(a) Principle of ac generator working Mark labelled diagram Derivation of the expression for induced emf (b) Calculation of potential difference (a) The AC Generator works on the principle of electromagnetic induction. When the magnetic flux through a coil changes, an emf is induced in it. As the coil rotates in magnetic field the effective area of the loop, (i.e. $A \cos heta)$ exposed to the magnetic field keeps on changing, hence magnetic flux changes and an emf is induced. When a coil is rotated with a constant angular speed ' $\omega$ ', the angle ' $heta$ ' between the magnetic field vector $\overline{B}$ and the area vector $\overline{A}$ , of the coil at any instant ' $t$ ' equals $\omega t$ ; (assuming $heta = 0^{\circ}$ at $t=0$ ) As a result, theeffective area of the coil exposed to the magnetic field changes with time; The flux at any instant ' $t$ ' is given by $\phi_B = N B A \cos heta = N B A \cos \omega t$ $herefore$ The induced emf $e = -N \; \frac{d\phi}{dt}$ $= -N B A \; \frac{d\phi}{dt} (\cos \omega t)$ $e = N B A \omega \sin \;\omega t$ (b) Potential difference developed between the ends of the wings ' $e$ ' $= B l v$ Given Velocity $v = 900 ext{km}/ ext{hour}$ $= 250 ext{m}/ ext{s}$Wing span $(l) = 20 ext{m}$Vertical component of Earth's magnetic field $B_v = B_H an \delta$ $= 5 imes 10^{-4} ( an 30^{\circ}) ext{Tesla}$ $herefore$ Potential difference $= 5 imes 10^{-4} ( an 30^{\circ}) imes 20 imes 250$ $= \frac{5 \times 20 \times 250 \times 10^{-4}}{\sqrt{3}}$ $= 1.44 ext{ volt}$ OR (a) Identification of the device $X$ Expression for reactance (b) Graphs of voltage and current with time (c) Variation of reactance with frequency (Graphical variation) (d) Phasor Diagram (a) $X$ : capacitor Reactance $X_c = \frac{1}{\omega C} = \frac{1}{2\pi v C}$ (b) (c) Reactance of the capacitor varies in inverse proportion to the frequency i.e., $X_c \propto \frac{1}{v}$