- When circuit element is X, the current is in phase with the applied emf, this implies that X is pure resistance.
When circuit element Y is connected, the current leads the voltage by $\frac{\pi}{2}$ so Y is pure capacitance.
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Resistance of X = R; $\text{I}=\frac{\text{V}}{\text{R}}$
$\sqrt{2}=\frac{\text{V}}{\text{R}} \ ...(\text{i})$
Reactance of $\text{Y, X}_{\text{C}}=\frac{1}{\omega\text{C}},\text{I}=\frac{\text{V}}{\text{X}_{\text{C}}}\Rightarrow\sqrt{2\text{C}}=\frac{\text{V}}{\text{X}_{\text{C}}} \ ...(\text{ii})$
This implies $\text{X}_{\text{C}}=\text{R}$
When R and C are connected in series across same voltage source, then
Impedance $\text{Z}=\sqrt{\text{R}^2+\text{X}^2_{\text{C}}}=\sqrt{\text{R}^2+\text{R}^2}=\sqrt{2}\text{R} \ ...(\text{iii})$
$\therefore$ Current in circuit $\text{I}=\frac{\text{V}}{\text{Z}}=\frac{\text{V}}{\sqrt{2}\text{R}}$
From (1), $\frac{\text{V}}{\text{R}}=\sqrt{2},\therefore\text{I}=\frac{1}{\sqrt{2}}\times\sqrt{2}=1\text{A}$
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$\text{Z}=\sqrt{\text{R}^2+\Big(\frac{1}{\omega\text{C}}\Big)^2}$
i.e., impedance decreases with increase of angular frequency $\omega$
When $\omega=0,\text{Z}=\infty$
When $\omega=0,\text{Z = R}$
The graph of impedance Z versus $\omega$ is shows in fig.
