Solution:

Ans. A transformer is based on the principle of mutual induction which states that due to continuous change in the current in the primary coil, an emf gets induced across the secondary coil.

1

Electric power generated at the power station, is stepped-up to very high voltages by means of a step-up transformer and transmitted to a distant place. At receiving end, it is stepped down by a step-down transformer.

[CBSE Marking Scheme, 2012]

[AI Q. 2. An athlete peddles a stationary tricycle whose pedals are attached to a coil having 100 turns each of area $0\cdot 1{\mathrm{m}}^2$. The coil, lying in the $X-Y$ plane, is rotated, in this plane, at the rate of $50\mathrm{rpm}$, about the $\mathrm{Z}$-axis, in a region where a uniform magnetic field, $\overrightarrow{B}=(0.01)\hat{k}$ tesla, is present. Find the

(i) maximum emf

(ii) average emf generated in the coil over one completerevolution. A [CBSE SQP 2013]

Ans. (i) The maximum emf ’ ${\epsilon }_0$ ’ generated in the coil is,

$$\begin{array}{}\text{(1)}& {\epsilon }_0& =NBA\omega & =NBA2\pi f& (\because 50\mathrm{rpm}=\frac{50}{60}\text{ revolution }=\frac{5}{6})& =[100\times 0.01\times 0.1\times 2\pi \frac{(5)}{6}]\mathrm{V}& =\frac{\pi }{6}\mathrm{V}=0.52\mathrm{V}\end{array}$$

(ii) The average emf generated in the coil over one complete revolution $=0$ [AI Q. 1. (a) What is the principle of transformer?

(b) Explain how laminating the core a transformer helps to reduce eddy current losses in it

(c) Why the primary and secondary coils of a transformer are preferably wound on the same core

U[CBSE SQP 2018-19] Ans. (a) Try yourself Similar to Q. 2 (a) Long Answer Type Questions.

(b) Laminations are thin, making the resistance higher. Eddy currents are confined within each thin lamination. This reduces the net eddy current.

(c) For maximum sharing of magnetic flux and magnetic flux per turn to be the same in both primary and secondary.

[CBSE Marking Scheme 2018]

Long Answer Type Questions

(5 marks each)

AI Q. 1. (i) Describe, with the help of a suitable diagram, the working principle of a step-up transformer. Obtain the relation between input and output voltages in terms of the number of turns of primary and secondary windings and the currents in the input and output circuits. (ii) Given the input current $15\mathrm{A}$ and the input voltage of $100\mathrm{V}$ for a step-up transformer having $\mathbf{9}\mathbf{0}$ efficiency, find the output power and the voltage in the secondary if the output current is $3\mathrm{A}$.

R [Foreign I, II, III 2017]

Ans. (i) Diagram

Principle

Relation between voltage, number of turns, and Currents

(ii) Input power

Output power

Output voltage

(i)

(ii)

Working principle

Whenever current in one coil changes an emf gets induced in the neighbouring coil (Principle of mutual induction)

Voltage across secondary

$$V_s=e_s=-N_s\frac{d\varphi }{dt}$$

Oltage across primary

$$\begin{array}{rlr}{V}_p& =e_p=-N_p\frac{d\varphi }{dt}\frac{V_s}{V_p}& =\frac{N_s}{N_p}\end{array}$$

$$(\text{ Here, }{N}_s>N_p$$

In an ideal transformer

Power Input $=$ Power output

$$\therefore \frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}$$

(ii) Input power, $P_i=I_i\times V_i=15\times 100=1500\mathrm{W}1/2$ Power output, $P_0=P_i\times \frac{90}{100}=1350\mathrm{W}$

$$\Rightarrow I_0{V}_0=1350\mathrm{W}$$

Output voltage, $V_0=\frac{1350}{3}\mathrm{V}=450\mathrm{V}$

[CBSE Marking Scheme, 2017]