91Ó°ÊÓ

We are given that $75Ω$resistor , a $0.12H$inductor, and a $30mF$capacitor are there in $RLC$circuit. We are given that frequency of power line source equal to $60Hz$and voltage $120V$.

We need to find peak current.

Firstly we will find inductive reactance ,$X_L=2\mathrm{Ï€´Ú³¢}=2\mathrm{Ï€}×60×0.12=45.2$$Ω$

then we will capacitive reactance , $X_C=\frac{1}{2\mathrm{Ï€´Ú°ä}}=\frac{1}{2\mathrm{Ï€}×60×30×{10}^{-3}}=0.084$$Ω$

We will calculate impedance by putting the value of $R,X_L,X_C$

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}=87.5$

Peak current = $I=\frac{V_0}{Z}=\frac{120}{87.5}$$=1.37A$

We are given that $75Ω$resistor ,a $0.12H$inductor, and a $30mF$capacitor are there in $RLC$circuit. We are given that frequency of power line source equal to $60Hz$and voltage $120V$.

We need to find phase angle .

Firstly we will find inductive reactance ,$X_L=2\mathrm{Ï€´Ú³¢}=2\mathrm{Ï€}×60×0.12=45.2$$Ω$

then we will capacitive reactance, $X_C=\frac{1}{2\mathrm{Ï€´Ú°ä}}=\frac{1}{2\mathrm{Ï€}×60×30×{10}^{-3}}=0.084$$Ω$

Phase angle is given by ,

$\mathrm{ϕ}={\tan }^{-1}\left[\frac{X_L-X_C}{R}\right]={\tan }^{-1}\left[\frac{45.116}{75}\right]={\tan }^{-1}\left(0.601\right)=0.54°$

We are given that $75Ω$resistor, a $0.12H$inductor, and a $30mF$capacitor are there in $RLC$circuit. We are given that frequency of power line source equal to $60Hz$and voltage $120V$.

We need to find average power loss.

Firstly we need to find , $I_{rms}=\frac{E_{rms}}{Z}$

then we can find ,$E_{rms}=\frac{E_0}{\sqrt{2}}=84.8$$V$

So from here we get, $I_{rms}=\frac{84.8}{87.5}=0.96$$A$

Power factor , $\cos \left(0.54\right)=0.85$

Average power loss $=I_{rms}{E}_{rms}\cos \mathrm{ϕ}=84.8×0.96×0.85=69.19watt$