(i) Principle of working: It works on the principle of mutual induction i.e. if two coils are inductively coupled and when current or magnetic flux is changed through one of the two coils, then induced e.m.f. is produced in the other coil. (ii) Turns ratio is $K = \frac{n_s}{n_p} = \frac{E_s}{E_p}$ A transformer with a primary winding of 1000 turns and secondary winding of 100 turns has a turns ratio of $1000:100$ or $10:1$. Therefore, 100 volts applied to primary winding will produce a secondary voltage of 10 volts. (iii) $E_s I_s = E_p I_p$ $\text{ (Input Power = Output Power) }$ $\Rightarrow \frac{E_s}{E_p} = \frac{I_p}{I_s}$ $\Rightarrow \frac{I_p}{I_s} = \frac{n_s}{n_p} = K$ (iv) $e_p = 220 \text{V}; e_s = 110 \text{V}, e_s I_s = 550 \text{W}$ Now, $e_p I_p = e_s I_s$ $I_p = \frac{e_s I_s}{e_p} = \frac{550}{220} = 2.5 \text{A}$ OR (a) Meaning of Mutual Inductance Expression (b) Proof Diagram (a) Mutual Inductance is the property of a pair of coils due to which an emf induced in one of the coils due to the change in the current in the other coil. $\text{ Mathematically } e_2 = \frac{M d i_1}{d t}$ $\therefore M = -\frac{e_2}{\frac{d i_1}{d t}}$ Let a current $I_2$ flows through the outer circular coil. Then $B_2 = \frac{\mu I_2}{2 r_2}$ $\therefore \phi_1 = \pi r^2 B_2 = \frac{\mu \pi r_1^2}{2 r_2} I_2 = M_{12} I_2$ Thus $M_{12} = \frac{\mu \pi r_1^2}{2 r_2} I_2 = M_{21}$ (b) Flux at any time ' $t$ '. $\phi_B = B A \cos \theta = B A \cos \omega t$ From Faraday's Law, induced emf $e = -N \frac{d \phi_B}{d t} = N B A \frac{d}{d t}(\cos \omega t)$ Thus the instantaneous value of emf is $E = N B A \omega \sin \omega t$ For maximum value of emf $\sin \omega t = \pm 1$ i.e., $e_0 = N B A \omega = 2 \pi f N B A$