Unsolved Problems BL Theraja Vol-1-Tutorial Problems No. 1.1

Last Updated on November 16, 2023 by Electricalvolt

In this article, we will solve the Unsolved Problems from BL Theraja Vol-1-Tutorial Problems No. 1.1 for better understanding of electrical resistivity.

Problem-1 Calculate the resistance of 100 m length of a wire having a uniform cross-sectional area of 0.1 mm2 if the wire is made of manganin having a resistivity of 50 × 10−8 Ω-m. If the wire is drawn out to three times its original length, by how many times would you expect its resistance to be increased ?

Solution:

R = ρ L / A               (1)

Where,
R = resistance of the conductor (ohms, Ω)
ρ = resistivity of the conductor material (ohm metre, Ω-m)
L = length of conductor (m)
A = cross-sectional area of conductor (m2)

According to the given data, we have,
L= 100 m,        A=0.1 mm2,     ρ= 50 x 10-8 Ω m
R = 50×10-8 ………………………………. (from equation 1)
= 50×10-8x102x107
R = 50×10-8x109
R = 500 Ω

Now, according to question if the wire is drawn out to three times its original length
The new length becomes,

L’ = 3L

As we know that Volume can be defined as, (Area of cross section multiplied by Length), and here the volume will remain same and hence this can be expressed as

V= AxL = A’x3L
A= 3A’
A’=A/3

Therefore, from equation 1,

Thus, the resistance will increase by 9 times.

Problem-2 : A cube of a material of side 1 cm has a resistance of 0.001 Ω between its opposite faces. If the same volume of the material has a length of 8 cm and a uniform cross-section, what will be the resistance of this length?

Solution:

R = ρ L / A                                                                               (1)

Where,

R = resistance of the conductor (ohms, Ω)
ρ = resistivity of the conductor material (ohm metre, Ω m)
L = length of conductor (m)
A = cross-sectional area of conductor (m2)

According to the given data,

L= 1 cm,          R= 0.001 Ω                                                                  (2)

R =  ρ

As Volume is same and we know volume can be calculated by multiplying Area of cross section (A) and Length (L)

Hence,

R’ = 0.064 Ω

Problem-3 : A lead wire and an iron wire are connected in parallel. Their respective specific resistances are in the ratio 49 : 24. The former carries 80 percent more current than the latter and the latter is 47 percent longer than the former. Determine the ratio of their cross-sectional area.

According to the data given in question,

And now question is saying former (I1) carried 80 % more current than the latter (I2)
That means,

If, I2 = 1 A
Then, I1 = 1.8 A

Similarly for the length,
If, L1 = 1
Then, L2 = 1.47

Now,

And as the lead wire and iron wire connected in parallel. By using current division rule,

I1 = V1/ R1 and I1 = V2/R 2 (2)

By equating above,

Problem-4 : A rectangular metal strip has the following dimensions : x = 10 cm, y = 0.5 cm, z = 0.2 cm Determine the ratio of resistances Rx, Ry, and Rz between the respective pairs of opposite faces.

Solution:

As we know the Relation between resistance to its resistivity and dimensions can be calculated by using formula,

R = ρ L / A

According to the question,

Problem-5 : The resistance of a conductor 1 mm2 in cross-section and 20 m long is 0.346 Ω. Determine the specific resistance of the conducting material.

Solution:

R = ρ L / A               (1)

Where,

R = resistance of the conductor (ohms, Ω)
ρ = resistivity of the conductor material (ohm metre, Ω m)
L = length of conductor (m)
A = cross-sectional area of conductor (m2)

According to the given data, we have,

L= 20 m,          A=1 mm2,        R= 0.346Ω

By using equation 1,

R = ρ L / A
ρ  = RA / L

Specific resistance of the conducting material is, ρ  = 1.73 x 10-6 Ω m

Problem-6 : When a current of 2 A flows for 3 micro-seconds in a copper wire, estimate the number of electrons crossing the cross-section of the wire.

Solution:

Given,

Current through the wire is, I = 2 A
Time taken for this current flow, t = 3 µ s
t = 3×10-6 s

Assume,

n = number of electrons that crossed the cross section of wire in time t
e = Charge of the electron
So, the total charge in wire can be calculated as

q = ne                                      (where, q is total charge)                      (1)

Putting the values in above formula,

Therefore, the number of electrons crossing the cross-section of the wire is 3.75 x 1013.

Satyadeo Vyas, M.Tech,M.B.A. is an electrical engineer and has more than 36 years of industrial experience in the operation, maintenance, and commissioning of electrical and instrumentation projects. He has good knowledge of electrical, electronics, and instrumentation engineering.