Alternating Current

$D.C.$ ammeter measure average current in $AC$ current, average current is zero for complete cycle. Hence reading will be zero.

For resonant frequency to remain same $LC$ should be const. $LC=$ const

$ \Rightarrow LC = L' \times 2C \Rightarrow L' = {L \over 2}$

Laminated core provide less area of cross-section for the current to flow. Because of this, resistance of the core increases and current decreases thereby decreasing the eddy current losses.

When the capacitor is completely charged, the total energy in the $L.C$ circuit is with the capacitor and that energy is $E = {1 \over 2}{{{Q^2}} \over C}$

When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get ${E \over 2} = {1 \over 2}{{Q{'^2}} \over C}$ where $Q'$ is the charge on plate of the capacitor

$\therefore$ ${1 \over 2} \times {1 \over 2}{{{Q^2}} \over C} = {1 \over 2}{{Q{'^2}} \over C}$

$ \Rightarrow Q' = {Q \over {\sqrt 2 }}$

The impedance triangle for resistance $\left( R \right)$ and inductor $(L)$ connected in series is shown in the figure.



Power factor $\cos \phi = {R \over {\sqrt {{R^2} + {\omega ^2}{L^2}} }}$

${N_p} = 140,\,\,{N_s} = 280,\,\,{I_p} = 4A,\,\,{I_s} = ?$

For a transformer ${{{I_s}} \over {{I_p}}} = {{{N_p}} \over {{N_s}}}$

$ \Rightarrow {{{I_s}} \over 4} = {{140} \over {280}} \Rightarrow {I_s} = 2A$