A circuit element X when connected to an a.c. supply of peak voltage 100V given a peak current of 5A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by  If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

v = 100 sin $\omega t$                                                             

$i_0=\frac{100}{R_x}=5\Rightarrow R_x=20\Omega$               

i = $5sin\left(\omega t-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$  

 i = 100 5sin $\omega t$                                                                           

$5=\frac{100}{R_y}\Rightarrow R_y=\frac{100}{5}=20\Omega$

When x and y both are connected in series :-

v = 100sin  $\omega t$

tanθ = 1 = q = 45°

R₀ =   $\sqrt[]{R_x^2+R_y^2}=20\sqrt[]{2}\Omega$

$\therefore \stackrel{o}{{\mathcal{l}}_0}=\frac{v_0}{R_0}=\frac{100}{20\sqrt[]{2}}=\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$\sqrt[]{2}$}\right.A.$

$\therefore {\mathcal{l}}_{ms}=\frac{{\mathcal{l}}_0}{\sqrt[]{2}}$

$=\frac{5}{\sqrt[]{2}\cdot \sqrt[]{2}}$

$=\frac{5}{2}A=\frac{5}{2}A$