

Ques. Express the relationship between current and drift velocity of electrons in a conductor. Also use the relationship to demonstrate the resistance of a conductor changing with the rise in temperature. [Compartment  2013]

Ans. The relationship between current and drift velocity of electrons in a conductor can be expressed by, l = Anev_d
Here,

• l = current, 
• A = area of conductor,
• n = number density of electrons
• v_d = drift velocity

Now, with the temperature increasing in a metallic conductor, the resistance increases too and therefore, drift velocity decreases.

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Long Answer Questions [3 marks Questions]

Ques. In a given metal, two identical slabs are attached together in two different ways, as per what is shown in the image below. Determine the ratio of the resistances of these two combinations. [Delhi 2010]

Ans. Assume that l and A are the length and cross-section of both slabs.

Thus, R = ρl/A

R₁ = \(\frac{\rho 2l}{A} = 2R\)

R₂ = \(\frac{\rho l}{2A} = \frac{R}{2}\)

Therefore, it can be said,

\(\frac{R_1}{R_2} = \frac{2R}{R/2}\)

= 4

Hence, the ratio of resistances of these two combinations can be expressed as:

R₁ : R₂ = 4 :1

Ques. Consider potential difference V has been applied across the ends of copper wire that has length (l) and diameter (D). Determine the effect on drift velocity of electrons assuming: 

1. V is doubled
2. l is doubled
3. D is doubled

Ans. (a) Assuming that V is doubled, V_d:

Therefore, assuming that V is doubled, drift velocity gets doubled as well.
(b) If l is doubled, then drift velocity becomes halved.
(c) Now since V is independent of D, drift velocity will have to remain unchanged.

Ques. Determine resistivity of a given material that has resistance of 2 Ω; area of cross-section as 25cm² and length of 15 cm.

Ans. According to the question,

• R = 2 Ω
• l = 15 cm = 0.15 m
• A = 25 cm² = 0.25 m²

Thus, the Resistivity formula can be denoted as:

\(\rho = \frac{RA}{l}\)

After replacing the values, we get:

⇒ \(\rho = \frac{2 \times 0.25}{0.15} = 3.333 \Omega m\)

Ques. Assuming that the resistance of a wire is 5 ohms (with 50^oC) and 6 ohms (with 100^oC). Determine the resistance of the wire if kept at 0^oC?

Ans. The Resistance of the wire should change linearly with temperature.

Thus, R = a(T) + b (let T be in the degree scale)

Now, after using the equation, we get, R (50) = 5

R (100) = 6

⇒ b = 4

⇒ a = 1/50

Thus, R = T/50 + 4

Placing R(0) for resistance at 0°, we can obtain:

⇒ R (0) = 4 ohms

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Very Long Answer Questions [5 marks Questions]

Ques. What is the relaxation time of free electrons that are seen to drift in a conductor. Also, determine how is it related to the drift velocity of free electrons? Finally use the relation to derive the expression for the electrical resistivity of the material. [All India 2012]

Ans. The relaxation time of free electrons that are seen to drift in a conductor can be defined as the average time passed between two successive collisions. The relation between τ and vd can be denoted by:

Now, assume length 'l' and cross-sectional area ‘A’ and electron density n.

Thus, the flow of current via the conductor is:

\(I = -neAv_d = neA \left(\frac{eE}{m}\tau\right)= \frac{ne^2EA}{m}\tau\)

Now, considering that V iequals potential difference applied across the two ends, then the electric field becomes, (E) = V/l. Therefore, current becomes:

Ques. Two metallic wires, each having the same material, also has the same length alongside cross-sectional area that is in the ratio 1: 2. They are both connected:

1. in series
2. in parallel.

​Given the data, now compare the electron’s drift velocities in two wires for both cases. [All India 2008]

Ans. As per the given question, A1 : A2 = 1 : 2

⇒ \(\frac{A_1}{A_2} = \frac{1}{2}\)

(i) Now, when the two wires are connected in a series combination, the current in both wires A and B will be same​
\(\therefore\) I_A = I_B

neA_1vd1 = neA_2vd2

Thus, \(\frac{vd_1}{vd_2} = \frac{A_2}{A_1} \quad \therefore \frac{vd_1}{vd_2} = \frac{2}{1} \implies vd_1 : vd_2 =2:1 \)

(ii) Now, when the two wires are connected in a parallel combination, then potential difference across A and B is going to be the same:

V_A = V_B ⇒ [∵ V = IR]

Thus, V = neA_d.\(\rho \frac{l}{A} \quad [\because V =ne\rho v_d]\)

Hence, \(ne\rho lv_{d1} = neA\rho v_{d2}\)

v_d1 = v_d2

Thus, \(\frac{vd_1}{vd_2} = \frac{1}{1} \implies v_{d1} : v_{d2} = 1:1\)

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