91Ó°ÊÓ

We are given that the resistor of $25$$Ω$, a$0.10H$ inductor , a $100mF$ capacitor are there in $RLC$circuit . We are given that$I_{rms}$ current of $2.5A$ and frequency $60Hz$ of source.

We need to find $E_{rms}$.

Firstly we will find inductive reactance , $X_L=2\mathrm{Ï€´Ú³¢}=2\mathrm{Ï€}×60×0.10=37.6$$Ω$

similarly we will find capacitive reactance ,$X_C=\frac{1}{2\mathrm{Ï€´Ú°ä}}=\frac{1}{2\mathrm{Ï€}×60×100×{10}^{-3}}=0.26$$Ω$

Impedance , $Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}=\sqrt{2019.27}=44.9$$Ω$

We also know that , $I_{rms}=\frac{E_{rms}}{Z}$$⇒E_{rms}=2.5×44.9=112.3V$

We are given that the resistor of $25Ω$, a $0.10H$ inductor , a $100mF$ capacitor are there in $RLC$circuit . We are given that$I_{rms}$current of $2.5A$and frequency $60Hz$ of source.

We need to find phase angle .

Firstly we will find inductive reactance , $X_L=2\mathrm{Ï€´Ú³¢}=2\mathrm{Ï€}×60×0.10=37.6$$Ω$

Similarly we will find capacitive reactance , $X_C=\frac{1}{2\mathrm{Ï€´Ú°ä}}=\frac{1}{2\mathrm{Ï€}×60×100×{10}^{-3}}=0.26$$Ω$

Phase angle is given by , $\mathrm{ϕ}={\tan }^{-1}\left[\frac{X_L-X_C}{R}\right]={\tan }^{-1}\left(37.34\right)=1.54°$

We are given that the resistor of$25$ , a$0.10H$ inductor , a $100mF$ capacitor are there in $RLC$circuit . We are given that$I_{rms}$ current of $2.5A$ and frequency$60Hz$ of source.

We need to find average power loss.

We will calculate power factor here which is given by ,

$\cos \left(\mathrm{Ï•}\right)=\cos \left(1.54\right)=0.026$

then average power lose is given by ,

average power loss =$I_{rms}{E}_{rms}\cos \left(\mathrm{ϕ}\right)=2.5×112.3×0.026=7.29watt$