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 $\frac{\pi }{2}.$  If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

Correct Option - 4


Detailed Solution:

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 $\omega t$

tan = 1 = 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.$

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

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

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

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