Physics Alternating Current

X_C  = 1/ ωC

On increasing w; XC decreases so Z decreases, current in circuit increases.

$As{V}^2=V_R^2+C_C^2$

$?V_R=\sqrt[]{V^2-V_C^2}=\sqrt[]{{10}^2-8^2}=6V$

$\tan ?\theta =\frac{X_C}{X_R}=\frac{V_C}{V_R}=\frac{8}{6}=\frac{4}{3}$  

Now,

Reactance of the capacitor is $X=\frac{1}{\omega C}$

$C^{\text{'}}=3C$

${\omega }^{\text{'}}=3\omega$

$X^{\text{'}}=\frac{1}{C^{\text{'}}{\omega }^{\text{'}}}=\frac{1}{9\omega C}$

$X^{\text{'}}=\frac{X}{9}$

No power consumed without any resistance.

At resonance V_L = V_C

_{$\Rightarrow cos?=\frac{V_R}{\sqrt[]{V_R^2+{\left(V_L-V_C\right)}^2}}=1$}

$X_C=\frac{1}{\omega C}$

On increasing $\omega$ $$ XC decreases so Z decreases, current in circuit increases.

$\frac{H_{D.C.}}{H_{A.C.}}=\frac{I^{2}R}{I_{rms}^{2}R}=1$

$E=\sqrt[]{V_R^2+{\left(V_L-V_C\right)}^2}$

$E=\sqrt[]{{\left(60\right)}^2+{\left(100-20\right)}^2}$

E = 100V

-> $Z=\sqrt[]{R^2+{\left(X_L-X_C\right)}^2}$

$Z=\sqrt[]{{\left(120\right)}^2+{\left(10-100\right)}^2}=150\Omega$

$\omega =\frac{1}{\sqrt[]{LC}}=\frac{1}{\sqrt[]{10^{-1}*10^{-4}}}=\sqrt[]{10^5}$

$?\omega =2\pi f$

$\Rightarrow f=\frac{10^3}{2\pi \sqrt[]{10}}=50Hz$