Question

In the shown circuit, $R_1=10\Omega ,L=\frac{\sqrt{3}}{10}H,R_2=20\Omega ,C=\frac{\sqrt{3}}{2}milli-farad$ and $t$ is time in seconds. Then at the instant current through $R_{1}is10\sqrt{2}A$ ;

find the current through resistor $R_2$ in amperes.

Answer 1

Given , $V=200\sqrt{2}\sin \left(100t\right)volt$.

thus, $V_0=200\sqrt{2},\omega =100$

For $R_1-L$ branch : $X_L=\omega L=100\left(\frac{\sqrt{3}}{10}\right)=10\sqrt{3}\Omega$, $R_1=10\Omega$

$\therefore \tan {\varphi }_1=\frac{X_L}{R_1}=\frac{10\sqrt{3}}{10}=\sqrt{3}\Rightarrow {\varphi }_1={60}^o$

Hence the current $I_1$ lags voltage by ${60}^o$.

For $R_2-C$ branch$:X_C=\frac{1}{\omega C}=\frac{1}{100(\frac{\sqrt{3}}{2}\times {10}^{-3})}=\frac{20}{\sqrt{3}}\omega$ and $R_2=20\Omega$

$\therefore \tan {\varphi }_2=\frac{X_C}{R_2}=\frac{1}{\sqrt{3}}\Rightarrow {\varphi }_2={30}^o$

The phase difference between $I_1$ and $I_2$ is $\varphi ={\varphi }_1+{\varphi }_2=60+30={90}^o$

The maximum current from source is $I=\frac{V_0}{\sqrt{R_1^2+{\omega }^2{L}^2}}=\frac{200\sqrt{2}}{{10}^2+(10\sqrt{3}{)}^2}=10\sqrt{2}A$.

Hence $I=I_1+I_2$

$10\sqrt{2}=10\sqrt{2}+I_2\Rightarrow I_2=0A$