Question

A series$RLC$circuit consists of a $25\Omega$resistor, a $0.10H$inductor, and a$100mF$capacitor. It draws a$2.5A$$rms$current when attached to a $60Hz$source. What are (a) the $emf$$E_{rms}$, (b) the phase angle$\varphi$and (c) the average power loss?

Short answer

a) The$emf{E}_{rms}$is$112.3V$

b) The Phase angle $\varphi$is $1.54°$

c ) The average power loss is $7.29watt$

Step-by-step solution

Part(a) Step 1: Given information 

We are given that the resistor of $25$$\Omega$, 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 $E_{rms}$.

Part(a) Step 2: Simplify 

Firstly we will find inductive reactance , $X_L=2\mathrm{\pi fL}=2\mathrm{\pi }\times 60\times 0.10=37.6$$\Omega$

similarly we will find capacitive reactance ,$X_C=\frac{1}{2\mathrm{\pi fC}}=\frac{1}{2\mathrm{\pi }\times 60\times 100\times {10}^{-3}}=0.26$$\Omega$

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

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

Part(b) Step 1: Given information 

We are given that the resistor of $25\Omega$, 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 .

Part(b) Step 2: Simplify 

Firstly we will find inductive reactance , $X_L=2\mathrm{\pi fL}=2\mathrm{\pi }\times 60\times 0.10=37.6$$\Omega$

Similarly we will find capacitive reactance , $X_C=\frac{1}{2\mathrm{\pi fC}}=\frac{1}{2\mathrm{\pi }\times 60\times 100\times {10}^{-3}}=0.26$$\Omega$

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

Part(c) Step 1: Given information 

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.

Part(c) Step 2: Simplify

We will calculate power factor here which is given by ,

$\cos \left(\varphi \right)=\cos \left(1.54\right)=0.026$

then average power lose is given by ,

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