(a) Labelled diagram of AC generator Expression for instantaneous value of induced emf. (b) Calculation of maximum value of current When the coil is rotated with constant angular speed the angle $\theta$ between the magnetic field and area vector of the coil, at instant $t$, is given by $\theta = \omega t,$ Therefore, magnets flux, $(\phi_{B})$, at this instant, is $\phi_{B} = B A \cos \omega t$ Induced emf $e = -N \frac{d\phi_B}{dt}$ $e = N B A \omega \sin \omega t$ $e = e_0 \sin \omega t$ where $e_0 = N B A \omega$ (b) Maximum value of emf $e_0 = N B A \omega$ $= 20 \times 200 \times 10^{-4} \times 3 \times 10^{-4} \times 50\,\mathrm{V}$ $= 600\,\mathrm{mV}$ Maximum induced current $i_0 = \frac{e_0}{R} = \frac{600}{R}\,\mathrm{mA}$ OR (a) Labelled diagram of a step up transformer Derivation of ratio of secondary and primary voltage (b) Calculation of number of turns in the secondary (a) When ac voltage is applied to primary coil the resulting current produces an alternating magnetic flux, which also links the secondary coil. The induced emf, in the secondary coil, having Ns turns, is $e_s = -N_s \frac{d\phi}{dt}$ This flux, also induces an emf, called back emf, in the primary coil. $e_s = -N_s \frac{d\phi}{dt}$ But $e_p = V_p$ and $e_s = V_s$ $\Rightarrow \frac{V_s}{V_p} = \frac{N_s}{N_p}$ For an ideal transformer $l_p V_p = i_s V_s$ $\Rightarrow \frac{V_s}{V_p} = \frac{i_p}{i_s}$ (b)$\frac{N_s}{N_p} = \frac{V_s}{V_p}$$\frac{N_s}{3000} = \frac{220}{2200}$$\therefore N_s = 300$