# NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (Part V)

Jul 26, 2017 13:45 IST

NCERT Exemplar Solutions for CBSE Class 12 Physics – Chapter 3 (Current Electricity) are available here. In this article, you will find solutions for long answer type questions (i.e. question number 3.28 to question number 3.31).  Solutions of question number 3.1 to 3.27 are already available in Part I, Part II, Part III & Part IV. These questions and solutions are important for CBSE Class 12th Physics board exam 2018 and other engineering & medical entrance exams like NEET, UPSEE, WBJEE, JEE Main etc.

NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (from question number 3.22 to 3.27) are given below:

Question 3.28: Two cells of voltage 10V and 2V and internal resistances 10Ω and 5Ω respectively, are connected in parallel with the positive end of 10V battery connected to negative pole of 2V battery (Fig 3.8). Find the effective voltage and effective resistance of the combination.

Solution 3.28:

Applying Kirchhoff’s junction rule, we have

I1 = I + I2

Kirchhoff’s loop rule gives:

10 = IR + 10I1....(i)

2 = 5I2 – RI = 5 (I1 – I) – RI

4 = 10 I1 – 10I – 2 RI..... (ii)

Subtracting …(i) & …(ii), we have

6 = 3RI + 10I or, 2 = I (R + 10/3)

2 = (R+ Reff) I Comparing with Veff = (R + Reff) I and Veff = 2 V

Reff = 10/3 ῼ

Question 3.29: A room has AC run for 5 hours a day at a voltage of 220V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions? [ρcu = 1.7 × 10–8 Ωm , ρAl = 2.7 × 10–8 Ωm]

Solution 3.29:

Power consumption = 2units/hour = 2KW = 2000J/s

I = P/V = 2000/220 = 9 A

Power loss in wire = RI2 J/s = ρ (l/A) I2 = 1.7 × 10‒8 × [10/(π × 10‒6) × 81 J/s = 4 J/s

Fractional loss due to the joule heating in first wire = (4/2000) × 100 = 0.2 %

Power loss in Al wire = 4 (ρAl/ρCu) = 1.6 × 4 = 6.4 J/s = 0.32%.

Question 3.30: In an experiment with a potentiometer, VB = 10V. R is adjusted to be 50Ω (Fig. 3.9). A student wanting to measure voltage E1 of a battery (approx. 8V) finds no null point possible. He then diminishes R to 10Ω and is able to locate the null point on the last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

Solution 3.30:

Let R’ be the resistance of the potentiometer wire.

(10 × R’)/(50 + R’) < 8

⇒ 10 R’ < 400 + 8 R

2R’ < 400 or R’ < 200 Ω.

(10 × R’)/(10 + R’) > 8 ⇒ R’ > 40

[10 × (3/4) R’]/[10 + R’] < 8 ⇒ 7.5 R’  < 80 + 8R

R’ > 160 ⇒160 < R’ < 200

Any R’ between 160W and 200W will achieve.

Potential drop across 400 cm of wire > 8V.

ϕ × 400 cm > 8V (ϕ → potential gradient)

ϕ × 4 m > 8V

ϕ > 2 V/m

Potential drop across 300 cm of wire < 8V.

ϕ × 300 cm < 8V

ϕ × 3 m < 8V

ϕ < 2(2/3) V/m

Thus, 2(2/3) V/m > ϕ > 2 V/m.

Question 3.31:

(a) Consider circuit in Fig 3.10. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity?

(b) Electrons give up energy at the rate of RI2 per second to the thermal energy. What time scale would one associate with energy in problem (a)? n = no of electron/volume = 1029/m3, length of circuit = 10 cm, cross-section = A = (1mm)2.

Solution 3.31:

(a)

Using Ohm’s law, current, I = 6V/6Ω = 1 A,

I = n e A vd

vd = [1/(1029 × 1.6 × 10‒19 × 10‒6)] = [(1/1.6) × 10‒4] m/s

Kinetic energy = (&frac12;) [me vd2 × nAl] = (&frac12;) [(9.1 × 10‒31) × (1/2.56) × 0‒8 × 1029 × 10‒6 × 10‒1

= 2 × 10‒17 Joule.

(b) Ohmic loss = RI2 = 6 × 12 = 6 J/s

All of KE of electrons would be lost in (2×10‒17)/6 seconds = 10‒17 seconds.

NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 3: Current Electricity

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