Solution:

Ans. (a) Principle of working

(b) Defining efficiency

1

(c) Any two factors

(d) Calculating the current drawn

$1 / 2+1 / 2$

(a) A tansformer works on the principle of mutual induction.

(Alternatively-an emf is induced in the secondary coil when the magnetic flux, linked with it changes with time due to a (time) changing magnetic flux linked with the primary coil).

(b) The efficiency of a transformer equals the ratio of the output power to the input power.

Alternatively

Efficiency $=\frac{\text { output power }}{\text { inputpower }}$

$$ \text { Efficiency }=\frac{V_{s} I_{s}}{V_{p} I_{p}} $$

(c) (i) Eddy current losses

(ii) joule heat losses

(iii) hysteresis losses

(iv) magnetic flux leakage losses

(Any two)

(d) We have

$$ \begin{array}{rlrl} & \frac{V_{s} I_{s}}{V_{p} I_{p}} & =90 %=0.9 \ & \therefore \quad \frac{22 I_{s}}{220 I_{p}} & =0.9 \ & \text { or } \quad \frac{I_{s}}{I_{p}} & =\frac{0.9}{0.1}=9 \ \therefore \quad & I_{p} & =\frac{I_{s}}{9}=\frac{\left(\frac{22}{440}\right)}{9} \mathrm{~A} \ & & =\frac{1}{180} \mathrm{~A} \ & =0.0056 \mathrm{~A} \end{array} $$

[CBSE Marking Scheme, 2018]

(AI Q. 3. (i) Draw a labelled diagram of an $a c$ generator. Obtain the expression for the emf induced in the rotating coil of $N$ turns each of cross-sectional area $A$, in the presence of a magnetic field $\vec{B}$.

(ii) A horizontal conducting rod $10 \mathrm{~m}$ long extending from east to west is falling with a speed $5.0 \mathrm{~ms}^{-1}$ at right angles to the horizontal component of the Earth’s magnetic field, $0.3 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$. Find the instantaneous value of the emf induced in the rod. U [O.D. Set III 2017]

Ans. (i) Considee a coil rotateng in a ceniform magnetic fold-The fux associaied wetb the coil $\phi . N B A \cos \theta$

[Topper’s Answer 2017]